Fast replacing values in dataframe in R

问题:

I have a dataframe of 150,000 rows with 2,000 columns containing values, some being negatives. I am replacing those negative values by 0, but it is extremely slow to do so (~60min or more).

df[df < 0] = 0

where df[,1441:1453] looks like (all columns/values numeric):

  V1441 V1442 V1443 V1444 V1445 V1446 V1447 V1448 V1449 V1450 V1451 V1452 V1453
1     3     1     0     4     4    -2     0     3    12     5    17    34    27
2     0     1     0     7     0     0     0     1     0     0     0     0     0
3     0     2     0     1     2     3     6     1     2     1    -6     3     1
4     1     2     3     6     1     2     1    -6     3     1    -4     1     0
5     1     2     1    -6     3     1    -4     1     0     0     1     0     0
6     1     0     0     1     0     0     0     0     0     0     1     2     2

Is there a way to speed up such process, eg the way I am doing it is utterly slow, and there is faster approach for this ? Thanks.

回答1:

Try transforming your df to a matrix.

df <- data.frame(a=rnorm(1000),b=rnorm(1000))
m <- as.matrix(df)
m[m<0] <- 0
df <- as.data.frame(m)

回答2:

Both your original approach and the current answer create an object the same size as m (or df) when creating m<0 (the matrix approach is quicker because there is less internal copying with [<- compared with [<-.data.frame

You can use lapply and replace, then you are only looking at a vector or length (nrow(df)) each time and not copying so much

df <- as.data.frame(lapply(df, function(x){replace(x, x <0,0)})

The above code should be quite effiicent.

If you use data.table, then most of the memory (and) time inefficiency of the data.frame approach is removed. It would be ideal for a large data situation like yours.

library(data.table)
# this really shouldn't be 
DT <- lapply(df, function(x){replace(x, x <0,0)})
# change to data.table
setattr(DT, 'class', c('data.table','data.frame'))
# or 
# DT <- as.data.table(df, function(x){replace(x, x <0,0)})

You could set keys on all the columns and then replacing by reference for key values less than 0