PA1_template.md
In AlfonsoRReyes/RepDataPeerAssessment1: Reproducible Research Assignment 1. Developed as a package
Reproducible Research, Assignment 1
Alfonso R. Reyes
A. Loading and preprocessing the data
We start by downloading the raw data using the link provided by the instructor. We use a function that will download the zip file, unpack it and place it in an indicated directory. The function is called downloadZip.
1. Load the data
Downloading and unpacking raw data file
library(RepDataPeerAssessment1)
library(rprojroot)
cat("Setting up the project folders:\n")
project.data <- find_package_root_file('data')
project.extdata <- find_package_root_file('inst/extdata')
project.R <- find_package_root_file('R')
project.data
project.R
project.extdata
Setting up the project folders:
[1] "/home/superuser/git.projects/RepDataPeerAssessment1/data"
[1] "/home/superuser/git.projects/RepDataPeerAssessment1/R"
[1] "/home/superuser/git.projects/RepDataPeerAssessment1/inst/extdata"
downloadZip <- function(fileUrl, outDir="./data") {
# function to download zipped file and unpack
temp <- tempfile()
download.file(fileUrl, temp, mode = "wb")
unzip(temp, exdir = outDir)
}
fileUrl <- "https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
cat("Unpacking the raw data file:\n")
Unpacking the raw data file:
outDir <- project.extdata # folder for raw data
downloadZip(fileUrl, outDir = outDir) # download and unpack zip file
\
Create the RData file
More RData files may be generated during this assignment. They will be placed under the folder data.
Saving the raw dataset to the data folder
Reads the CSV raw data file from inst/extdata and save it under data. Then remove it from memory.
# save the dataset
activity.raw <- read.csv(paste(project.extdata, "activity.csv", sep = "/"))
save(activity.raw, file=paste(project.data, "activity.raw.RData", sep = "/"))
head(activity.raw)
steps date interval
1 NA 2012-10-01 0
2 NA 2012-10-01 5
3 NA 2012-10-01 10
4 NA 2012-10-01 15
5 NA 2012-10-01 20
6 NA 2012-10-01 25
Convert the variable date from as.factor to as.date
activity <- activity.raw
activity$date <- as.Date(activity.raw$date)
str(activity)
'data.frame': 17568 obs. of 3 variables:
$ steps : int NA NA NA NA NA NA NA NA NA NA ...
$ date : Date, format: "2012-10-01" "2012-10-01" ...
$ interval: int 0 5 10 15 20 25 30 35 40 45 ...
# save the dataset
save(activity, file=paste(project.data, "activity.RData", sep = "/"))
head(activity)
steps date interval
1 NA 2012-10-01 0
2 NA 2012-10-01 5
3 NA 2012-10-01 10
4 NA 2012-10-01 15
5 NA 2012-10-01 20
6 NA 2012-10-01 25
Confirm it has been saved:
rm(activity) # remove variable from memory
load(paste(project.data, "activity.RData", sep = "/")) # load data file
cat("Checking dataset has the structure we want\n\n")
str(activity)
head(activity)
# file.exists(paste(project.data, "activity.RData", sep = "/")) # we could use this too
Checking dataset has the structure we want
'data.frame': 17568 obs. of 3 variables:
$ steps : int NA NA NA NA NA NA NA NA NA NA ...
$ date : Date, format: "2012-10-01" "2012-10-01" ...
$ interval: int 0 5 10 15 20 25 30 35 40 45 ...
steps date interval
1 NA 2012-10-01 0
2 NA 2012-10-01 5
3 NA 2012-10-01 10
4 NA 2012-10-01 15
5 NA 2012-10-01 20
6 NA 2012-10-01 25
\
2. Process and transform
Process/transform the data (if necessary) into a format suitable for the analysis.
Basic sanity check
- Show first 6 rows of data frame
- Show dimensions of the data frame
- Show variable names
- Show summary
library(RepDataPeerAssessment1) #load my package
data("activity")
head(activity)
steps date interval
1 NA 2012-10-01 0
2 NA 2012-10-01 5
3 NA 2012-10-01 10
4 NA 2012-10-01 15
5 NA 2012-10-01 20
6 NA 2012-10-01 25
Show dimensions
dim(activity)
[1] 17568 3
Names of the variables
names(activity)
[1] "steps" "date" "interval"
Summary
suma <- summary(activity)
suma
steps date interval
Min. : 0.00 Min. :2012-10-01 Min. : 0.0
1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
Median : 0.00 Median :2012-10-31 Median :1177.5
Mean : 37.38 Mean :2012-10-31 Mean :1177.5
3rd Qu.: 12.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
Max. :806.00 Max. :2012-11-30 Max. :2355.0
NA's :2304
Notice that we have NA's :2304 .
B. What is the mean total number of steps taken per day?
We will ignore the NAs in this part of the assignment.
1. Calculate the total number of steps taken per day
# get only observations that are not NA
complete <- complete.cases(activity)
activity.cases <- activity[complete, ]
activity.NAs <- activity[!complete, ] # NAs
activity.NAs.not <- activity.cases
cat("# of observations:\t", dim(activity.cases)[1], "\n")
cat("# of NAs:\t\t", dim(activity.NAs)[1], "\n")
# of observations: 15264
# of NAs: 2304
plot(seq(1:nrow(activity.cases)), activity.cases$steps)
Histogram of total number of steps each day
byDate.steps.total <- aggregate(activity.cases$steps,
by = list(activity.cases$date), sum)
# rename the variable to something meaningful
names(byDate.steps.total) <- c("Day", "total.steps")
# byDate.steps.total
hist(byDate.steps.total$total.steps)
Find the mean and the median total number of steps per day
mean.0 <- mean(byDate.steps.total$total.steps)
mean.0
[1] 10766.19
median.0 <- median(byDate.steps.total$total.steps)
median.0
[1] 10765
How many unique intervals are there?
