如何将多个变量的重复测量值传播为宽格式？

Question

我正在尝试获取长格式的列并将它们扩展为宽格式，如下所示。我想使用 tidyr 通过我正在投资的数据处理工具来解决这个问题，但为了使这个答案更通用，请提供其他解决方案。

这是我所拥有的：

library(dplyr); library(tidyr)

set.seed(10)
dat <- data_frame(
    Person = rep(c("greg", "sally", "sue"), each=2),
    Time = rep(c("Pre", "Post"), 3),
    Score1 = round(rnorm(6, mean = 80, sd=4), 0),
    Score2 = round(jitter(Score1, 15), 0),
    Score3 = 5 + (Score1 + Score2)/2
)

##   Person Time Score1 Score2 Score3
## 1   greg  Pre     80     78   84.0
## 2   greg Post     79     80   84.5
## 3  sally  Pre     75     74   79.5
## 4  sally Post     78     78   83.0
## 5    sue  Pre     81     78   84.5
## 6    sue Post     82     81   86.5

所需的宽格式：

  Person Pre.Score1 Pre.Score2 Pre.Score3  Post.Score1 Post.Score2 Post.Score3
1   greg         80         78       84.0           79          80        84.5
2  sally         75         74       79.5           78          78        83.0
3    sue         81         78       84.5           82          81        86.5

我可以通过为每个分数做这样的事情来做到这一点：

spread(dat %>% select(Person, Time, Score1), Time, Score1) %>% 
    rename(Score1_Pre = Pre, Score1_Post = Post)

然后使用_join，但这似乎很冗长，并且必须有更好的方法。

相关问题：
tidyr wide to long with two repeated measures
Is it possible to use spread on multiple columns in tidyr similar to dcast?

Answer 1

编辑：我正在更新这个答案，因为 pivot_wider 已经存在了一段时间，并解决了这个问题和 cmets 中的问题。你现在可以做

pivot_wider(
    dat, 
    id_cols = 'Person', 
    names_from = 'Time', 
    values_from = c('Score1', 'Score2', 'Score3'), 
    names_glue = '{Time}.{.value}'
)

得到想要的结果。

---

原来的答案是

dat %>% 
  gather(temp, score, starts_with("Score")) %>% 
  unite(temp1, Time, temp, sep = ".") %>% 
  spread(temp1, score)

Answer 2

使用reshape2：

library(reshape2)
dcast(melt(dat), Person ~ Time + variable)

生产：

Using Person, Time as id variables
  Person Post_Score1 Post_Score2 Post_Score3 Pre_Score1 Pre_Score2 Pre_Score3
1   greg          79          78        83.5         83         81       87.0
2  sally          82          81        86.5         75         74       79.5
3    sue          78          78        83.0         82         79       85.5

Answer 3

使用data.table 包中的dcast。

library(data.table)#v1.9.5+
dcast(setDT(dat), Person~Time, value.var=paste0("Score", 1:3))
#     Person Score1_Post Score1_Pre Score2_Post Score2_Pre Score3_Post Score3_Pre
#1:   greg          79         80          80         78        84.5       84.0
#2:  sally          78         75          78         74        83.0       79.5
#3:    sue          82         81          81         78        86.5       84.5

或reshape 来自baseR

reshape(as.data.frame(dat), idvar='Person', timevar='Time',direction='wide')

---

更新

从开发版本tidyr_0.8.3.9000 或CRAN 版本tidyr_1.0.0，我们可以将pivot_wider 用于多个值列

library(tidyr)
library(stringr)
dat %>%
     pivot_wider(names_from = Time, values_from = str_c("Score", 1:3))
# A tibble: 3 x 7
#  Person Score1_Pre Score1_Post Score2_Pre Score2_Post Score3_Pre Score3_Post
#   <chr>       <dbl>       <dbl>      <dbl>       <dbl>      <dbl>       <dbl>
#1 greg           80          79         78          80       84          84.5
#2 sally          75          78         74          78       79.5        83  
#3 sue            81          82         78          81       84.5        86.5

Answer 4

我为自己做了一个基准测试并在这里发布以防有人感兴趣：

代码

设置是从 OP、三个变量、两个时间点中选择的。但是，数据框的大小从 1,000 到 100,000 行不等。

library(magrittr)
library(data.table)
library(bench)

f1 <- function(dat) {
    tidyr::gather(dat, key = "key", value = "value", -Person, -Time) %>% 
        tidyr::unite("id", Time, key, sep = ".") %>%
        tidyr::spread(id, value)
}

f2 <- function(dat) {
    reshape2::dcast(melt(dat, id.vars = c("Person", "Time")), Person ~ Time + variable)
}

f3 <- function(dat) {
    dcast(melt(dat, id.vars = c("Person", "Time")), Person ~ Time + variable)
}

create_df <- function(rows) {
    dat <- expand.grid(Person = factor(1:ceiling(rows/2)),
                       Time = c("1Pre", "2Post"))
    dat$Score1 <- round(rnorm(nrow(dat), mean = 80, sd = 4), 0)
    dat$Score2 <- round(jitter(dat$Score1, 15), 0)
    dat$Score3 <- 5 + (dat$Score1 + dat$Score2)/2
    return(dat)
}

结果

如您所见，reshape2 比 tidyr 快一点，可能是因为 tidyr 的开销更大。重要的是，data.table 的行数超过 10,000 行。

press(
    rows = 10^(3:5),
    {
        dat <- create_df(rows)
        dat2 <- copy(dat)
        setDT(dat2)
        bench::mark(tidyr     = f1(dat),
                    reshape2  = f2(dat),
                    datatable = f3(dat2),
                    check = function(x, y) all.equal(x, y, check.attributes = FALSE),
                    min_iterations = 20
        )
    }
)
#> Warning: Some expressions had a GC in every iteration; so filtering is
#> disabled.
#> # A tibble: 9 x 11
#>   expression   rows      min     mean   median      max `itr/sec` mem_alloc
#>   <chr>       <dbl> <bch:tm> <bch:tm> <bch:tm> <bch:tm>     <dbl> <bch:byt>
#> 1 tidyr        1000    5.7ms   6.13ms   6.02ms  10.06ms    163.      2.78MB
#> 2 reshape2     1000   2.82ms   3.09ms   2.97ms   8.67ms    323.       1.7MB
#> 3 datatable    1000   3.82ms      4ms   3.92ms   8.06ms    250.      2.78MB
#> 4 tidyr       10000  19.31ms  20.34ms  19.95ms  22.98ms     49.2     8.24MB
#> 5 reshape2    10000  13.81ms   14.4ms   14.4ms   15.6ms     69.4    11.34MB
#> 6 datatable   10000  14.56ms  15.16ms  14.91ms  18.93ms     66.0     2.98MB
#> 7 tidyr      100000 197.24ms 219.69ms 205.27ms 268.92ms      4.55   90.55MB
#> 8 reshape2   100000 164.02ms 195.32ms 176.31ms 284.77ms      5.12  121.69MB
#> 9 datatable  100000  51.31ms  60.34ms  58.36ms 113.69ms     16.6    27.36MB
#> # ... with 3 more variables: n_gc <dbl>, n_itr <int>, total_time <bch:tm>

^{由reprex package (v0.2.1) 于 2019 年 2 月 27 日创建}