用tidyr传播两列数据帧

ljo*_*jos 12 r dplyr tidyr

我有一个如下所示的数据框:

  a b
1 x 8
2 x 6
3 y 3
4 y 4
5 z 5
6 z 6
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我想把它变成这个:

  x y z
1 8 3 5
2 6 4 6
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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)
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回报

   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
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我究竟做错了什么?

seb*_*n-c 13

虽然我知道你在追求tidyr,base但在这种情况下有一个解决方案:

unstack(df, b~a)
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它也快一点:

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
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受欢迎的需求,更大的东西

我没有包括data.table解决方案,因为我不确定通过引用是否会成为问题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)))
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即使在100万,倒塌也更快.值得注意的是,split解决方案也非常快.

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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  • `stack/unstack`通常比较慢.此基准是否基于更大的数据集? (2认同)

Dat*_*neR 8

不知何故这样吗?

df <- data.frame(ind = rep(1:min(table(df$a)), length(unique(df$a))), df)
df %>% spread(a, b) %>% select(-ind)
  ind x y z
1   1 8 3 5
2   2 6 4 6
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Jaa*_*aap 6

你可以做到这一点dcast,并rowid包,以及:

dat <- dcast(setDT(df), rowid(a) ~ a, value.var = "b")[,a:=NULL]
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这使:

> dat
   x y z
1: 8 3 5
2: 6 4 6
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旧解决方案:

# create a sequence number by group
setDT(df)[, r:=1:.N, by = a]
# reshape to wide format and remove the sequence variable
dat <- dcast(df, r ~ a, value.var = "b")[,r:=NULL]
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这使:

> dat
   x y z
1: 8 3 5
2: 6 4 6
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nic*_*ola 5

另一个base答案(看起来也很快):

data.frame(split(df$b,df$a))
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