用于将一列中的值替换为另一列中的缺失值的R代码

Question

I have a data set named one with four columns: D1 , D2 , D3 and D4 . D1 is the id. D2 has seven levels ( a , b , c , d , e , f , g ). D3 has missing data, which I want to fill by matching conditions from columns D2 and D4 . I am selecting values from column D4 corresponding to four levels ( a , c , d , e ) of column D2 and then replacing the missing values of column D3 with those from D4 .

D1  D2  D3  D4
1   a   .   5
2   c   12  6
3   e   .   3
4   b   .   7
5   f   .   8
6   e   .   9
7   e   11  8
8   c   .   3
9   c   52  5
10  a   .   6
11  b   4   7
12  f   .   2
13  f   .   10
14  d   .   12
15  d   .   13
16  e   .   24
17  a   1   54
18  b   2   19
19  c   5   21

I have following solution but it is not working. Any suggestion or help? Thanks.

index <- with(one, D2 %in% c('a','c','d','e'))
one$D4[index] <- one$D3[index]
one

Answer 1

Assuming that you actually do have "." in the data, and that the data are read in as characters instead of numbers/NAs, the following solution should be easier to understand than the with() call:

d <- read.table(header=T, stringsAsFactors=F, text=
"D1  D2  D3  D4
1   a   .   5
2   c   12  6
3   e   .   3
4   b   .   7
5   f   .   8
6   e   .   9
7   e   11  8
8   c   .   3
9   c   52  5
10  a   .   6
11  b   4   7
12  f   .   2
13  f   .   10
14  d   .   12
15  d   .   13
16  e   .   24
17  a   1   54
18  b   2   19
19  c   5   21"
)

indices <- d$D2 %in% c("a","c","d","e") & d$D3 == "."
d$D3[ indices ] <- d$D4[ indices ]

And if you actually do have NAs instead of the "." characters you could easily just use is.na(d$D3) as your vector indices.

Answer 2

Another way is to use na.strings when reading the table and then using ifelse . Slightly verbose but easy to understand !

d <- read.table(header=T, stringsAsFactors=F, na.strings=".", text=
                  "D1  D2  D3  D4
1   a   .   5
2   c   12  6
3   e   .   3
4   b   .   7
5   f   .   8
6   e   .   9
7   e   11  8
8   c   .   3
9   c   52  5
10  a   .   6
11  b   4   7
12  f   .   2
13  f   .   10
14  d   .   12
15  d   .   13
16  e   .   24
17  a   1   54
18  b   2   19
19  c   5   21"
)


d$D3 <- ifelse(is.na(d$D3) & (d$D2 == 'a' | d$D2 == 'c' | d$D2 == 'd' | d$D2 == 'e'), d$D4, d$D3)