本文介绍了使用 tidyr 传播两列数据框的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个如下所示的数据框:
I have a data frame that looks like this:
a b
1 x 8
2 x 6
3 y 3
4 y 4
5 z 5
6 z 6
我想把它变成这样:
x y z
1 8 3 5
2 6 4 6
但是打电话
library(tidyr)
df <- data.frame(
a = c("x", "x", "y", "y", "z", "z"),
b = c(8, 6, 3, 4, 5, 6)
)
df %>% spread(a, b)
返回
x y z
1 8 NA NA
2 6 NA NA
3 NA 3 NA
4 NA 4 NA
5 NA NA 5
6 NA NA 6
我做错了什么?
推荐答案
虽然我知道你在追求 tidyr
,但 base
在这种情况下有一个解决方案:
While I'm aware you're after tidyr
, base
has a solution in this case:
unstack(df, b~a)
它也快了一点:
Unit: microseconds
expr min lq mean median uq max neval
df %>% spread(a, b) 657.699 679.508 717.7725 690.484 724.9795 1648.381 100
unstack(df, b ~ a) 309.891 335.264 349.4812 341.9635 351.6565 639.738 100
应大众需求,推出更大的产品
我没有包含 data.table
解决方案,因为我不确定按引用传递是否会成为 microbenchmark
的问题.
By popular demand, with something bigger
I haven't included the data.table
solution as I'm not sure if pass by reference would be a problem for microbenchmark
.
library(microbenchmark)
library(tidyr)
library(magrittr)
nlevels <- 3
#Ensure that all levels have the same number of elements
nrow <- 1e6 - 1e6 %% nlevels
df <- data.frame(a=sample(rep(c("x", "y", "z"), length.out=nrow)),
b=sample.int(9, nrow, replace=TRUE))
microbenchmark(df %>% spread(a, b), unstack(df, b ~ a), data.frame(split(df$b,df$a)), do.call(cbind,split(df$b,df$a)))
即使是 100 万,unstack 也更快.值得注意的是,split
解决方案也非常快.
Even on 1 million, unstack is faster. Notably, the split
solution is also very fast.
Unit: milliseconds
expr min lq mean median uq max neval
df %>% spread(a, b) 366.24426 414.46913 450.78504 453.75258 486.1113 542.03722 100
unstack(df, b ~ a) 47.07663 51.17663 61.24411 53.05315 56.1114 102.71562 100
data.frame(split(df$b, df$a)) 19.44173 19.74379 22.28060 20.18726 22.1372 67.53844 100
do.call(cbind, split(df$b, df$a)) 26.99798 27.41594 31.27944 27.93225 31.2565 79.93624 100
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