We want to know how many unique intervals there are because we will need later to calculate the maximum steps per interval and we need this number to verify our count of intervals is correct.
unique(activity.cases$interval)
[1] 0 5 10 15 20 25 30 35 40 45 50 55 100 105
[15] 110 115 120 125 130 135 140 145 150 155 200 205 210 215
[29] 220 225 230 235 240 245 250 255 300 305 310 315 320 325
[43] 330 335 340 345 350 355 400 405 410 415 420 425 430 435
[57] 440 445 450 455 500 505 510 515 520 525 530 535 540 545
[71] 550 555 600 605 610 615 620 625 630 635 640 645 650 655
[85] 700 705 710 715 720 725 730 735 740 745 750 755 800 805
[99] 810 815 820 825 830 835 840 845 850 855 900 905 910 915
[113] 920 925 930 935 940 945 950 955 1000 1005 1010 1015 1020 1025
[127] 1030 1035 1040 1045 1050 1055 1100 1105 1110 1115 1120 1125 1130 1135
[141] 1140 1145 1150 1155 1200 1205 1210 1215 1220 1225 1230 1235 1240 1245
[155] 1250 1255 1300 1305 1310 1315 1320 1325 1330 1335 1340 1345 1350 1355
[169] 1400 1405 1410 1415 1420 1425 1430 1435 1440 1445 1450 1455 1500 1505
[183] 1510 1515 1520 1525 1530 1535 1540 1545 1550 1555 1600 1605 1610 1615
[197] 1620 1625 1630 1635 1640 1645 1650 1655 1700 1705 1710 1715 1720 1725
[211] 1730 1735 1740 1745 1750 1755 1800 1805 1810 1815 1820 1825 1830 1835
[225] 1840 1845 1850 1855 1900 1905 1910 1915 1920 1925 1930 1935 1940 1945
[239] 1950 1955 2000 2005 2010 2015 2020 2025 2030 2035 2040 2045 2050 2055
[253] 2100 2105 2110 2115 2120 2125 2130 2135 2140 2145 2150 2155 2200 2205
[267] 2210 2215 2220 2225 2230 2235 2240 2245 2250 2255 2300 2305 2310 2315
[281] 2320 2325 2330 2335 2340 2345 2350 2355
So, there are 288 unique intervals, with the first interval being 0 and the last 2355.
C. What is the average daily activity pattern?
1. Make a time series plot (i.e. type = "l")
Make a time series plot (i.e. type = "l") of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all days (y-axis)
byDate.steps.mean <- aggregate(activity.cases$steps,
by = list(activity.cases$date), mean)
# rename the variable to something meaningful
names(byDate.steps.mean) <- c("Day", "mean.steps")
plot(byDate.steps.mean$Day, byDate.steps.mean$mean.steps, type = "l")
2. Which interval contain the maximum # of steps
Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?
Find day and interval with maximum number of steps
Let's find the maximum number of steps:
max.steps <- max(activity.cases$steps)
max.steps
[1] 806
Now, let's find which interval holds the maximum number of steps:
index <- which(activity.cases$steps == max.steps)
whole_row <- activity.cases[index, ]
whole_row
steps date interval
16492 806 2012-11-27 615
The interval with the maximum number of steps is 615.
Grouping by interval
byInterval <- aggregate(activity.cases$steps,
by = list(activity.cases$interval),
max)
names(byInterval) <- c("interval", "steps.max")
sorted <- byInterval[order(-byInterval$steps.max), ] # order by steps, descending
cat("These are the top 10 intervals with more activity\n\n")
head(sorted, 10)
These are the top 10 intervals with more activity
interval steps.max
76 615 806
109 900 802
71 550 794
89 720 789
104 835 786
114 925 785
193 1600 785
200 1635 785
141 1140 783
107 850 781
\
Find what are the intervals with more activity
We will plot the number maximum of steps in logarithmic scale:
plot(sorted$interval, sorted$steps.max,
xlim = c(0, 2500), ylim = range(1:1000),
panel.first = grid(lty = 1))
plot(sorted$interval, sorted$steps.max, log = "y",
xlim = c(0, 2500), ylim = range(1:1000),
panel.first = grid(lty = 1))
Warning in xy.coords(x, y, xlabel, ylabel, log): 19 y values <= 0 omitted
from logarithmic plot
And zooming in in a non-logarithmic plot, we can see that the maximum values are those around 800.
# non-logarithmic plot
plot(sorted$interval, sorted$steps.max,
xlim = range(0:2500), ylim = range(700:825),
panel.first = grid(lty = 1))
What we can confirm is that the maximum activity occurs between the 500 and 2000 intervals.
D. Imputing missing values
1. Calculate the number of missing values
Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)
# get only observations that are NA
complete <- complete.cases(activity)
activity.missing <- activity[!complete, ]
nrow(activity.missing)
[1] 2304
2. Complete missing values
Devise a strategy for filling in all of the missing values in the dataset. The strategy does not need to be sophisticated. For example, you could use the mean/median for that day, or the mean for that 5-minute interval, etc.
