In NiklasTR/bmi713:

Overview:

I load a package collection in the background.

library(tidyverse)
require(knitr)
# Set so that long lines in R will be wrapped:
opts_chunk$set(tidy.opts=list(width.cutoff=80),tidy=TRUE)

Question 1 (4 pts)

1.2 Getting Help (2 pts)

Please state what the arguments to the "sum" function in R?

It may be useful to use some of R's built in help functionality.

# Get help on the function sum

# ANSWER 1.2
# The arguments of the sum function are:
# * A vector with either numeric, logical (True/False = 1/0) or complex numbers
# * An na.rm argument. This argument is a logical and decides if vectors with a 
#missing value are going to produce a missing value as overall result or if NAs 
#are dropped during summation.  

1.3 Using Help (2 pts)

vector_example <- c(5,6,7,8) 
vector_example

Using the R documentation for help if needed, please write one to two lines of code to create a vector of numbers ranging from 50 to 100 (inclusive) and report the standard deviation of this vector. Credit will NOT be given if the entire vector is manually entered like the example below:

my_vector <- c(50:100)
sd(my_vector)

# ANSWER 1.3
# The vector can be generated using the colon operator 
# The standard deviation is about 14.87

set.seed(42)
# Create random vector
random_vector <- runif(50, min=0, max=50)
# Get standard deviation of vector
sd(random_vector)

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Question 2 (12 pts)

Basic R syntax

2.1 Indexing (1 pts)

Each element in our vector, vector_example is stored in a position that can be addressed by its index. Different programming languages use different starting indices. Write the command that will index the first value in vector_example.

# Index first element 
vector_example[1]

# ANSWER 2.1
#R starts indexing at 1 (different from many other programming languages like Python)

2.2 Booleans & subsetting (4 pts)

Now write a code that will:

1) Create variable c which stores whether each number in vector_example is even or odd (hint logical vector)

2) Using vector c, create two separate vectors, Evens_example and Odds_example, which store the even and odd values of vector_example, respectively.

Your answer should make use of of an ifelse statement and appropriate subsetting.

# create vector C
# elegant alternative: c <- as.logical(vector_example%%2)
c <- ifelse(vector_example%%2, 1, 0)

# create Evens_example
Evens_example <- vector_example[!c]

# create Odds_example
Odds_example <- vector_example[c]

#ANSWER 2.2
#I calculate the modulo 2 for each number in the vector. 
#Uneven elements have a residual of one. 
#By using a logical transformation, I turn the residual value of 1 into a TRUE statement. 
#This statement hence indicates an uneven number at its respective index.
#Next I use c as an index-vector to return even and uneven numbers. 
#To return even numbers I use the logical negating variable "!".

2.3 Understanding Syntax (2 pts)

In our first lecture, we worked through of data normalization in which a random 5x20 example matrix needed to be normalized by column (scaling values so that they are between 0 and 1). Now write a similar normalization function from so that it normalizes by rows instead of columns.

Remember: check your logic for the normalization procedure as we corrected in the example in class.

# Create row-wise normalization function
normalize_row_wise <- function(input_matrix){
require(magrittr) 
require(dplyr)

input_matrix %>% t() %>% 
  #I use my function and insert two transpositions, here
  as.data.frame() %>% 
  mutate_all(funs(.-min(.))) %>% 
  # I subtract the minimal value from all values (thereby creating a 0 value field per column)
  mutate_all(funs(./max(.))) %>% 
  # Afterwards I scale all values by the maximum value, making the latter equal to 1.  
  as.matrix() %>% t() %>% #and here
    return()
}

#It was not clear to me wether we have to generate a "classic" matrix normalization 
#or if we should scale every row between 0 and 1. English is not my first language.
# normalize_simple <- function(input_matrix){
#   return((input_matrix-min(input_matrix))/max(input_matrix-min(input_matrix)))
# }

2.4 Adding conditionals (5 pts)

Write a normalization function (column-wise) that takes an input matrix as an argument and returns nothing if the normalized matrix is equal to the input matrix but otherwise returns normalized matrix.

Is a return statement needed in an R function (look back to the "normalize" functions from lecture and in 2.4)?

# Create column-wise normalization function 
column_wise <- function(input_matrix){
require(magrittr) 
require(dplyr)

output_matrix <- input_matrix %>%
  as.data.frame() %>% 
  mutate_all(funs(.-min(.))) %>% 
  # I subtract the minimal value from all values (thereby creating a 0 value field per column)
  mutate_all(funs(./max(.))) %>% 
  # Afterwards I scale all columns by their maximum value, making the latter equal to 1.
  as.matrix() %>% 
  unname() #helper to remove all names that have been introduced during conversions

if(!identical(input_matrix, output_matrix)){return(output_matrix)} else{return()}
}

# ANSWER 2.4
# The function is very similar to the function above with the only difference 
#that I did not introduce a transposition twice. The output of the function is controlled 
#by a conditional statement which depends on the result of identical(). Since identical() is 
#very strict, I have to remove all object attributes such as name from the 
#output matrix before comparison with unname().
#A return statement is not necessary in R functions, as R will always 
#return the last modified object in the function envionment. In general, a return() 
#argument is considered good style. In my implementation it is necessary to control the output. 

Question 3 (5 pts)

a) What is one reason why using the "assign" function in R is stylistically a bad choice?

b) Which two other operators are most commonly utilized for variable assignment?

c) Which of these two operators is better to use stylistically?

Write a short piece of code that utilizes both possible operators most commonly utilized for variable assignment at least once where swapping all occurence of each operator for the other results in a different output or generate an error.

