Primary exercises
- Manually created factor.
In a study participants were asked whether their sport activity is none
, oncePerWeek
, severalPerWeek
or daily
.
Build a proper factor for the responses below and store it in a variable w
.
Print the factor.
Write the code to count the numbers of occurrences of each level and print the counts.
severalPerWeek, none, none, oncePerWeek, oncePerWeek, oncePerWeek, oncePerWeek, ?, none, none
v <- c( "severalPerWeek", "none", "none", "oncePerWeek", "oncePerWeek", "oncePerWeek", "oncePerWeek", NA, "none", "none" )
w <- factor( v, levels = c( "none", "oncePerWeek", "severalPerWeek", "daily" ) )
w
[1] severalPerWeek none none oncePerWeek oncePerWeek oncePerWeek
[7] oncePerWeek <NA> none none
Levels: none oncePerWeek severalPerWeek daily
fct_count( w )
# A tibble: 5 × 2
f n
<fct> <int>
1 none 4
2 oncePerWeek 4
3 severalPerWeek 1
4 daily 0
5 <NA> 1
- A factor with a random content.
Read help about the function sample
.
Then study and try the following lines of code to understand the results.
Next, understand why an error is generated and use replace
argument to generate a vector with 100 samples.
Store this vector in a variable v
and build a factor w
from it.
Finally, count the numbers of occurrences of each level in w
.
Ensure, that levels are in order provided in the variable lvs
.
lvs <- c( "none", "oncePerWeek", "severalPerWeek", "daily" )
sample( lvs, 3 )
[1] "severalPerWeek" "none" "daily"
sample( lvs, 3 )
[1] "oncePerWeek" "none" "severalPerWeek"
sample( lvs, 3 )
[1] "severalPerWeek" "none" "oncePerWeek"
sample( lvs, 100 )
Error in sample.int(length(x), size, replace, prob): cannot take a sample larger than the population when 'replace = FALSE'
v <- sample( lvs, 100, replace = TRUE )
w <- factor( v, levels = lvs )
w
[1] severalPerWeek oncePerWeek none oncePerWeek none severalPerWeek
[7] none oncePerWeek none severalPerWeek severalPerWeek daily
[13] oncePerWeek oncePerWeek oncePerWeek daily daily none
[19] daily none oncePerWeek none severalPerWeek none
[25] severalPerWeek daily none oncePerWeek oncePerWeek daily
[31] none severalPerWeek severalPerWeek none none severalPerWeek
[37] oncePerWeek daily daily oncePerWeek oncePerWeek none
[43] severalPerWeek oncePerWeek none oncePerWeek none daily
[49] daily severalPerWeek none daily none oncePerWeek
[55] daily none daily daily oncePerWeek daily
[61] daily daily daily none none severalPerWeek
[67] none severalPerWeek daily severalPerWeek daily oncePerWeek
[73] oncePerWeek daily none severalPerWeek daily severalPerWeek
[79] none daily oncePerWeek daily severalPerWeek daily
[85] severalPerWeek oncePerWeek oncePerWeek oncePerWeek none daily
[91] none none severalPerWeek oncePerWeek none severalPerWeek
[97] none daily none oncePerWeek
Levels: none oncePerWeek severalPerWeek daily
fct_count( w )
# A tibble: 4 × 2
f n
<fct> <int>
1 none 29
2 oncePerWeek 24
3 severalPerWeek 20
4 daily 27
- Reordering factor levels.
When a factor is shown on an axis of a plot, the order is given by its levels.
The factor w
from the previous exercise will be then shown in this order: none
, oncePerWeek
, severalPerWeek
, daily
.
But for a picture in a manuscript the following order might be needed: daily
, severalPerWeek
, oncePerWeek
, none
.
Apply to w
one of the fct_
functions from the tidyverse
library to produce a factor w2
with the requested order.
Show the levels of w2
.
Again show the number of elements of each level in w2
and compare it with the table of the previous exercise.
w2 <- fct_relevel( w, c( "daily", "severalPerWeek", "oncePerWeek", "none" ) )
levels( w2 )
[1] "daily" "severalPerWeek" "oncePerWeek" "none"
fct_count( w2 )
# A tibble: 4 × 2
f n
<fct> <int>
1 daily 27
2 severalPerWeek 20
3 oncePerWeek 24
4 none 29