There are several ways we can complete the missing values and lots of R packages available to perform the imputation. We will make it the easier way. This is the strategy:
Motivation
Since there is no way to compare in a dataset with missing values how a imputation method performs, after several attempts on resolving this, I found in the literature that the imputation method selection is done based on a dataset without any missing values first.
-
Using the dataset under analysis, we proceed to generate a new one but without any missing values.
-
We apply several imputation methods on the new dataset but this time the missing values are entered under control, the amount and randomness which could be MCAR or MAR, among several others.
-
The result of applying these methods will give us a performance plot of the methods used versus the Root Mean Squared Error (RMSE). We will chose the imputation method with the lowest RMSE.
-
After selecting the imputation method, we go back to our original dataset with missing data and apply the imputation method with the best performance, as selected above.
-
The result will be the data frame with imputed data.
Since this topic needs not to be sophisticated at this stage, we will use a pretty straight forward R package called imputTestbench. It is available in CRAN. We will describe the steps in detail below.
2.1. Generate a new dataset but without missing values
ok <- complete.cases(activity$steps)
steps.clean <- activity$steps[ok]
length(steps.clean)
[1] 15264
2.2. Apply several imputation methods
library(imputeTestbench)
itb <- impute_errors(steps.clean,
missPercentFrom = 0, missPercentTo = 10,
interval = 1, blckper = TRUE, blck = 10)
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
itb
$Parameter
[1] "rmse"
$MissingPercent
[1] 0 1 2 3 4 5 6 7 8 9 10
$na.approx
[1] 0.000000 5.454556 8.464277 9.868567 11.428547 12.319045 12.893618
[8] 14.459085 28.757653 21.950026 23.074516
$na.interp
[1] 0.000000 5.454556 8.464277 9.868567 11.428547 12.319045 12.893618
[8] 14.459085 15.907699 16.514481 17.861687
$na.interpolation
[1] 0.000000 5.454556 8.464277 9.868567 11.428547 12.319045 12.893618
[8] 14.459085 15.907699 16.514481 17.861687
$na.locf
[1] 0.000000 7.017167 10.559172 12.729527 15.313857 16.609814 16.649949
[8] 19.293431 32.467342 26.404211 28.383793
$na.mean
[1] 0.00000 10.77232 17.33313 18.81955 23.24894 25.02165 26.14002
[8] 28.41363 31.77541 33.91116 35.10665
2.3. Performance plots
plot_errors(itb)
From the boxplot we can see that the methods with the lowest error are na.interp and na.interpolation. These two methods are included in the package _. We will be able to see this clearly with the line type plot:
plot_errors(dataIn = itb, plotType = "line")
Both interpolation methods are overlapping which means that there is not significant difference between applying any of them. Also we can see that the worst performing imputing methods are the Last Observation Carried Forward (LOCF) and the mean.
This how it looks four of the different imputation simulations at 10% missing data rate.
plot_impute(steps.clean,
methods = c("na.mean", "na.locf", "na.interp"),
missPercent = 10)
2.4. Apply selected imputation method to the original dataset
We bring up the original data frame and get the vectors for the steps variable.
The function used is na.interp from the package 'forecast'. So, the imputation will be performed with na.interp.
library(forecast)
steps <- activity$steps
summary(activity$steps)
Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
0.00 0.00 0.00 37.38 12.00 806.00 2304
se.0 <- sd(activity$steps,
na.rm = TRUE) / sqrt(sum(!is.na(activity$steps)))
cat("SE(before imp.) = ", se.0, "\n\n")
SE(before imp.) = 0.9064973
steps.imp.interp <- na.interp(steps) # interpolating the NA values
summary(steps.imp.interp)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.00 0.00 0.00 32.48 0.00 806.00
se.1 <- sd(steps.imp.interp,
na.rm = TRUE) / sqrt(sum(!is.na(steps.imp.interp)))
cat("SE(after imp.) = ", se.1, "\n")
SE(after imp.) = 0.7933427
What we can see is that there is an improvement in the Standard Error (SE) from 0.91 to 0.79.
\
4. Make a histogram (with missing data filled in)
Make a histogram of the total number of steps taken each day and Calculate and report the mean and median total number of steps taken per day. Do these values differ from the estimates from the first part of the assignment? What is the impact of imputing missing data on the estimates of the total daily number of steps?
First, we put together the activity data frame with the imputed values.
activity.imp <- activity
activity.imp$steps <- as.numeric(steps.imp.interp)
summary(activity.imp)
steps date interval
Min. : 0.00 Min. :2012-10-01 Min. : 0.0
1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
Median : 0.00 Median :2012-10-31 Median :1177.5
Mean : 32.48 Mean :2012-10-31 Mean :1177.5
3rd Qu.: 0.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
Max. :806.00 Max. :2012-11-30 Max. :2355.0
str(activity.imp)
'data.frame': 17568 obs. of 3 variables:
$ steps : num 0 0 0 0 0 0 0 0 0 0 ...
$ date : Date, format: "2012-10-01" "2012-10-01" ...
$ interval: int 0 5 10 15 20 25 30 35 40 45 ...
byDate.steps.total.1 <- aggregate(activity.imp$steps,
by = list(activity.imp$date), sum)
# rename the variable to something meaningful
names(byDate.steps.total.1) <- c("Day", "total.steps")
summary(byDate.steps.total.1$total.steps)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0 6778 10400 9354 12810 21190
hist(byDate.steps.total.1$total.steps)
Calculating the mean and the media for the imputed steps
mean.1 <- mean(byDate.steps.total.1$total.steps)
median.1 <- median(byDate.steps.total.1$total.steps)
mean.1
[1] 9354.23
median.1
[1] 10395
And we find that there is a difference between mean and the media with the dataset with missing data and the dataset with imputed values.
mean.0 - mean.1
[1] 1411.959
median.0 - median.1
[1] 370
The change in the mean is 13 percent and of the median is 3 percent.
