Clever Geek Handbook

Replace values ​​in data.frame column based on values ​​in another 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 values week.born

must be 23

. For two

all values week.born

must be 19

. For one

all values week.born

must be 24

.

What's the best way to do this?

+3


source to share


5 answers


I would create another data.frame containing the display and then do a simple connection:

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 comparison test 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
+4


source


One solution:

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


Hello

+2


source


You can do:

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

Or basic R

with ave

:

df$week.born = with(df, ave(week.born, id, FUN=function(u) u[!is.na(u)][1]))
+2


source


@ cho7tom is ok if you only have a few groups, otherwise you might prefer to have a lookup table and create a join to that lookup table week.born

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 a binary join from data.table 

library(data.table)
setkey(setDT(df))[lkp]
+2


source


When matching multiple combinations like this, the function mapvalues

from the package plyr

is simple:

library(plyr)
df$week.born <- mapvalues(df$id, c("one", "two", "three"), c(23, 19, 24))
0


source






More articles:


All Articles