CODEWITHSUNDEEP
rdataframe
luciano
luciano

Reputation: 13792

Replace values in a data.frame column based on values in a different column

I have this data.frame:

df <- data.frame(id = rep(c("one", "two", "three"), each = 10), week.born = NA)
df$week.born[c(5,15,28)] <- c(23,19,24)

df 

  id week.born
1    one        NA
2    one        NA
3    one        NA
4    one        NA
5    one        23
6    one        NA
7    one        NA
8    one        NA
9    one        NA
10   one        NA
11   two        NA
12   two        NA
13   two        NA
14   two        NA
15   two        19
16   two        NA
17   two        NA
18   two        NA
19   two        NA
20   two        NA
21 three        NA
22 three        NA
23 three        NA
24 three        NA
25 three        NA
26 three        NA
27 three        NA
28 three        24
29 three        NA
30 three        NA

For one all week.born values should be 23. For two all week.born values should be 19. For one all week.born values should be 24.

Whats the best way to do this?

Upvotes: 3

Views: 2742

Answers (5)

Sam Firke
Sam Firke

Reputation: 23004

When mapping just a few combinations like this, the mapvalues function from the plyr package is simple:

library(plyr)
df$week.born <- mapvalues(df$id, c("one", "two", "three"), c(23, 19, 24))

Upvotes: 0

faidherbard
faidherbard

Reputation: 1035

I would create another data.frame containing the mapping and then do a simple join:

require(dplyr)
map <- data.frame(id=c("one","two","three"), new.week.born=c(23,19,24))
left_join(df, map, by="id")

# id week.born new.week.born
# 1    one        NA            23
# 2    one        NA            23
# ...
# 16   two        NA            19
# 17   two        NA            19
# 18   two        NA            19
# 19   two        NA            19
# 20   two        NA            19
# 21 three        NA            24
# 22 three        NA            24
# 23 three        NA            24
# ...

See benchmark below.

library(microbenchmark)
library(dplyr) # v 0.4.1
library(data.table) # v 1.9.5

df <- data.frame(id = rep(c("one", "two", "three"), each = 1e6))
df2 <- copy(df)
map <- data.frame(id=c("one","two","three"), new.week.born=c(23,19,24))

dplyr_join <- function() { 
  left_join(df, map, by="id")
}

r_merge <- function() {
  merge(df, map, by="id")
}

data.table_join <- function() {
  setkey(setDT(df2))[map]
}

Unit: milliseconds
              expr         min         lq       mean     median         uq       max neval
      dplyr_join()   409.10635   476.6690   910.6446   489.4573   705.4021  2866.151    10
         r_merge() 41589.32357 47376.0741 55719.1752 50133.0918 54636.3356 83562.931    10
 data.table_join()    94.14621   132.3788   483.4220   225.3309  1051.7916  1416.946    10

Upvotes: 6

jangorecki
jangorecki

Reputation: 16697

@cho7tom is OK if you have just few groups, otherwise you may prefer to have a lookup table and make a join to that table to lookup week.born value based on id.

base R

df <- data.frame(id = rep(c("one", "two", "three"), each = 10))
lkp <- data.frame(id=c("one","two","three"), week.born=c(23,19,24))
merge(df, lkp, by="id")

Or using binary join from data.table

library(data.table)
setkey(setDT(df))[lkp]

Upvotes: 2

Colonel Beauvel
Colonel Beauvel

Reputation: 31161

You can do:

library(data.table)
setDT(df)[,week.born:=week.born[!is.na(week.born)][1], by=id]

Or base R using ave:

df$week.born = with(df, ave(week.born, id, FUN=function(u) u[!is.na(u)][1]))

Upvotes: 3

cho7tom
cho7tom

Reputation: 1070

One solution is:

df$week.born[df$id == "one"] <- 23
df$week.born[df$id == "two"] <- 19
df$week.born[df$id == "three"] <- 24

Regards

Upvotes: 2

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