\
E. Are there differences in activity patterns between weekdays and weekends?
1. Create a new factor variable
Create a new factor variable in the dataset with two levels – “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.
activity.imp$week <- ifelse(weekdays(activity.imp$date) %in% c("Saturday", "Sunday"), "weekend", "weekday")
# View(activity.imp)
activity.imp$week <- as.factor(activity.imp$week)
str(activity.imp)
'data.frame': 17568 obs. of 4 variables:
$ steps : num 0 0 0 0 0 0 0 0 0 0 ...
$ date : Date, format: "2012-10-01" "2012-10-01" ...
$ interval: int 0 5 10 15 20 25 30 35 40 45 ...
$ week : Factor w/ 2 levels "weekday","weekend": 1 1 1 1 1 1 1 1 1 1 ...
2. Make a panel
Make a panel plot containing a time series plot (i.e. type = "l") of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all weekday days or weekend days (y-axis). See the README file in the GitHub repository to see an example of what this plot should look like using simulated data.
activity.imp.1 <- activity.imp
byInterval <- aggregate(activity.imp.1$steps,
by = list(activity.imp.1$interval, activity.imp.1$week), mean)
names(byInterval) <- c("interval", "week", "steps.mean")
byInterval
interval week steps.mean
1 0 weekday 2.02222222
2 5 weekday 0.40000000
3 10 weekday 0.15555556
4 15 weekday 0.17777778
5 20 weekday 0.08888889
6 25 weekday 1.31111111
7 30 weekday 0.62222222
8 35 weekday 1.02222222
9 40 weekday 0.00000000
10 45 weekday 1.60000000
11 50 weekday 0.35555556
12 55 weekday 0.00000000
13 100 weekday 0.37777778
14 105 weekday 0.00000000
15 110 weekday 0.17777778
16 115 weekday 0.40000000
17 120 weekday 0.00000000
18 125 weekday 1.31111111
19 130 weekday 1.97777778
20 135 weekday 0.00000000
21 140 weekday 0.20000000
22 145 weekday 0.20000000
23 150 weekday 0.31111111
24 155 weekday 0.00000000
25 200 weekday 0.00000000
26 205 weekday 0.00000000
27 210 weekday 1.24444444
28 215 weekday 0.00000000
29 220 weekday 0.00000000
30 225 weekday 0.15555556
31 230 weekday 0.00000000
32 235 weekday 0.26666667
33 240 weekday 0.00000000
34 245 weekday 0.00000000
35 250 weekday 1.82222222
36 255 weekday 1.11111111
37 300 weekday 0.00000000
38 305 weekday 0.00000000
39 310 weekday 0.00000000
40 315 weekday 0.00000000
41 320 weekday 0.00000000
42 325 weekday 0.73333333
43 330 weekday 1.02222222
44 335 weekday 0.44444444
45 340 weekday 0.35555556
46 345 weekday 0.08888889
47 350 weekday 0.00000000
48 355 weekday 0.00000000
49 400 weekday 0.11111111
50 405 weekday 1.11111111
51 410 weekday 1.88888889
52 415 weekday 0.00000000
53 420 weekday 0.40000000
54 425 weekday 0.00000000
55 430 weekday 2.82222222
56 435 weekday 0.13333333
57 440 weekday 3.31111111
58 445 weekday 0.77777778
59 450 weekday 1.93333333
60 455 weekday 0.57777778
61 500 weekday 0.00000000
62 505 weekday 1.84444444
63 510 weekday 3.53333333
64 515 weekday 1.88888889
65 520 weekday 3.77777778
66 525 weekday 2.31111111
67 530 weekday 2.46666667
68 535 weekday 7.13333333
69 540 weekday 18.26666667
70 545 weekday 21.20000000
71 550 weekday 45.08888889
72 555 weekday 50.33333333
73 600 weekday 37.08888889
74 605 weekday 58.02222222
75 610 weekday 62.91111111
76 615 weekday 68.68888889
77 620 weekday 57.26666667
78 625 weekday 53.75555556
79 630 weekday 59.48888889
80 635 weekday 42.73333333
81 640 weekday 49.80000000
82 645 weekday 48.97777778
83 650 weekday 42.06666667
84 655 weekday 53.88888889
85 700 weekday 44.75555556
86 705 weekday 44.91111111
87 710 weekday 55.28888889
88 715 weekday 62.06666667
89 720 weekday 56.44444444
90 725 weekday 52.31111111
91 730 weekday 58.80000000
92 735 weekday 48.44444444
93 740 weekday 55.75555556
94 745 weekday 74.11111111
95 750 weekday 60.02222222
96 755 weekday 59.08888889
97 800 weekday 72.93333333
98 805 weekday 62.86666667
99 810 weekday 126.75555556
100 815 weekday 160.97777778
101 820 weekday 177.75555556
102 825 weekday 162.88888889
103 830 weekday 175.24444444
104 835 weekday 202.88888889
105 840 weekday 192.77777778
106 845 weekday 161.71111111
107 850 weekday 166.77777778
108 855 weekday 154.82222222
109 900 weekday 148.53333333
110 905 weekday 109.24444444
111 910 weekday 79.40000000
112 915 weekday 72.88888889
113 920 weekday 89.71111111
114 925 weekday 79.66666667
115 930 weekday 49.68888889
116 935 weekday 29.82222222
117 940 weekday 24.15555556
118 945 weekday 35.68888889
119 950 weekday 34.46666667
120 955 weekday 14.82222222
121 1000 weekday 32.46666667
