Creating a loop to add labels to colums: library(Hmisc) - r

Question

I have a dataset which looks something like this:
Year Country Matchcode P H
1 2000 France 0001 1213 1872
2 2001 France 0002 1234 2345
3 2000 UK 0003 1726 2234
4 2001 UK 0004 6433 9082
I have another dataset which looks something like this:
Indicator Code Indicator Name
P Power
H Happiness
I would like to add info in the second column of the second dataset (Power, Happiness) as a label to the abbreviation used in the first dataset with a loop, but I don't know exactly how to write the loop.
This is how far I got:
library(Hmisc)
for i in df2[,1]{
if (df1[,i] == df2[i,]){
label(df1[,i]) <- df2[i,2]
}}
But this merely checks whether names are the same and does not search for it.
Could anyone guide further?
Desired output:
Year Country Matchcode P(label=Power) H(label=Happiness)
1 2000 France 0001 1213 1872
2 2001 France 0002 1234 2345
3 2000 UK 0003 1726 2234
4 2001 UK 0004 6433 9082

Answer 1

If you specifically want to use a loop, this approach gives the output you describe:
df <- data.frame(Year = c(2000, 2001, 2000, 2001),
Country = c("France", "France", "UK","UK"),
Matchcode = c("0001", "0002", "0003", "0004"),
P = c(1213, 1234, 1726, 6433),
H = c(1872, 2345, 2234, 9082))
lookup <- data.frame(code = c ("P", "H"),
label = c("Power", "Happiness"),
stringsAsFactors = FALSE)
for (i in 1:length(colnames(df))) {
if(!is.na(match(colnames(df), lookup$code)[i])) {
Hmisc::label(df[[i]]) <- lookup$label[(match(colnames(df), lookup$code))[i]]
}
}
This works:
Hmisc::label(df[4])
# P
# "Power"
It also checks out in the RStudio viewer:
Like several of the other answerers and commenters, I had originally thought you wanted to append the "label = " text to the column names. For anyone wanting that, this is the (loop) code.
for (i in 1:length(colnames(df))) {
if(!is.na(match(colnames(df), lookup$code)[i])) {
colnames(df)[i] <- paste0(colnames(df)[i],
"(label=",
lookup$label[(match(colnames(df), lookup$code))[i]],
")")
}
}

Answer 2

It's not clear to me at all what you're trying to do with Hmisc::label but I think you're misinterpreting the role & function of Hmisc::label.
Consider the following:
Let's construct a sample data.frame consisting of 2 rows and 2 columns.
df <- setNames(data.frame(matrix(0, ncol = 2, nrow = 2)), c("a", "b"))
df
# a b
#1 0 0
#2 0 0
We extract the column names. Note that cn is a character vector.
cn <- colnames(df)
cn
#[1] "a" "b"
We now set a Hmisc::label for cn.
label(cn) <- "label for cn"
cn
#label for cn
#[1] "a" "b"
We inspect the attributes of cn
attributes(cn)
#$label
#[1] "label for cn"
#
#$class
#[1] "labelled" "character"
We now assign cn to the column names of df.
colnames(df) <- cn
df
# a b
#1 0 0
#2 0 0
Note how the label attribute is not stored as part of the column names.

Answer 3

Here's a dplyr solution:
# example datasets
df = read.table(text = "
Year Country Matchcode P H
1 2000 France 0001 1213 1872
2 2001 France 0002 1234 2345
3 2000 UK 0003 1726 2234
4 2001 UK 0004 6433 9082
", header=T)
df2 = read.table(text = "
IndicatorName IndicatorCode
P Power
H Happiness
", header=T)
library(dplyr)
data.frame(original_names = names(df)) %>% # get original names
left_join(df2, by=c("original_names"="IndicatorName")) %>% # join names that should be updated
mutate(new_names = ifelse(is.na(IndicatorCode), original_names, paste0(original_names,"(label=",IndicatorCode,")"))) %>% # if there is a match update the name
pull(new_names) -> list_new_names # get column of new names and store it in a vector
# update names
names(df) = list_new_names
# check new names
df
# Year Country Matchcode P(label=Power) H(label=Happiness)
# 1 2000 France 1 1213 1872
# 2 2001 France 2 1234 2345
# 3 2000 UK 3 1726 2234
# 4 2001 UK 4 6433 9082

