Hello,
A one-liner could be
df1 <- read.table(text="
cola colb colc cold cole
1 NA 5 9 NA 17
2 NA 6 NA 14 NA
3 3 NA 11 15 19
4 4 8 12 NA 20
", header=TRUE)
df2 <- read.table(text="
cola colb colc cold cole
1 1.4 0.8 0.02 1.6 0.6
", header=TRUE)
sapply(names(df1), function(nm) {df1[[nm]][is.na(df1[[nm]])] <-
df2[[nm]]; df1[[nm]]})
Avoid loops, use *apply.
Hope this helps,
Rui Barradas
Em 11-07-2012 15:11, paulalou escreveu:
Hi,
I have two dataframes:
The first, df1, contains some missing data:
cola colb colc cold cole
1 NA 5 9 NA 17
2 NA 6 NA 14 NA
3 3 NA 11 15 19
4 4 8 12 NA 20
The second, df2, contains the following:
cola colb colc cold cole
1 1.4 0.8 0.02 1.6 0.6
I'm wanting all missing data in df1$cola to be replaced by the value of
df2$cola. Then the missing data in df1$colb to be replaced with the
corresponding value in df2$colb etc.
I can get this to work column by column with single input lines but as my
original dataset is a lot larger I'm wanting a create a loop but can't work
out how.
The single line command is:
df1$cola[is.na(df1$cola)]<-df2$cola
I've tried a replace function within a loop but get error messages:
list<-colnames(df1)
for (i in list) {
r<-replace(df1$i,df1$i[is.na(df1$i)],df2$i)
}
with error messages of:
Warning messages:
1: In is.na(mymat$snp) :
is.na() applied to non-(list or vector) of type 'NULL'
Can anyone help me with this?
Thanks
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