122 1005 weekday 14.62222222
123 1010 weekday 33.42222222
124 1015 weekday 40.80000000
125 1020 weekday 25.15555556
126 1025 weekday 28.37777778
127 1030 weekday 27.22222222
128 1035 weekday 19.26666667
129 1040 weekday 18.88888889
130 1045 weekday 22.13333333
131 1050 weekday 18.68888889
132 1055 weekday 19.00000000
133 1100 weekday 17.51111111
134 1105 weekday 21.13333333
135 1110 weekday 8.84444444
136 1115 weekday 12.86666667
137 1120 weekday 20.40000000
138 1125 weekday 20.20000000
139 1130 weekday 28.31111111
140 1135 weekday 43.53333333
141 1140 weekday 38.95555556
142 1145 weekday 41.97777778
143 1150 weekday 43.97777778
144 1155 weekday 48.24444444
145 1200 weekday 47.20000000
146 1205 weekday 61.15555556
147 1210 weekday 71.00000000
148 1215 weekday 62.91111111
149 1220 weekday 40.26666667
150 1225 weekday 40.13333333
151 1230 weekday 55.31111111
152 1235 weekday 26.42222222
153 1240 weekday 18.44444444
154 1245 weekday 24.28888889
155 1250 weekday 26.77777778
156 1255 weekday 47.62222222
157 1300 weekday 18.95555556
158 1305 weekday 20.42222222
159 1310 weekday 18.80000000
160 1315 weekday 10.17777778
161 1320 weekday 29.46666667
162 1325 weekday 37.33333333
163 1330 weekday 26.06666667
164 1335 weekday 19.95555556
165 1340 weekday 19.91111111
166 1345 weekday 33.04444444
167 1350 weekday 19.26666667
168 1355 weekday 28.22222222
169 1400 weekday 39.48888889
170 1405 weekday 32.62222222
171 1410 weekday 26.31111111
172 1415 weekday 38.55555556
173 1420 weekday 22.75555556
174 1425 weekday 25.75555556
175 1430 weekday 25.91111111
176 1435 weekday 10.84444444
177 1440 weekday 9.26666667
178 1445 weekday 18.51111111
179 1450 weekday 36.04444444
180 1455 weekday 32.44444444
181 1500 weekday 26.86666667
182 1505 weekday 30.24444444
183 1510 weekday 25.22222222
184 1515 weekday 26.73333333
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library(lattice)
# xyplot(y ~ x | panel, data = dataset, type = "o")
xyplot(steps.mean ~ interval | as.factor(week),
data = byInterval,
type = "l",
layout=c(1,2))
AlfonsoRReyes/RepDataPeerAssessment1 documentation built on May 5, 2019, 4:53 a.m.
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Reproducible Research, Assignment 1
Alfonso R. Reyes
A. Loading and preprocessing the data
We start by downloading the raw data using the link provided by the instructor. We use a function that will download the zip file, unpack it and place it in an indicated directory. The function is called downloadZip.
1. Load the data
Downloading and unpacking raw data file
library(RepDataPeerAssessment1)
library(rprojroot)
cat("Setting up the project folders:\n")
project.data <- find_package_root_file('data')
project.extdata <- find_package_root_file('inst/extdata')
project.R <- find_package_root_file('R')
project.data
project.R
project.extdata
Setting up the project folders:
[1] "/home/superuser/git.projects/RepDataPeerAssessment1/data"
[1] "/home/superuser/git.projects/RepDataPeerAssessment1/R"
[1] "/home/superuser/git.projects/RepDataPeerAssessment1/inst/extdata"
downloadZip <- function(fileUrl, outDir="./data") {
# function to download zipped file and unpack
temp <- tempfile()
download.file(fileUrl, temp, mode = "wb")
unzip(temp, exdir = outDir)
}
fileUrl <- "https://d396qusza40orc.cloudfront.net/repdata%2Fdata%2Factivity.zip"
cat("Unpacking the raw data file:\n")
Unpacking the raw data file:
outDir <- project.extdata # folder for raw data
downloadZip(fileUrl, outDir = outDir) # download and unpack zip file
\
Create the RData file
More RData files may be generated during this assignment. They will be placed under the folder data.
Saving the raw dataset to the data folder
Reads the CSV raw data file from inst/extdata and save it under data. Then remove it from memory.
# save the dataset
activity.raw <- read.csv(paste(project.extdata, "activity.csv", sep = "/"))
save(activity.raw, file=paste(project.data, "activity.raw.RData", sep = "/"))
head(activity.raw)
steps date interval
1 NA 2012-10-01 0
2 NA 2012-10-01 5
3 NA 2012-10-01 10
4 NA 2012-10-01 15
5 NA 2012-10-01 20
6 NA 2012-10-01 25
Convert the variable date from as.factor to as.date
activity <- activity.raw
activity$date <- as.Date(activity.raw$date)
str(activity)
'data.frame': 17568 obs. of 3 variables:
$ steps : int NA NA NA NA NA NA NA NA NA NA ...
$ date : Date, format: "2012-10-01" "2012-10-01" ...
$ interval: int 0 5 10 15 20 25 30 35 40 45 ...