Answer 4

This would work. Find the corresponding text using %in%, and use paste0 to generate the label.
colnames(df1)[4:5] <- paste0(colnames(df1)[4:5], '(label=', df2$V2[colnames(df1)[4:5] %in% df2$V1], ')')
df1
Year Country Matchcode P(label=Power) H(label=Happiness)
1 2000 France 1 1213 1872
2 2001 France 2 1234 2345
3 2000 UK 3 1726 2234
4 2001 UK 4 6433 9082
Data used
df1 <- read.table(text="Year Country Matchcode P H
1 2000 France 0001 1213 1872
2 2001 France 0002 1234 2345
3 2000 UK 0003 1726 2234
4 2001 UK 0004 6433 9082", header=T, stringsAsFactors=F)
df2 <- read.table(text="
P Power
H Happiness", header=F, stringsAsFactors=F)

Answer 5

If you still stick with Hmisc, you can modify the 'print' function to handle the extra information provided by the labels, or rather (and less harmfull) says to R that your data has to be printed using the labels. You can achieve this by creating a new data frame class for which the print function behaves differently.
The 'print' trick is not necessary with Rstudio that natively uses the labels together with the column names.
df1 = read.table(text = "
Year Country Matchcode P H
1 2000 France 0001 1213 1872
2 2001 France 0002 1234 2345
3 2000 UK 0003 1726 2234
4 2001 UK 0004 6433 9082 ", header=T)
df2 = read.table(text = "
var lab
P Power
H Happiness", header=T, stringsAsFactors=FALSE)
## Set the labels of the columns in df1 accordingly to df2
library(Hmisc)
for (i in 1:ncol(df1)) {
lab <- df2[df2$var==colnames(df1)[i],2]
if (length(lab!=0)) label(df1[[i]]) <- lab
}
# A print' function dedicated to 'truc' objects
# Mainly it is the code from the original 'print' except for dimnames[[2L]]
print.truc <- function (x, ..., digits = NULL, quote = FALSE, right = TRUE,
row.names = TRUE)
{
n <- length(row.names(x))
if (length(x) == 0L) {
cat(sprintf(ngettext(n, "data frame with 0 columns and %d row",
"data frame with 0 columns and %d rows"), n), "\n",
sep = "")
}
else if (n == 0L) {
print.default(names(x), quote = FALSE)
cat(gettext("<0 rows> (or 0-length row.names)\n"))
}
else {
m <- as.matrix(format.data.frame(x, digits = digits,
na.encode = FALSE))
if (!isTRUE(row.names))
dimnames(m)[[1L]] <- if (isFALSE(row.names))
rep.int("", n)
else row.names
dimnames(m)[[2L]] <- purrr::map(1:ncol(x),
function(i) {
z <- attributes(x[[i]])$label
if (length(z)!=0) z else colnames(x)[i]
})
print(m, ..., quote = quote, right = right)
}
invisible(x)
}
# Says that 'df1' is an 'enhanced' data frame
class(df1) <- c("truc",class(df1))
# Print as enhanced
print(df1)
# Eyra Country Matchcode Power Happiness
#1 2000 France 1 1213 1872
#2 2001 France 2 1234 2345
#3 2000 UK 3 1726 2234
#4 2001 UK 4 6433 9082
# Print using standard way
print(as.data.frame(df1))
# Year Country Matchcode P H
#1 2000 France 1 1213 1872
#2 2001 France 2 1234 2345
#3 2000 UK 3 1726 2234
#4 2001 UK 4 6433 9082

Answer 6

No need for a loop with Hmisc, can do this in one line using the option self = FALSE in the label command.
label(df1[, df2$IndicatorName], self = FALSE) <- df2$IndicatorCode
Ie.
library(Hmisc, warn.conflicts = FALSE, quietly = TRUE)
df1 = read.table(text = "
Year Country Matchcode P H
1 2000 France 0001 1213 1872
2 2001 France 0002 1234 2345
3 2000 UK 0003 1726 2234
4 2001 UK 0004 6433 9082
", header=T)
df2 = read.table(text = "
IndicatorName IndicatorCode
P Power
H Happiness
", header=T)
label(df1[, df2$IndicatorName], self = FALSE) <- df2$IndicatorCode
sapply(df1, label)
#> Year Country Matchcode P H
#> "" "" "" "Power" "Happiness"
Created on 2020-09-14 by the reprex package (v0.3.0)