I have a data.frame with columns that needs to be labeled.
df <- structure(list(q1 = c("5", "6", "5", "5", "7", "5", "5", "5",
"5", "6", "5", "6", "6", "6", "7", "6", "5", "6", "5", "6", "6",
"5", "7", "5", "6", "6", "5", "6", "6", "5", "5", "5", "5", "5",
"5", "5", "4", "5", "5", "4", "4", "5", "4", "4", "5", "4", "5",
"5", "4", "5"), q2 = c("2", "2", "1", "1", "2", "1", "1", "2",
"1", "1", "1", "2", "1", "1", "2", "1", "2", "1", "2", "2", "2",
"1", "2", "2", "2", "2", "2", "1", "1", "2", "2", "2", "2", "2",
"2", "1", "2", "1", "1", "1", "1", "2", "1", "1", "1", "1", "2",
"2", "1", "2"), q3 = c("3", "3", "3", "3", "3", "3", "3", "3",
"3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3",
"3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3", "3",
"3", "2", "2", "2", "2", "2", "2", "2", "2", "2", "2", "2", "2",
"2", "2", "2")), row.names = c(NA, -50L), class = c("tbl_df",
"tbl", "data.frame"), na.action = structure(c(`71` = 71L, `78` = 78L,
`96` = 96L, `250` = 250L, `393` = 393L, `488` = 488L, `644` = 644L,
`847` = 847L, `862` = 862L, `1083` = 1083L, `1120` = 1120L, `1149` = 1149L,
`1322` = 1322L, `1357` = 1357L), class = "omit"))
Each column needs a different label. Those labels are in separated objects as shown below.
Q1_Label <- c("12 y", "13 y", "14 y", "15 y", "16 y", "17 y", "18 y")
Q2_Label <- c("Female", "Male" )
Q3_Label <- c("9th", "10th", "11th", "12th", "Ung"
How can I label the data frame columns using the character objects in the less line of code as possible?
Below is a code that tries to do that but I can not get the name of the sapply structure.
Thanks in advance for your help.
a_df <- sapply(X = df, FUN = function(x) factor(x,
levels = 1:length(table(x)),
labels = get(paste(toupper(names(x)), "_Label", sep = "")) # This line is where I get the problem
))
We can use put all the labels in a list and change the values in column using factor.
df[] <- Map(function(x, y) factor(x, labels = y[1:length(unique(x))]),
df,mget(ls(pattern = "Q\\d+_Label")))
df
# A tibble: 50 x 3
# q1 q2 q3
# <fct> <fct> <fct>
# 1 13 y Male 10th
# 2 14 y Male 10th
# 3 13 y Female 10th
# 4 13 y Female 10th
# 5 15 y Male 10th
# 6 13 y Female 10th
# 7 13 y Female 10th
# 8 13 y Male 10th
# 9 13 y Female 10th
#10 14 y Female 10th
# … with 40 more rows