# save the dataset
save(activity, file=paste(project.data, "activity.RData", sep = "/"))
head(activity)
steps date interval
1 NA 2012-10-01 0
2 NA 2012-10-01 5
3 NA 2012-10-01 10
4 NA 2012-10-01 15
5 NA 2012-10-01 20
6 NA 2012-10-01 25
Confirm it has been saved:
rm(activity) # remove variable from memory
load(paste(project.data, "activity.RData", sep = "/")) # load data file
cat("Checking dataset has the structure we want\n\n")
str(activity)
head(activity)
# file.exists(paste(project.data, "activity.RData", sep = "/")) # we could use this too
Checking dataset has the structure we want
'data.frame': 17568 obs. of 3 variables:
$ steps : int NA NA NA NA NA NA NA NA NA NA ...
$ date : Date, format: "2012-10-01" "2012-10-01" ...
$ interval: int 0 5 10 15 20 25 30 35 40 45 ...
steps date interval
1 NA 2012-10-01 0
2 NA 2012-10-01 5
3 NA 2012-10-01 10
4 NA 2012-10-01 15
5 NA 2012-10-01 20
6 NA 2012-10-01 25
\
2. Process and transform
Process/transform the data (if necessary) into a format suitable for the analysis.
Basic sanity check
- Show first 6 rows of data frame
- Show dimensions of the data frame
- Show variable names
- Show summary
library(RepDataPeerAssessment1) #load my package
data("activity")
head(activity)
steps date interval
1 NA 2012-10-01 0
2 NA 2012-10-01 5
3 NA 2012-10-01 10
4 NA 2012-10-01 15
5 NA 2012-10-01 20
6 NA 2012-10-01 25
Show dimensions
dim(activity)
[1] 17568 3
Names of the variables
names(activity)
[1] "steps" "date" "interval"
Summary
suma <- summary(activity)
suma
steps date interval
Min. : 0.00 Min. :2012-10-01 Min. : 0.0
1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
Median : 0.00 Median :2012-10-31 Median :1177.5
Mean : 37.38 Mean :2012-10-31 Mean :1177.5
3rd Qu.: 12.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
Max. :806.00 Max. :2012-11-30 Max. :2355.0
NA's :2304
Notice that we have NA's :2304 .
B. What is the mean total number of steps taken per day?
We will ignore the NAs in this part of the assignment.
1. Calculate the total number of steps taken per day
# get only observations that are not NA
complete <- complete.cases(activity)
activity.cases <- activity[complete, ]
activity.NAs <- activity[!complete, ] # NAs
activity.NAs.not <- activity.cases
cat("# of observations:\t", dim(activity.cases)[1], "\n")
cat("# of NAs:\t\t", dim(activity.NAs)[1], "\n")
# of observations: 15264
# of NAs: 2304
plot(seq(1:nrow(activity.cases)), activity.cases$steps)
Histogram of total number of steps each day
byDate.steps.total <- aggregate(activity.cases$steps,
by = list(activity.cases$date), sum)
# rename the variable to something meaningful
names(byDate.steps.total) <- c("Day", "total.steps")
# byDate.steps.total
hist(byDate.steps.total$total.steps)
Find the mean and the median total number of steps per day
mean.0 <- mean(byDate.steps.total$total.steps)
mean.0
[1] 10766.19
median.0 <- median(byDate.steps.total$total.steps)
median.0
[1] 10765
How many unique intervals are there?
We want to know how many unique intervals there are because we will need later to calculate the maximum steps per interval and we need this number to verify our count of intervals is correct.
unique(activity.cases$interval)
[1] 0 5 10 15 20 25 30 35 40 45 50 55 100 105
[15] 110 115 120 125 130 135 140 145 150 155 200 205 210 215
[29] 220 225 230 235 240 245 250 255 300 305 310 315 320 325
[43] 330 335 340 345 350 355 400 405 410 415 420 425 430 435
[57] 440 445 450 455 500 505 510 515 520 525 530 535 540 545
[71] 550 555 600 605 610 615 620 625 630 635 640 645 650 655
[85] 700 705 710 715 720 725 730 735 740 745 750 755 800 805
[99] 810 815 820 825 830 835 840 845 850 855 900 905 910 915
[113] 920 925 930 935 940 945 950 955 1000 1005 1010 1015 1020 1025
[127] 1030 1035 1040 1045 1050 1055 1100 1105 1110 1115 1120 1125 1130 1135
[141] 1140 1145 1150 1155 1200 1205 1210 1215 1220 1225 1230 1235 1240 1245
[155] 1250 1255 1300 1305 1310 1315 1320 1325 1330 1335 1340 1345 1350 1355
[169] 1400 1405 1410 1415 1420 1425 1430 1435 1440 1445 1450 1455 1500 1505
[183] 1510 1515 1520 1525 1530 1535 1540 1545 1550 1555 1600 1605 1610 1615
[197] 1620 1625 1630 1635 1640 1645 1650 1655 1700 1705 1710 1715 1720 1725
[211] 1730 1735 1740 1745 1750 1755 1800 1805 1810 1815 1820 1825 1830 1835
[225] 1840 1845 1850 1855 1900 1905 1910 1915 1920 1925 1930 1935 1940 1945
[239] 1950 1955 2000 2005 2010 2015 2020 2025 2030 2035 2040 2045 2050 2055
[253] 2100 2105 2110 2115 2120 2125 2130 2135 2140 2145 2150 2155 2200 2205
[267] 2210 2215 2220 2225 2230 2235 2240 2245 2250 2255 2300 2305 2310 2315
[281] 2320 2325 2330 2335 2340 2345 2350 2355
So, there are 288 unique intervals, with the first interval being 0 and the last 2355.
