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:
> df[,2] [1] 2 5 8 Is there a way to get the index (2) from the column label ('B')?
- 2See @matthewdowle's answer here for the best solution: stackoverflow.com/a/9277935/636656Ari B. Friedman– Ari B. Friedman2014-07-14 16:20:39 +00:00Commented Jul 14, 2014 at 16:20
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10 Answers
you can get the index via grep and colnames:
grep("B", colnames(df)) [1] 2 or use
grep("^B$", colnames(df)) [1] 2 to only get the columns called "B" without those who contain a B e.g. "ABC".
4 Comments
IRTFM
Your original example's advantages could be demonstrated in code if you showed its use in something like df[ , grep("^B", colnames(df)) ], i.e, returning the dataframe columns starting with "B". Feel free to use in a further edit if you agree.
IRTFM
Or even df[ , grep("^[BC]", colnames(df)) ], i.e., the columns that start with either B or C.
Henrik
@Dwin: As @aix already said, the asker wants the index. But I also usually use
grep the way you describe it.
user989762
@Henrik. Thank you so much. This must be the single most useful command to work with dplyr and variables!
The following will do it:
which(colnames(df)=="B") 
6 Comments
Henrik
The problem with
grep is also the advantage, namely that it uses regular expressions (so you can search for any pattern in your colnames). To just get the colnames "B" use "^B$" as the pattern in grep. ^ is the metacharacter for the beginning and $ for the end of a string.
nico
You don't even need
which. You can directly use df[names(df)=="B"]
NPE
@nico The question is to get the index of the column.
Panos Kalatzantonakis
"Which" worked for me in every case. I couldn't get a column with the name "fBodyAcc-meanFreq()-Z" using grep.
Steve
@Kabamaru: Grep will work as long as you escape the metacharacters. For the example you gave, this will work:
grep("^fBodyAcc-meanFreq\\()-Z$",colnames(df)) or also grep("^fBodyAcc-meanFreq\\(\\)-Z$",colnames(df)). |
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 
2 Comments
lillemets
A more concise way to do this is
cbind(names(df)).
Gregor Thomas
@lillemets if brevity is your goal,
t(t(names(df))) saves you 2 characters ;)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()) 
1 Comment
Dimitrios Zacharatos
i think this is the best answer because it can be generalized
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 
Comments
match("B", names(df)) Can work also if you have a vector of names.
Comments
To generalize @NPE's answer slightly:
which(colnames(dat) %in% var) where var is of the form
c("colname1","colname2",...,"colnamen") returns the indices of whichever column names one needs.
Comments
Use t function:
t(colnames(df)) [,1] [,2] [,3] [,4] [,5] [,6] [1,] "var1" "var2" "var3" "var4" "var5" "var6" 
Comments
Here is an answer that will generalize Henrik's answer.
df=data.frame(A=rnorm(100), B=rnorm(100), C=rnorm(100)) numeric_columns<-c('A', 'B', 'C') numeric_index<-sapply(1:length(numeric_columns), function(i) grep(numeric_columns[i], colnames(df))) 
2 Comments
Gregor Thomas
That
sapply is a long way to write match(numeric_columns, names(df)) --- unless you really need the regex power rather than exact string matching.
Jimmy TwoCents
thanks @GregorThomas...not super familar with match. In this case it is quite a bit shorter, but I like the sapply because it's a little more explicit what is going on...to each their own i guess (havem't benchmarked any performance differences)
#I wanted the column index instead of the column name. This line of code worked for me:
which (data.frame (colnames (datE)) == colnames (datE[c(1:15)]), arr.ind = T)[,1] #with datE being a regular dataframe with 15 columns (variables) data.frame(colnames(datE)) #> colnames.datE. #> 1 Ce #> 2 Eu #> 3 La #> 4 Pr #> 5 Nd #> 6 Sm #> 7 Gd #> 8 Tb #> 9 Dy #> 10 Ho #> 11 Er #> 12 Y #> 13 Tm #> 14 Yb #> 15 Lu which(data.frame(colnames(datE))==colnames(datE[c(1:15)]),arr.ind=T)[,1] #> [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 
