How to replace some values in a row based on condition in R

Question

i am new to R and hence going to ask a not-so-difficult question. I have a df which is about 200 columns and 1000 rows. It looks a bit like this

enter image description here

I need to calculate for each if the number of 2s is more than 0s in each row or vice-versa. If 2 is more than convert 0s in the same row to -1 or if 0 is more than convert 2 to -1.

I tried:

ifelse((rowSums(b=="0") > rowSums(b=="2")) , apply(b, 1 , function(b) b[b== "2" , ] <- "-1"), apply(b, 1 , function(b) b[b== "0" , ] <- "-1"))

but it gives the error:

Error in b[b == "2", ] <- "-1" : incorrect number of subscripts on matrix

Any help and suggestion are most welcome!

Do not post your data as an image, please read the info on how to give a reproducible example — Jaap
– Jaap, Commented Jun 26, 2016 at 16:49

akrun · Accepted Answer · 2016-06-26 17:27:47Z

We can use apply

t(apply(b, 1, FUN = function(x) {
            if(sum(x==2) > sum(x==0)) replace(x, x==0, -1)
            else if (sum(x==0) > sum(x==2)) replace( x, x==2, -1) 
            else x}))
#  ind1 ind2 ind3 ind4 ind5 ind6 ind7 ind8 ind9 ind10 ind11 ind12 ind13 ind14 ind15 ind16 ind17 ind18 ind19 ind20
#M8    -1    2    2    2   -1    2    2    1    1    -1     1     1     1     1     1     1     1     1     1     2
#M9     2    2    2    2    2    2    2   -1   -1     2     1     1     1     1     1     1     1     1    -1     1
#M17    1    1   -1    1    1    1    1    1    1     1     2     2     2     2     2    -1    -1    -1    -1     2
#M19    0   -1    0    0    0    0   -1    0    0     0     1    -1     1    -1    -1     1     1     1     1     1

Or we can do this based on rowSums

 i1 <- rowSums(b == 0) > rowSums(b == 2)
 b[b==0 & !i1] <- -1
 b[b==2 & i1] <- -1
 b
 #   ind1 ind2 ind3 ind4 ind5 ind6 ind7 ind8 ind9 ind10 ind11 ind12 ind13 ind14 ind15 ind16 ind17 ind18 ind19 ind20
 #M8    -1    2    2    2   -1    2    2    1    1    -1     1     1     1     1     1     1     1     1     1     2
 #M9     2    2    2    2    2    2    2   -1   -1     2     1     1     1     1     1     1     1     1    -1     1
 #M17    1    1   -1    1    1    1    1    1    1     1     2     2     2     2     2    -1    -1    -1    -1     2
 #M19    0   -1    0    0    0    0   -1    0    0     0     1    -1     1    -1    -1     1     1     1     1     1

data

 b <- structure(list(ind1 = c(0, 2, 1, 0), ind2 = c(2, 2, 1, 2), 
 ind3 = c(2, 
 2, -1, 0), ind4 = c(2, 2, 1, 0), ind5 = c(0, 2, 1, 0), ind6 = c(2, 
 2, 1, 0), ind7 = c(2, 2, 1, -1), ind8 = c(1, 0, 1, 0), ind9 = c(1, 
 0, 1, 0), ind10 = c(0, 2, 1, 0), ind11 = c(1, 1, 2, 1), ind12 = c(1, 
 1, 2, -1), ind13 = c(1, 1, 2, 1), ind14 = c(1, 1, 2, -1), ind15 = c(1, 
 1, 2, -1), ind16 = c(1, 1, 0, 1), ind17 = c(1, 1, -1, 1), ind18 = c(1, 
 1, -1, 1), ind19 = c(1, 0, 0, 1), ind20 = c(2, 1, 2, 1)), 
 .Names = c("ind1", 
 "ind2", "ind3", "ind4", "ind5", "ind6", "ind7", "ind8", "ind9", 
 "ind10", "ind11", "ind12", "ind13", "ind14", "ind15", "ind16", 
 "ind17", "ind18", "ind19", "ind20"), row.names = c("M8", "M9", 
 "M17", "M19"), class = "data.frame")

Just wondering: How did you transform the data in the image to a dataframe? I suppose it wasn't by typing everything .... (but I might be wrong)
@ProcrastinatusMaximus It is the old way of typing and no magic here :-)

mkearney · Accepted Answer · 2016-06-26 22:19:10Z

0

apply(data, 1, function(x) {
  if (sum(x == 2, na.rm = TRUE) > sum(x == 0, na.rm = TRUE)) {
    x[x == 0] <- -1
  } else if {sum(x == 0, na.rm = TRUE) > sum(x == 0, na.rm = TRUE)) {
    x[x == 2] <- -1
  } 
  x
})

answered Jun 26, 2016 at 22:19

mkearney

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