C. What is the average daily activity pattern?
1. Make a time series plot (i.e. type = "l")
Make a time series plot (i.e. type = "l") of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all days (y-axis)
byDate.steps.mean <- aggregate(activity.cases$steps,
by = list(activity.cases$date), mean)
# rename the variable to something meaningful
names(byDate.steps.mean) <- c("Day", "mean.steps")
plot(byDate.steps.mean$Day, byDate.steps.mean$mean.steps, type = "l")
2. Which interval contain the maximum # of steps
Which 5-minute interval, on average across all the days in the dataset, contains the maximum number of steps?
Find day and interval with maximum number of steps
Let's find the maximum number of steps:
max.steps <- max(activity.cases$steps)
max.steps
[1] 806
Now, let's find which interval holds the maximum number of steps:
index <- which(activity.cases$steps == max.steps)
whole_row <- activity.cases[index, ]
whole_row
steps date interval
16492 806 2012-11-27 615
The interval with the maximum number of steps is 615.
Grouping by interval
byInterval <- aggregate(activity.cases$steps,
by = list(activity.cases$interval),
max)
names(byInterval) <- c("interval", "steps.max")
sorted <- byInterval[order(-byInterval$steps.max), ] # order by steps, descending
cat("These are the top 10 intervals with more activity\n\n")
head(sorted, 10)
These are the top 10 intervals with more activity
interval steps.max
76 615 806
109 900 802
71 550 794
89 720 789
104 835 786
114 925 785
193 1600 785
200 1635 785
141 1140 783
107 850 781
\
Find what are the intervals with more activity
We will plot the number maximum of steps in logarithmic scale:
plot(sorted$interval, sorted$steps.max,
xlim = c(0, 2500), ylim = range(1:1000),
panel.first = grid(lty = 1))
plot(sorted$interval, sorted$steps.max, log = "y",
xlim = c(0, 2500), ylim = range(1:1000),
panel.first = grid(lty = 1))
Warning in xy.coords(x, y, xlabel, ylabel, log): 19 y values <= 0 omitted
from logarithmic plot
And zooming in in a non-logarithmic plot, we can see that the maximum values are those around 800.
# non-logarithmic plot
plot(sorted$interval, sorted$steps.max,
xlim = range(0:2500), ylim = range(700:825),
panel.first = grid(lty = 1))
What we can confirm is that the maximum activity occurs between the 500 and 2000 intervals.
D. Imputing missing values
1. Calculate the number of missing values
Calculate and report the total number of missing values in the dataset (i.e. the total number of rows with NAs)
# get only observations that are NA
complete <- complete.cases(activity)
activity.missing <- activity[!complete, ]
nrow(activity.missing)
[1] 2304
2. Complete missing values
Devise a strategy for filling in all of the missing values in the dataset. The strategy does not need to be sophisticated. For example, you could use the mean/median for that day, or the mean for that 5-minute interval, etc.
There are several ways we can complete the missing values and lots of R packages available to perform the imputation. We will make it the easier way. This is the strategy:
Motivation
Since there is no way to compare in a dataset with missing values how a imputation method performs, after several attempts on resolving this, I found in the literature that the imputation method selection is done based on a dataset without any missing values first.
-
Using the dataset under analysis, we proceed to generate a new one but without any missing values.
-
We apply several imputation methods on the new dataset but this time the missing values are entered under control, the amount and randomness which could be MCAR or MAR, among several others.
-
The result of applying these methods will give us a performance plot of the methods used versus the Root Mean Squared Error (RMSE). We will chose the imputation method with the lowest RMSE.
-
After selecting the imputation method, we go back to our original dataset with missing data and apply the imputation method with the best performance, as selected above.
-
The result will be the data frame with imputed data.
Since this topic needs not to be sophisticated at this stage, we will use a pretty straight forward R package called imputTestbench. It is available in CRAN. We will describe the steps in detail below.
2.1. Generate a new dataset but without missing values
ok <- complete.cases(activity$steps)
steps.clean <- activity$steps[ok]
length(steps.clean)
[1] 15264
2.2. Apply several imputation methods
library(imputeTestbench)
itb <- impute_errors(steps.clean,
missPercentFrom = 0, missPercentTo = 10,
interval = 1, blckper = TRUE, blck = 10)
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
Warning in obs - pred: longer object length is not a multiple of shorter
object length
itb
$Parameter
[1] "rmse"
$MissingPercent
[1] 0 1 2 3 4 5 6 7 8 9 10
$na.approx
[1] 0.000000 5.454556 8.464277 9.868567 11.428547 12.319045 12.893618
[8] 14.459085 28.757653 21.950026 23.074516
$na.interp
[1] 0.000000 5.454556 8.464277 9.868567 11.428547 12.319045 12.893618
[8] 14.459085 15.907699 16.514481 17.861687
$na.interpolation
[1] 0.000000 5.454556 8.464277 9.868567 11.428547 12.319045 12.893618
[8] 14.459085 15.907699 16.514481 17.861687
$na.locf
[1] 0.000000 7.017167 10.559172 12.729527 15.313857 16.609814 16.649949
[8] 19.293431 32.467342 26.404211 28.383793
$na.mean
[1] 0.00000 10.77232 17.33313 18.81955 23.24894 25.02165 26.14002
[8] 28.41363 31.77541 33.91116 35.10665
2.3. Performance plots
plot_errors(itb)
From the boxplot we can see that the methods with the lowest error are na.interp and na.interpolation. These two methods are included in the package _. We will be able to see this clearly with the line type plot:
plot_errors(dataIn = itb, plotType = "line")
Both interpolation methods are overlapping which means that there is not significant difference between applying any of them. Also we can see that the worst performing imputing methods are the Last Observation Carried Forward (LOCF) and the mean.
