How to identify columns in a dataframe and replace them with new columns-CodePudding

I have this data frame

df <- data.frame(c(1:5), c(6:10), rep(1,length(5)), c(11:15), rep(4,length(5)))

I want to find all columns which have numbers that are all equal, eg. (all 1 or all 4).

After finding these columns, I would like to replace these columns with others that has 1 in the first row and 0 in all the others rows like this:

I used this to find the columns

i1 <- sapply(df, function(x) length(unique(x)) >1)

I do not know how to replace those columns with the new ones.

Thank you

CodePudding user response：

Welcome to SO!
You can try this in R base, probably there are more elegant ways:

# get the columns which have the same value (all values in columns == to their mean)
vec <- sapply(df, function(x) all(x == mean(x)))

# split df in two, one with the columns to replaces, and the other columns
df_no <- df[,!vec]
df_yes <- df[,vec]

# replace with 1,0,0 ...
df_yes <- sapply(df_yes, function (x) x = c(1, rep(0,length(x)-1)) )

# put together
df <- cbind(df_no, df_yes)
  
# order alphabetically the output if needed
df[order(names(df))]

  a  b c  d e
1 1  6 1 11 1
2 2  7 0 12 0
3 3  8 0 13 0
4 4  9 0 14 0
5 5 10 0 15 0

with data:

df <- data.frame(a=c(1:5), b=c(6:10),  c=rep(1,length(5)),d= c(11:15), e=rep(4,length(5)))

CodePudding user response：

Here's an apply solution:

df <- setNames(
    data.frame(c(1:5), c(6:10),  rep(1,length(5)), c(11:15), rep(4,length(5))),
    letters[1:5])

df_new <- apply(df, 2, function(col) 
    if(length(unique(col)) == 1) c(1, rep(0, length(col)-1)) else col)

df_new
     a  b c  d e
[1,] 1  6 1 11 1
[2,] 2  7 0 12 0
[3,] 3  8 0 13 0
[4,] 4  9 0 14 0
[5,] 5 10 0 15 0

Note that the output is now a matrix, which may actually be a better fit if your data really is all the same type across all columns; however, if you want a data.frame, as.data.frame(df_new) will do that for you.