Get column index from label in a data frame

Question

Say we have the following data frame:

> df
  A B C
1 1 2 3
2 4 5 6
3 7 8 9

We can select column \'B\' from its index:

&g         


        

Answer 1

match("B", names(df))

Can work also if you have a vector of names.

Answer 2

Following on from chimeric's answer above:

To get ALL the column indices in the df, so i used:

which(!names(df)%in%c()) 

or store in a list:

indexLst<-which(!names(df)%in%c())

Answer 3

you can get the index via grep and colnames:

grep("B", colnames(df))
[1] 2

or use

grep("^B$", colnames(df))
[1] 2

to only get the columns called "B" without those who contain a B e.g. "ABC".

Answer 4

This seems to be an efficient way to list vars with column number:

cbind(names(df)) 

Output:

     [,1]
[1,] "A" 
[2,] "B" 
[3,] "C" 

Sometimes I like to copy variables with position into my code so I use this function:

varnums<- function(x) {w=as.data.frame(c(1:length(colnames(x))),
          paste0('# ',colnames(x)))
names(w)= c("# Var/Pos")
w}
varnums(df)

Output:

# Var/Pos
# A         1
# B         2
# C         3

Answer 5

Use t function:

t(colnames(df))

     [,1]   [,2]   [,3]   [,4]   [,5]   [,6]  
[1,] "var1" "var2" "var3" "var4" "var5" "var6"

Answer 6

I wanted to see all the indices for the colnames because I needed to do a complicated column rearrangement, so I printed the colnames as a dataframe. The rownames are the indices.

as.data.frame(colnames(df))

1 A
2 B
3 C