将r替换为r中每个组的相同列的另一行中的值

Question

I need to replace the NA's of each row with non NA's values of different row for a given column for each group

let say sample data like:

id   name
 1     a
 1     NA
 2     b
 3     NA
 3     c
 3     NA

desired output:

id   name
 1     a
 1     a
 2     b
 3     c
 3     c
 3     c

Is there a way to perform this in r ?

Answer 1

We can use data.table to do this. Convert the 'data.frame' to 'data.table' ( setDT(df1) ). Grouped by 'id', we replace the 'name' with the non-NA value in 'name'.

library(data.table)#v1.9.5+
setDT(df1)[, name:= name[!is.na(name)][1L] , by = id]
df1
#   id name
#1:  1    a
#2:  1    a
#3:  2    b
#4:  3    c
#5:  3    c
#6:  3    c

NOTE: Here I assumed that there is only a single unique non-NA value within each 'id' group.

Or another option would be to join the dataset with the unique rows of the data after we order by 'id' and 'name'.

 setDT(df1)
 df1[unique(df1[order(id, name)], by='id'), on='id', name:= i.name][]
 #   id name
 #1:  1    a
 #2:  1    a
 #3:  2    b
 #4:  3    c
 #5:  3    c
 #6:  3    c

NOTE: The on is only available with the devel version of data.table . Instructions to install the devel version are here

data

df1 <- structure(list(id = c(1L, 1L, 2L, 3L, 3L, 3L), name = c("a", 
NA, "b", NA, "c", NA)), .Names = c("id", "name"),
class = "data.frame",    row.names = c(NA, -6L))

Answer 2

Here is an approach using dplyr . From the data frame x we group by id and replace NA with the relevant values. I am assuming one unique value of name per id .

x <- data.frame(id = c(1, 1, 2, rep(3,3)), 
 name = c("a", NA, "b", NA, "c", NA), stringsAsFactors=F)

require(dplyr)
x %>%
  group_by(id) %>%
  mutate(name = unique(name[!is.na(name)]))

Source: local data frame [6 x 2]
Groups: id

#  id name
#1  1    a
#2  1    a
#3  2    b
#4  3    c
#5  3    c
#6  3    c

Answer 3

Base R

d<-na.omit(df)
transform(df,name=d$name[match(id,d$id)])

again assuming one unique value of name per id (forces it)