# Using Operator 1
y = 5
plot(x = 1,y = 3)
y

# Replace w/ Operator 2 and get error/different output
y <- 5
plot(x <-  1,y <- 3)
y

# Using Operator 2
x <- 3
c(mean(x <- 1:15), x)

# Replace w/ Operator 1 and get error/different output
x = 3
c(mean(x = 1:15), x)


# ANSWER 3a
#Assign sets "a value to a name in an envionment" (R Documentation).
#This environment, however, is most of the time not specified. 
#This leads to an assignment of the name and value pair in the general envionment, 
#even if assign() is used in a seperate function or loop. 
#Therefore, if you set x <- 3 in your script and you would run an unrelated third-party 
#function afterwards that uses the name "x" and assign() to assign a value, 
#you could not expect x==3 to be TRUE. 
#Moreover, assign breaks the magrittr pipe operator %>% which constructs 
#object-overwriting helper functions to proccess a pipe. 

# ANSWER 3b
# Values can be assigned with <- and = 

# ANSWER 3c
# <- should be preferred acc. to the Google Style Guide for R 

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Question 4 (10 pts)

For Loops and Apply

Find the drive time for each car in the cars dataset. (Hint: Time = Distance/Speed)

# Cars is a built-in dataset
# You can use head() to view the dataset
head(cars)
times_elegant <- cars %>% 
  mutate(time = dist/speed)

head(times_elegant, 20)
#ANSWER
#I use the mutate() function from the dplyr package to introduce 
#a rowwise operation to the dataframe

4a. Using For (3)

Using a For Loop, find and report the time for each car.

# Calculate drive time for each car using for loop
times_loop <- c()
for(ii in 1:nrow(cars)){
  times_loop[[ii]] <- cars$dist[ii]/cars$speed[ii]
}

# Display the drive times
times_loop

4b. Now using apply (3)

Using Apply, find and report the time for each car.

# Calculate drive time using apply()
times_apply <- apply(cars, 1, function(x){x[2]/x[1]})
# Display the drive times
times_apply

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Question 5 (7 pts)

5a. Scope Example

If the below block of code were to be executed, what are the resulting values of a and b?

a <- c(3,6,2,3,6,4,1,7,6,3,9)
b <- 0.5
function_1 <- function(a,b){
  b <- lapply(a,function(x) x/b)
  b
}
a = b <- function_1(a,b)

a <- c(3,6,2,3,6,4,1,7,6,3,9)
b <- 0.5
function_1 <- function(a,b){
  b <- lapply(a,function(x) x/b)
  b
}
a = b <- function_1(a,b)

identical(a,b)

# ANSWER 5a
# a and b are identical lists of 11 numeric elements, representing the 
#initial values of a*2. The initial values in "a" are coerced to a list by 
#list-apply (aka lappy), which divides the values in "a" by 0.5 and stores them 
#in a list with the name "b". "b" then is assigned to "a". 

5b. A Questions On Scope

*a) How does variable assignment within a function affect a variable that has already been previously assigned a value outside a function (look back to question 5a and to lecture)? *

# ANSWER 5b.
# It does not effect the value outside the function in general (if assign() was not used). 
#The assigned objects outside the function are usually part of the global enviromnent 
#while objects inside the function are located in a separate envionment. 

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Question 6 (12 pts)

Conditionals

We explored 2 types of conditionals: if/else and switches.

Use the iris dataset for these exercises.

# Iris is a built-in dataset
# You can use head() to view the dataset
head(iris)

6a. If/else

i. Define a function using if/else and apply() that will return sepal length as long as petal length / petal width > 5, otherwise return NA. Use apply to run this on the iris dataset and report your results. (3)

# Create function using if/else
f_ifelse <- function(row_var){
  if(row_var[3]/row_var[4] > 5){return(row_var[1])}
  else{return(NA)}
}

# Call function using apply() and report results
apply(iris[,1:4], 1, f_ifelse)

#ANSWER
#The function is based on an if() and else() element 

ii. Instead of using apply, do the same thing using the ifelse() function. Run using the iris dataset, and report your results. (2)

# Use ifelse() and report results
iris[ifelse(iris$Petal.Length/iris$Petal.Width > 5, TRUE, NA),1]

6b. Switches (3)

Using switch, define a function that will return sepal dimensions (Length and width) for setosa flowers and return Petal demensions (Length and width) otherwise. Run this on iris and report your results.

# Create function using switch
# elegant_switch <- function(df, switcher){
#   switch(switcher, 
#          sepal = df %>% as.tibble %>%
#            filter(Species == "setosa") %>%
#            select(Sepal.Length, Sepal.Width) %>% 
#            as.data.frame(),
#          df %>% as.tibble %>%
#            filter(Species != "setosa") %>%
#            select(Petal.Length, Petal.Width) %>% 
#            as.data.frame())
# }

#Create another function using switch that needs apply to go over the df. 
apply_switch <- function(row_var, switcher){

  switch (switcher,
    sepal = if(row_var[5] == "setosa"){return(c(row_var[1], row_var[2]) %>% unlist())}
    else{return(c(NA, NA))},
    if(row_var[5] != "setosa"){return(c(row_var[3], row_var[4]) %>% unlist())}
    else{return(c(NA, NA))}
  )
}

# Call function using apply() and report results
switch_result <- apply(iris, 1, apply_switch, switcher = "sepal") %>% t() %>% as.tibble() %>%
  magrittr::set_colnames(c("my Variable 1", "my Variable 2"))
head(switch_result)

switch_result <- apply(iris, 1, apply_switch, switcher = "petal") %>% t() %>% as.tibble() %>% 
  magrittr::set_colnames(c("my Variable 1", "my Variable 2"))
head(switch_result)

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Question 7 (1 pt)

How long did this problem set take to complete?

Please report this number in hours. 4 hours

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Session Info

sessionInfo()

NiklasTR/bmi713 documentation built on May 29, 2019, 11:01 p.m.