dplyr 和 tidyr：将长格式转换为宽格式并排列列

我正在创建一个shiny app用户将在其中上传包含多个变量的 .csv 文件。使用dplyr， 我会select前四个变量，如下所示，并将它们从长格式转换。

DATA

df <- read.table(text = c("
Customer    Rate    Factor  Power
W1  6   TK1 5
W2  3   TK1 0
W3  1   TK1 0
W4  2   TK1 0
W5  4   TK1 0
W6  8   TK1 0
W7  5   TK1 0
W8  7   TK1 3
W1  6   TK2 0
W2  3   TK2 1
W3  1   TK2 0
W4  2   TK2 5
W5  4   TK2 0
W6  8   TK2 0
W7  5   TK2 0
W8  7   TK2 3
W1  6   TK3 0
W2  3   TK3 5
W3  1   TK3 1
W4  2   TK3 0
W5  4   TK3 0
W6  8   TK3 0
W7  5   TK3 0
W8  7   TK3 0
W1  6   TK4 0
W2  3   TK4 3
W3  1   TK4 0
W4  2   TK4 0
W5  4   TK4 0
W6  8   TK4 0
W7  5   TK4 0
W8  7   TK4 0
W1  6   TK5 1
W2  3   TK5 0
W3  1   TK5 5
W4  2   TK5 0
W5  4   TK5 1
W6  8   TK5 0
W7  5   TK5 0
W8  7   TK5 0
W1  6   TK6 0
W2  3   TK6 0
W3  1   TK6 0
W4  2   TK6 0
W5  4   TK6 0
W6  8   TK6 0
W7  5   TK6 5
W8  7   TK6 0
W1  6   TK7 0
W2  3   TK7 0
W3  1   TK7 0
W4  2   TK7 0
W5  4   TK7 0
W6  8   TK7 3
W7  5   TK7 3
W8  7   TK7 0
W1  6   TK8 0
W2  3   TK8 0
W3  1   TK8 1
W4  2   TK8 0
W5  4   TK8 0
W6  8   TK8 3
W7  5   TK8 0
W8  7   TK8 0
W1  6   TK9 0
W2  3   TK9 0
W3  1   TK9 0
W4  2   TK9 0
W5  4   TK9 5
W6  8   TK9 0
W7  5   TK9 0
W8  7   TK9 0
W1  6   TK10    0
W2  3   TK10    0
W3  1   TK10    0
W4  2   TK10    0
W5  4   TK10    0
W6  8   TK10    5
W7  5   TK10    0
W8  7   TK10    0
W1  6   TK11    0
W2  3   TK11    0
W3  1   TK11    0
W4  2   TK11    0
W5  4   TK11    0
W6  8   TK11    0
W7  5   TK11    0
W8  7   TK11    3
W1  6   TK12    0
W2  3   TK12    0
W3  1   TK12    0
W4  2   TK12    0
W5  4   TK12    0
W6  8   TK12    0
W7  5   TK12    0
W8  7   TK12    5"), header = T) 

我使用下面的代码将长格式转换为宽格式

从长到宽

library(dplyr)
library(tidyr)
df_wide <-  df %>%
  tidyr::spread(Factor, Power)

RESULT

> df_wide
  Customer Rate TK1 TK10 TK11 TK12 TK2 TK3 TK4 TK5 TK6 TK7 TK8 TK9
1       W1    6   5    0    0    0   0   0   0   1   0   0   0   0
2       W2    3   0    0    0    0   1   5   3   0   0   0   0   0
3       W3    1   0    0    0    0   0   1   0   5   0   0   1   0
4       W4    2   0    0    0    0   5   0   0   0   0   0   0   0
5       W5    4   0    0    0    0   0   0   0   1   0   0   0   5
6       W6    8   0    5    0    0   0   0   0   0   0   3   3   0
7       W7    5   0    0    0    0   0   0   0   0   5   3   0   0
8       W8    7   3    0    3    5   3   0   0   0   0   0   0   0

宽幅格式显示了Factor变量为TK1进而TK10

> levels(df$Factor)
 [1] "TK1"  "TK10" "TK11" "TK12" "TK2"  "TK3"  "TK4"  "TK5"  "TK6"  "TK7"  "TK8"  "TK9"

我希望 Factor 的级别从 TK1、TK2 到 TK12

我可以解决如下

df$Factor <- factor(df$Factor, levels = c("TK1", "TK2" , "TK3" , "TK4",  "TK5" , "TK6" , "TK7" , "TK8" , "TK9", "TK10", "TK11", "TK12"))

然而，因子变量的水平将是用户输入的函数。可能是 14、15 或 20。

QUESTION

有没有办法将 Factor 列的级别从最低到最高排列，而不管用户的输入如何？

我们可以将其更改为factor with levels指定的

df %>%
  mutate(Factor = factor(Factor, levels = paste0("TK", 1:12))) %>%
  spread(Factor, Power)

或者让它更有活力，我们extract将非数字和数字部分分成单独的列（“Factor1”、“Factor2”），将“Factor”更改为factor with levels由指定paste的序列min to max将“Factor2”中的值与“Factor1”中第一个字符值的值进行比较，删除“Factor1”和“Factor2”，并且spread.

library(tidyr)
res <- df %>%
         extract(Factor, into = c("Factor1", "Factor2"), "(\\D+)(\\d+)",
                                   remove = FALSE, convert=TRUE) %>% 
         mutate(Factor = factor(Factor, levels = paste0(Factor1[1], 
                              min(Factor2):max(Factor2)))) %>% 
         select(-Factor1, -Factor2) %>% 
         spread(Factor, Power)
head(res, 2)
#  Customer Rate TK1 TK2 TK3 TK4 TK5 TK6 TK7 TK8 TK9 TK10 TK11 TK12
#1       W1    6   5   0   0   0   1   0   0   0   0    0    0    0
#2       W2    3   0   1   5   3   0   0   0   0   0    0    0    0