loop to create a new variable based on other cases

问题:

I have a dataframe with three variables: ID, group, and nominated_ID. I want to know the group that nominated_ID belongs in.

I'm imagining that for each case, we take nominated_ID, find the case where it is equal to ID, and then set the nominated_Group variable in the original case equal to the group variable in the matched case. (If there is no match, set it to NA)

I wouldn't be surprised if this can be done without a loop, so I'm open-minded about the solution. Thanks so much for your help. Know that I did try to look for similar questions before posting.

回答1:

You can achieve this in one step without the use of cbind by directly allocating results to a column in your data.frame:

df$nominated_group <- with(df, group[match(nominated_ID, ID)])
df
  ID group nominated_ID nominated_group
1  9   Odd            9             Odd
2  5   Odd            8            <NA>
3  2  Even            4            Even
4  4  Even            9             Odd
5  3   Odd            2            Even

I used with as a convenient way of referring to the columns of df without having to repeatedly write df$.

回答2:

The following seems to work; there may be better ways

> df <- data.frame(ID = c(9, 5, 2, 4, 3), 
+                  group = c("Odd", "Odd", "Even", "Even", "Odd"),
+                  nominated_ID = c(9, 8, 4, 9, 2)                 )
> df
  ID group nominated_ID
1  9   Odd            9
2  5   Odd            8
3  2  Even            4
4  4  Even            9
5  3   Odd            2
> nominated_Group <- df[match(df$nominated_ID, df$ID), ]$group
> newDF <- cbind(df, nominated_Group)
> newDF
  ID group nominated_ID nominated_Group
1  9   Odd            9             Odd
2  5   Odd            8            <NA>
3  2  Even            4            Even
4  4  Even            9             Odd
5  3   Odd            2            Even

回答3:

You can do this in a syntactically compact way using transform, match and array indexing. Using @Henry's data-frame:

df <- transform( df, nominated_group = group[match(nominated_ID, ID)])

> df
  ID group nominated_ID nominated_group
1  9   Odd            9             Odd
2  5   Odd            8            <NA>
3  2  Even            4            Even
4  4  Even            9             Odd
5  3   Odd            2            Even

回答4:

Probably not the most "intuitive' way, but merging df against df also works if you use nominated_ID as the merge index for the first copy and ID as the by index for the second and keep all rows. You need to drop the second nominated_ID column and rearrange the order to get things to match the answers above:

merge(df,df, by.x=3, by.y=1, all.x=TRUE)[order(df$nominated_ID), c(2,3, 1, 4)]

  ID group.x nominated_ID group.y
5  4    Even            9     Odd
3  5     Odd            8    <NA>
2  2    Even            4    Even
1  3     Odd            2    Even
4  9     Odd            9     Odd