根据R或python中的条件替换另一列列表中的一个列列表中的值

Question

(for pythonistas, the below code is in R's format before I get some #hatehard)

This one has been frustrating me for a way too long.

I have 2 datasets

df1 <- data.frame(ID = c("Person.A", "Person.B", "Person.C", "Person.D", "Person.E", "Person.F"),
                  Aa = c(0,1,2,NA,1,1),
                  Ab = c(0,NA,2,1,1,1),
                  Ac = c(NA,NA,2,2,1,1),
                  no.match = c(0,1,2,2,1,2))

df2 <- data.frame(ID = c("Person.A", "Person.B", "Person.C", "Person.D", "Person.E"),
                  Ba = c(0,NA,2,1,1),
                  Bb = c(NA,1,2,2,1),
                  Bc = c(0,1,2,2,1))

I then merge these 2 datasets using merge(df1, df2, all.x = T, by = "ID" to get:

         ID Aa Ab Ac no.match Ba Bb Bc
1 Person.A  0  0 NA        0  0 NA  0
2 Person.B  1 NA NA        1 NA  1  1
3 Person.C  2  2  2        2  2  2  2
4 Person.D NA  1  2        2  1  2  2
5 Person.E  1  1  1        1  1  1  1
6 Person.F  1  1  1        2 NA NA NA

The actual datasets are much more complicated with lots of columns that have no matches in other columns. So I don't think I could do something that depends on the arrangement of the columns.

Columns Aa and Ba contain the same information; and columns Ab and Bb do as well, and so on, but column no.match does not contain a matching column.

I want to "map" the values from from the same row of Ba to Aa if Aa is NA and do the same for Ab and Bb , Ac and Bc , etc.

The result DF in this case would look like:

        ID Aa Ab Ac no.match Ba Bb Bc
1 Person.A  0  0  0      0    0 NA  0
2 Person.B  1  1  1      1    NA  1  1
3 Person.C  2  2  2      2    2  2  2
4 Person.D  1  1  2      2    1  2 NA
5 Person.E  1  1  1      1    1  1  1
6 Person.F  1  1  1      2    NA NA NA

Where element [4,2] was replaced by element [4,6] The rows and the columns need to match up.

I've tried an embarrassingly large number of things: apply , ifelse , iterating through a list of columns l1 = c('Aa','Ab','Ac'), l2 = c('Ba', 'Bb', 'Bc')

I can do the one-off: which(is.na(mdf$Aa)) <- mdf[which(is.na(mdf$Aa)), c("Ba")]

But how can I do this iteratively?

Thank you! (sorry for the long-windedness)

Answer 1

Here's one using data.table v1.9.5 - installation instructions here :

require(data.table) # v1.9.5+
cols1 = names(df1)[2:4]
cols2 = names(df2)[2:4]

foo <- function(x, y) {
    nas = is.na(x)
    x[nas] = y[nas]
    x
}
setDT(df1)[df2, c(cols1, cols2) := c(Map(foo, mget(cols1), 
                   mget(cols2)), mget(cols2)), on = "ID"]

> df1
#          ID Aa Ab Ac no.match Ba Bb Bc
# 1: Person.A  0  0  0        0  0 NA  0
# 2: Person.B  1  1  1        1 NA  1  1
# 3: Person.C  2  2  2        2  2  2  2
# 4: Person.D  1  1  2        2  1  2  2
# 5: Person.E  1  1  1        1  1  1  1
# 6: Person.F  1  1  1        2 NA NA NA

• setDT() converts df1 to a data.table by reference.

• setDT(df1)[df2, on = "ID"] performs a join. For each row of df2 , we find the matching rows in df1 and extract the columns corresponding to matching rows..

• On the matching rows, we update columns in cols1 and add new columns in cols2 by reference using the := operator. For updating columns, we extract the columns specified in cols1 and cols2 and replace NA s with the function foo() . For adding columns, we simply pull the columns cols2 , using mget() . We concatenate the two lists using c() .

If you're interested, have a look at the HTML vignettes to learn more.