我有这个 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
对于
one
,所有 week.born
值都应该是 23
。对于
two
,所有 week.born
值都应该是 19
。对于
one
,所有 week.born
值都应该是 24
。什么是最好的方法来做到这一点?
最佳答案
我会创建另一个包含映射的 data.frame,然后做一个简单的连接:
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
# ...
请参阅下面的基准。
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
关于根据不同列中的值替换 data.frame 列中的值,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/29807032/