2

I need to sum some columns in a data.frame with a rule that says, a column is to be summed to NA if more than one observation is missing NA if only 1 or less missing it is to be summed regardless.

Say I have some data like this,

dfn <- data.frame(
a  = c(3, 3, 0, 3),
b  = c(1, NA, 0, NA),
c  = c(0, 3, NA, 1))

dfn
  a  b  c
1 3  1  0
2 3 NA  3
3 0  0 NA
4 3 NA  1

and I apply my rule, and sum the columns with less then 2 missing NA. So I get something like this.

  a  b  c
1 3  1  0
2 3 NA  3
3 0  0 NA
4 3 NA  1
5 9 NA  4

I've played around with colSums(dfn, na.rm = FALSE) and colSums(dfn, na.rm = TRUE). In my real data there is more then three columns and also more then 4 rows. I imagine I can count the missing some way and use that as a rule?

Ricardo Oliveros-Ramos
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Eric Fail
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2 Answers2

5

I don't think you can do this with colSums alone, but you can add to its result using ifelse:

colSums(dfn,na.rm=TRUE) + ifelse(colSums(is.na(dfn)) > 1, NA, 0)
 a  b  c 
 9 NA  4
flodel
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James
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  • Works like a charm, I wasn't aware of the open `+ ifelse`. Thanks a lot! – Eric Fail Jan 18 '13 at 18:21
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    @EricFail In this context `ifelse` produces another vector of the same size as the result from `colSums`. You are just adding 2 vectors together. – James Jan 18 '13 at 18:24
  • I see, I keep getting impressed by how freely the function in R can be combined. Thank you! – Eric Fail Jan 18 '13 at 18:29
1

Nothing wrong with @James' Answer, but here's a slightly cleaner way:

colSums(apply(dfn, 2, function(col) replace(col, match(NA, col), 0)))
# a  b  c 
# 9 NA  4

match(NA, col) returns the index of the first NA in col, replace replaces it with 0 and returns the new column, and apply returns a matrix with all of the new columns.

Matthew Plourde
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