How to loop through and modify multiple data frames in R

Question

I have data frames A, B, C, ... and want to modify each data frame in the same way, e.g. re-ordering factors levels of a factor which is present in all of the data frames:

A = data.frame( x=c('x','x','y','y','z','z') )
B = data.frame( x=c('x','y','z') )
C = data.frame( x=c('x','x','x','y','y','y','z','z','z') )

A$x = factor( A$x, levels=c('z','y','x') )
B$x = factor( B$x, levels=c('z','y','x') )
C$x = factor( C$x, levels=c('z','y','x') )

This gets laborious if there are lots of data frames and/or lots of modifications to be done. How can I do it concisely, using a loop or something better? A straightforward approach like

for ( D in list( A, B, C ) ) {
D$x = factor( D$x, levels=c('z','y','x') )
}

does not work, because it doesn't modify the original data frames.

EDIT: added definitions of A, B, and C to make it reproducible.

Definitions of A, B, and C have been added so that you can run the code. — baixiwei
– baixiwei, Commented Nov 2, 2013 at 3:26
Thanks. I know it can annoying especially when situation is obvious but it is a good practice and makes our lives easier :) — zero323
– zero323, Commented Nov 2, 2013 at 6:31

gung - Reinstate Monica · Accepted Answer · 2013-11-02 03:27:37Z

4

One thing to note about R is that, with respect to assignment, <- is transitive, whereas = is not. Thus, if your data frames are all the same in this respect, you should be able to do something like this:

A$x <- B$x <- C$x <- factor( C$x, levels=c('z','y','x') )

answered Nov 2, 2013 at 3:27

gung - Reinstate Monica

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2 Comments

baixiwei Over a year ago

Thanks for the reply. Unfortunately it will not work for my case (I think) because the data frames all have different numbers of rows. I'll modify my example to make this clear.

gung - Reinstate Monica Over a year ago

This is why the reproducible example is necessary.

zero323 · Accepted Answer · 2013-11-02 14:59:39Z

3

If you don't need explicit loop you can use lapply:

ll <- lapply(
    list(A, B, C),
    function(df) {
        df$x <- factor(df$x, levels=c('z', 'y', 'x'))
        return(df)
    }
)

Since data is only copied you'll have to use list returned from lapply.

Edit

dfs <- list('A', 'B', 'C')
levels <- c('z', 'y', 'x')

l <- lapply(
    dfs,
    function(df) {
        # Get data frame by name
        df <- get(df)
        df$x <- factor(df$x, levels=levels)
        return(df)
    }
)


for ( i in 1:length(dfs)) {
    assign(dfs[[i]], l[[i]])
}

edited Nov 2, 2013 at 14:59

answered Nov 2, 2013 at 6:29

zero323

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3 Comments

IRTFM Over a year ago

If you don't put in return(df) you will not get back dataframe elements.

baixiwei Over a year ago

This is OK, but I would like a way to modify the original data frames, or more precisely, I want to continue referring to them by their original names. Is there an easy way to get that result using the output of this solution?

zero323 Over a year ago

I've posted an edit with example solution but I cannot say I like it.

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