Replace the NA value of a cell by the value of another column in the same dataframe

问题

I have a problem which seems to me quite simple but I don't manage to solve it by myself. I've searched for the solution on StackOverflow, I guess it has already been solved by someone but I haven't found it yet.

I have a data frame based upon the merger of 5 data frames, which looks like that :

id | mag1 | mag2 | mag3
1 | name | name | name
2 | NA | NA | name
3 | NA | name | NA

With mag2 and mag3 there always is a name which is filled (there is no row with an NA in mag1, mag2 and mag3). I would like to change the value of mag1 in order that it is never empty and that it takes the value of the next non-empty cell.

I have imagined to use this kind of code :

db$mag1[is.na(db$mag1)] <- db$mag2
db$mag1[is.na(db$mag1)] <- db$mag3

With this code, it seems to me that for instance, in the second line, the replacement with the value of db$mag2 will leave mag1 unchanged (NA) and that the replacement with db$mag3 will change its value to "name". The second line shouldn't be activated if there is a non-NA value in mag2.

Now, here is the error I got :

Warning message:
In db$mag[is.na(db$mag1)] <- db$mag2 :
   number of items to replace is not a multiple of replacement length

I guess there is a very simple error in my code line, but I don't manage to see it. Any idea?

回答1:

You have to use the logical index on both sides of the assignment <- so that the lengths are the same and corresponding elements are replaced.

 db$mag1[is.na(db$mag1)] <- db$mag3[is.na(db$mag1)]
 db
 #  id mag1 mag2 mag3
 #1  1 name name name
 #2  2 name <NA> name
 #3  3 <NA> name <NA>

data

 db <- structure(list(id = 1:3, mag1 = c("name", NA, NA), mag2 = c("name", 
 NA, "name"), mag3 = c("name", "name", NA)), .Names = c("id", 
 "mag1", "mag2", "mag3"), class = "data.frame", row.names = c(NA, 
 -3L))

来源：https://stackoverflow.com/questions/27548959/replace-the-na-value-of-a-cell-by-the-value-of-another-column-in-the-same-datafr

标签

dataframe

replace