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For the following data, I want each column to be replaced by (divide by) --> /corresponding length. (i.e. A/len, B/len, C/len,...)
... implies more columns, upto 70. As this has many columns, how one should proceed ?
A B C D E F ... len
2 4 5 7 8 8 5
5 8 3 1 0 4 6
8 9 3 9 6 2 12
2 6 2 6 7 8 10
1 2 4 2 9 5 20
812
To divide each column by a particular column, we can use division sign (/). For example, if we have a data frame called df that contains three columns say x, y, and z then we can divide all the columns by column z using the command df/df[,3].
To pick out single or multiple columns use the select() function. The select() function expects a dataframe as it's first input ('argument', in R language), followed by the names of the columns you want to extract with a comma between each name.
To access a specific column in a dataframe by name, you use the $ operator in the form df$name where df is the name of the dataframe, and name is the name of the column you are interested in.
The function unite() takes multiple columns and paste them together into one.
If your data frame df is exactly as you show, you can simply do
df[-ncol(df)] / df$len
If you have other columns to exclude, and you want them all included in the result, you can do something like
with(df, cbind(ID, df[!names(df) %in% c("ID", "len")]/len, len))
# ID A B C D E F len
# 1 1 0.4000000 0.800000 1.00 1.4000000 1.60 1.6000000 5
# 2 2 0.8333333 1.333333 0.50 0.1666667 0.00 0.6666667 6
# 3 3 0.6666667 0.750000 0.25 0.7500000 0.50 0.1666667 12
# 4 4 0.2000000 0.600000 0.20 0.6000000 0.70 0.8000000 10
# 5 5 0.0500000 0.100000 0.20 0.1000000 0.45 0.2500000 20
Also, as suggested by David in the comments, you can use data.table
library(data.table)
x <- c(1L, ncol(df))
setDT(df)[, names(df)[-x] := lapply(.SD, "/", df$len), .SDcols = -x]
which results in
# ID A B C D E F len
# 1: 1 0.4000000 0.800000 1.00 1.4000000 1.60 1.6000000 5
# 2: 2 0.8333333 1.333333 0.50 0.1666667 0.00 0.6666667 6
# 3: 3 0.6666667 0.750000 0.25 0.7500000 0.50 0.1666667 12
# 4: 4 0.2000000 0.600000 0.20 0.6000000 0.70 0.8000000 10
# 5: 5 0.0500000 0.100000 0.20 0.1000000 0.45 0.2500000 20
where df is
df <- read.table(text = "ID A B C D E F len
1 2 4 5 7 8 8 5
2 5 8 3 1 0 4 6
3 8 9 3 9 6 2 12
4 2 6 2 6 7 8 10
5 1 2 4 2 9 5 20", header = TRUE)
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