stratified: Sample from a 'data.frame' according to a stratification...

In dustinfife/fifer: A Biostatisticians Toolbox for Various Activities, Including Plotting, Data Cleanup, and Data Analysis

Description

The stratified function samples from a data.frame in which one of the columns can be used as a "stratification" or "grouping" variable. The result is a new data.frame with the specified number of samples from each group.

Usage

stratified(df, group, size, select = NULL, replace = FALSE, bothSets = FALSE)

Arguments

df	The source data.frame.
group	Your grouping variables. Generally, if you are using more than one variable to create your "strata", you should list them in the order of slowest varying to quickest varying. This can be a vector of names or column indexes.
size	The desired sample size. If size is a value between 0 and 1 expressed as a decimal, size is set to be proportional to the number of observations per group. If size is a single positive integer, it will be assumed that you want the same number of samples from each group. If size is a vector, the function will check to see whether the length of the vector matches the number of groups and use those specified values as the desired sample sizes. The values in the vector should be in the same order as you would get if you tabulated the grouping variable (usually alphabetic order); alternatively, you can name each value to ensure it is properly matched.
select	A named list containing levels from the "group" variables in which you are interested. The list names must be present as variable names for the input data.frame.
replace	Logical. Should the sampling be done with replacement?
bothSets	Logical. Should just the samples be returned, or a list with two items: the sampled subset and the unsampled subset?

Note

Slightly different sizes than requested

Because of how computers deal with floating-point arithmetic, and because R uses a "round to even" approach, the size per strata that results when specifying a proportionate sample may be slightly higher or lower per strata than you might have expected.

Author(s)

Ananda Mahto

Examples

# Generate a couple of sample data.frames to play with
set.seed(1)
dat1 <- data.frame(ID = 1:100,
              A = sample(c("AA", "BB", "CC", "DD", "EE"), 100, replace = TRUE),
              B = rnorm(100), C = abs(round(rnorm(100), digits=1)),
              D = sample(c("CA", "NY", "TX"), 100, replace = TRUE),
              E = sample(c("M", "F"), 100, replace = TRUE))
dat2 <- data.frame(ID = 1:20,
              A = c(rep("AA", 5), rep("BB", 10),
                    rep("CC", 3), rep("DD", 2)))
# What do the data look like in general?
summary(dat1)
summary(dat2)

# Let's take a 10% sample from all -A- groups in dat1
stratified(dat1, "A", .1)

# Let's take a 10% sample from only "AA" and "BB" groups from -A- in dat1
stratified(dat1, "A", .1, select = list(A = c("AA", "BB")))

# Let's take 5 samples from all -D- groups in dat1,
#   specified by column number
stratified(dat1, group = 5, size = 5)

# Let's take a sample from all -A- groups in dat1, 
#   where we specify the number wanted from each group
stratified(dat1, "A", size = c(3, 5, 4, 5, 2))

# Use a two-column strata: -E- and -D-
#   -E- varies more slowly, so it is better to put that first
stratified(dat1, c("E", "D"), size = .15)

# Use a two-column strata (-E- and -D-) but only interested in
#   cases where -E- == "M"
stratified(dat1, c("E", "D"), .15, select = list(E = "M"))

## As above, but where -E- == "M" and -D- == "CA" or "TX"
stratified(dat1, c("E", "D"), .15,
     select = list(E = "M", D = c("CA", "TX")))

# Use a three-column strata: -E-, -D-, and -A-
s.out <- stratified(dat1, c("E", "D", "A"), size = 2)

list(head(s.out), tail(s.out))

# How many samples were taken from each strata?
table(interaction(s.out[c("E", "D", "A")]))

# Can we verify the message about group sizes?
names(which(table(interaction(dat1[c("E", "D", "A")])) < 2))

names(which(table(interaction(s.out[c("E", "D", "A")])) < 2))

dustinfife/fifer documentation built on Oct. 31, 2020, 3:36 p.m.