This how it looks four of the different imputation simulations at 10% missing data rate.
plot_impute(steps.clean,
methods = c("na.mean", "na.locf", "na.interp"),
missPercent = 10)
2.4. Apply selected imputation method to the original dataset
We bring up the original data frame and get the vectors for the steps variable.
The function used is na.interp from the package 'forecast'. So, the imputation will be performed with na.interp.
library(forecast)
steps <- activity$steps
summary(activity$steps)
Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
0.00 0.00 0.00 37.38 12.00 806.00 2304
se.0 <- sd(activity$steps,
na.rm = TRUE) / sqrt(sum(!is.na(activity$steps)))
cat("SE(before imp.) = ", se.0, "\n\n")
SE(before imp.) = 0.9064973
steps.imp.interp <- na.interp(steps) # interpolating the NA values
summary(steps.imp.interp)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.00 0.00 0.00 32.48 0.00 806.00
se.1 <- sd(steps.imp.interp,
na.rm = TRUE) / sqrt(sum(!is.na(steps.imp.interp)))
cat("SE(after imp.) = ", se.1, "\n")
SE(after imp.) = 0.7933427
What we can see is that there is an improvement in the Standard Error (SE) from 0.91 to 0.79.
\
4. Make a histogram (with missing data filled in)
Make a histogram of the total number of steps taken each day and Calculate and report the mean and median total number of steps taken per day. Do these values differ from the estimates from the first part of the assignment? What is the impact of imputing missing data on the estimates of the total daily number of steps?
First, we put together the activity data frame with the imputed values.
activity.imp <- activity
activity.imp$steps <- as.numeric(steps.imp.interp)
summary(activity.imp)
steps date interval
Min. : 0.00 Min. :2012-10-01 Min. : 0.0
1st Qu.: 0.00 1st Qu.:2012-10-16 1st Qu.: 588.8
Median : 0.00 Median :2012-10-31 Median :1177.5
Mean : 32.48 Mean :2012-10-31 Mean :1177.5
3rd Qu.: 0.00 3rd Qu.:2012-11-15 3rd Qu.:1766.2
Max. :806.00 Max. :2012-11-30 Max. :2355.0
str(activity.imp)
'data.frame': 17568 obs. of 3 variables:
$ steps : num 0 0 0 0 0 0 0 0 0 0 ...
$ date : Date, format: "2012-10-01" "2012-10-01" ...
$ interval: int 0 5 10 15 20 25 30 35 40 45 ...
byDate.steps.total.1 <- aggregate(activity.imp$steps,
by = list(activity.imp$date), sum)
# rename the variable to something meaningful
names(byDate.steps.total.1) <- c("Day", "total.steps")
summary(byDate.steps.total.1$total.steps)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0 6778 10400 9354 12810 21190
hist(byDate.steps.total.1$total.steps)
Calculating the mean and the media for the imputed steps
mean.1 <- mean(byDate.steps.total.1$total.steps)
median.1 <- median(byDate.steps.total.1$total.steps)
mean.1
[1] 9354.23
median.1
[1] 10395
And we find that there is a difference between mean and the media with the dataset with missing data and the dataset with imputed values.
mean.0 - mean.1
[1] 1411.959
median.0 - median.1
[1] 370
The change in the mean is 13 percent and of the median is 3 percent.
\
E. Are there differences in activity patterns between weekdays and weekends?
1. Create a new factor variable
Create a new factor variable in the dataset with two levels – “weekday” and “weekend” indicating whether a given date is a weekday or weekend day.
activity.imp$week <- ifelse(weekdays(activity.imp$date) %in% c("Saturday", "Sunday"), "weekend", "weekday")
# View(activity.imp)
activity.imp$week <- as.factor(activity.imp$week)
str(activity.imp)
'data.frame': 17568 obs. of 4 variables:
$ steps : num 0 0 0 0 0 0 0 0 0 0 ...
$ date : Date, format: "2012-10-01" "2012-10-01" ...
$ interval: int 0 5 10 15 20 25 30 35 40 45 ...
$ week : Factor w/ 2 levels "weekday","weekend": 1 1 1 1 1 1 1 1 1 1 ...
2. Make a panel
Make a panel plot containing a time series plot (i.e. type = "l") of the 5-minute interval (x-axis) and the average number of steps taken, averaged across all weekday days or weekend days (y-axis). See the README file in the GitHub repository to see an example of what this plot should look like using simulated data.
activity.imp.1 <- activity.imp
byInterval <- aggregate(activity.imp.1$steps,
by = list(activity.imp.1$interval, activity.imp.1$week), mean)
names(byInterval) <- c("interval", "week", "steps.mean")
byInterval
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570 2325 weekend 0.56250000
571 2330 weekend 1.06250000
572 2335 weekend 11.00000000
573 2340 weekend 5.87500000
574 2345 weekend 1.62500000
575 2350 weekend 0.00000000
576 2355 weekend 0.00000000
library(lattice)
# xyplot(y ~ x | panel, data = dataset, type = "o")
xyplot(steps.mean ~ interval | as.factor(week),
data = byInterval,
type = "l",
layout=c(1,2))
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