如何清理R中的数据？

Question

目前，我的数据格式如下：

Var   year   Co-1      Co-2     Co-3  ...
A     2018     a         j        .  
A     2017     b         k        .
A     2016     c         l        .
B     2018     d         m        .
B     2017     e         n        .
B     2016     f         o        .
C     2018     g         p        .
C     2017     h         q        .
C     2016     i         r        .
.       .      .         .        .
.       .      .         .
.       .      .         .

我想将其转换为以下格式：

Company   year    A       B       C
Co-1      2018    a       d       g
Co-1      2017    b       e       h
Co-1      2016    c       f       i
Co-2      2018    j       m       p 
Co-2      2017    k       n       q
Co-2      2016    l       o       r
Co-3      2018    .       .
Co-3      2017    .       .
Co-3      2016    .       .
.
.
.

本质上的变化是：

1. 在第一列中多次插入公司名称，每年一个（2018,17,16）
2. 使每列标题中的变量为 A、B 和 C，而不是在第一列中有多个 AAA、BBB、CCC

通过这样做，我希望能够分别回归年份与 A、B 和 C 中的每一个，同时保持每个数据点的公司区别，因此我可以在完成的图表中按公司对数据点进行分组。

非常感谢！

Answer 1

以长格式获取数据，然后以宽但具有不同列的格式获取数据。

library(tidyr)

df %>%
  pivot_longer(cols = starts_with("Co")) %>%
  pivot_wider(names_from = Var, values_from = value)

# A tibble: 6 x 5
#   year name  A     B     C    
#  <int> <chr> <fct> <fct> <fct>
#1  2018 Co-1  a     d     g    
#2  2018 Co-2  j     m     p    
#3  2017 Co-1  b     e     h    
#4  2017 Co-2  k     n     q    
#5  2016 Co-1  c     f     i    
#6  2016 Co-2  l     o     r    

数据

df <- structure(list(Var = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 
3L, 3L), .Label = c("A", "B", "C"), class = "factor"), year = c(2018L, 
2017L, 2016L, 2018L, 2017L, 2016L, 2018L, 2017L, 2016L), `Co-1` = 
structure(1:9, .Label = c("a", "b", "c", "d", "e", "f", "g", "h", "i"), 
class = "factor"), `Co-2` = structure(1:9, .Label = c("j", 
"k", "l", "m", "n", "o", "p", "q", "r"), class = "factor")), class = "data.frame", 
row.names = c(NA, -9L))

Answer 2

我认为在这里简单地使用因子会容易得多。 标签的长度应该是唯一公司的数量，即"Var"值

df$Var <- factor(df$Var, labels=c("Pepsi", "Coke", "Sprite"))
names(df) <- c("company", "year", LETTERS[seq(names(df)[-(1:2)])])

或者在一个步骤中：

df <- setNames(transform(df, Var=factor(Var, labels=c("Pepsi", "Coke", "Sprite"))),
               c("company", "year", LETTERS[seq(names(df)[-(1:2)])]))
df
#   company year A B C
# 1   Pepsi 2018 a j .
# 2   Pepsi 2017 b k .
# 3   Pepsi 2016 c l .
# 4    Coke 2018 d m .
# 5    Coke 2017 e n .
# 6    Coke 2016 f o .
# 7  Sprite 2018 g p .
# 8  Sprite 2017 h q .
# 9  Sprite 2016 i r .

还产生更清洁的类：

sapply(df, class)
# company        year           A           B           C 
# "factor"   "integer" "character" "character" "character"