Question

我认为这个问题已被问过多次，但我找不到合适的词来找到可行的解决方案。

如何spread()基于多个键的多个值的数据框？

我正在使用的数据经过简化（我有更多的列需要扩展，但是只有两个键：给定度量的Id和time点），如下所示：

df <- data.frame(id = rep(seq(1:10),3), 
                 time = rep(1:3, each=10), 
                 x = rnorm(n=30), 
                 y = rnorm(n=30))

> head(df)
  id time           x           y
1  1    1 -2.62671241  0.01669755
2  2    1 -1.69862885  0.24992634
3  3    1  1.01820778 -1.04754037
4  4    1  0.97561596  0.35216040
5  5    1  0.60367158 -0.78066767
6  6    1 -0.03761868  1.08173157
> tail(df)
   id time           x          y
25  5    3  0.03621258 -1.1134368
26  6    3 -0.25900538  1.6009824
27  7    3  0.13996626  0.1359013
28  8    3 -0.60364935  1.5750232
29  9    3  0.89618748  0.0294315
30 10    3  0.14709567  0.5461084

我想要的是一个这样填充的数据框：

Id和每个测量变量中的每个值每time列一行。

Answer 1

您的输入数据框不整齐。您应该使用collect来做到这一点。

gather(df, key, value, -id, -time) %>%
  mutate(key = paste0(key, "time", time)) %>%
  select(-time) %>%
  spread(key, value)

Answer 2

最好使用dcast中的data.table或reshape中的base R来重塑多个值变量。

library(data.table)
out <- dcast(setDT(df), id ~ paste0("time", time), value.var = c("x", "y"), sep = "")
out
#    id     xtime1     xtime2      xtime3      ytime1      ytime2      ytime3
# 1:  1  0.4334921 -0.5205570 -1.44364515  0.49288757 -1.26955148 -0.83344256
# 2:  2  0.4785870  0.9261711  0.68173681  1.24639813  0.91805332  0.34346260
# 3:  3 -1.2067665  1.7309593  0.04923993  1.28184341 -0.69435556  0.01609261
# 4:  4  0.5240518  0.7481787  0.07966677 -1.36408357  1.72636849 -0.45827205
# 5:  5  0.3733316 -0.3689391 -0.11879819 -0.03276689  0.91824437  2.18084692
# 6:  6  0.2363018 -0.2358572  0.73389984 -1.10946940 -1.05379502 -0.82691626
# 7:  7 -1.4979165  0.9026397  0.84666801  1.02138768 -0.01072588  0.08925716
# 8:  8  0.3428946 -0.2235349 -1.21684977  0.40549497  0.68937085 -0.15793111
# 9:  9 -1.1304688 -0.3901419 -0.10722222 -0.54206830  0.34134397  0.48504564
#10: 10 -0.5275251 -1.1328937 -0.68059800  1.38790593  0.93199593 -1.77498807

使用reshape我们可以做到

# setDF(df) # in case df is a data.table now
reshape(df, idvar = "id", timevar = "time", direction = "wide")

Answer 3

使用tidyr（tidyr_0.8.3.9000）的改进版本，我们可以使用pivot_wider将多值列从长格式更改为宽格式

library(dplyr)
library(tidyr)
library(stringr)
df %>%
   mutate(time = str_c("time", time)) %>%
   pivot_wider(names_from = time, values_from = c("x", "y"), names_sep="")
# A tibble: 10 x 7
#      id  xtime1 xtime2  xtime3  ytime1 ytime2 ytime3
#   <int>   <dbl>  <dbl>   <dbl>   <dbl>  <dbl>  <dbl>
# 1     1 -0.256   0.483 -0.254  -0.652   0.655  0.291
# 2     2  1.10   -0.596 -1.85    1.09   -0.401 -1.24 
# 3     3  0.756  -2.19  -0.0779 -0.763  -0.335 -0.456
# 4     4 -0.238  -0.675  0.969  -0.829   1.37  -0.830
# 5     5  0.987  -2.12   0.185   0.834   2.14   0.340
# 6     6  0.741  -1.27  -1.38   -0.968   0.506  1.07 
# 7     7  0.0893 -0.374 -1.44   -0.0288  0.786  1.22 
# 8     8 -0.955  -0.688  0.362   0.233  -0.902  0.736
# 9     9 -0.195  -0.872 -1.76   -0.301   0.533 -0.481
#10    10  0.926  -0.102 -0.325  -0.678  -0.646  0.563

注意：数字不同，因为在创建样本数据集时没有设置种子

tidyr :: spread（）具有多个键和值

3 个答案: