我有一个这样的数据集 :
# install.packages(dplyr)
library(dplyr)
df <- tibble(period = c(201501,201502,201503,201504,201505,201506,201507,201508,201509,201510,201511,201512,201513),
sales = sample(1:100,13),
P1 = c(1,0,0,0,0,0,0,0,0,0,0,0,0),
P10 = c(0,0,0,0,0,0,0,0,0,1,0,0,0),
P11 = c(0,0,0,0,0,0,0,0,0,0,1,0,0),
P12 = c(0,0,0,0,0,0,0,0,0,0,0,1,0),
P13 = c(0,0,0,0,0,0,0,0,0,0,0,0,1),
P2 = c(0,1,0,0,0,0,0,0,0,0,0,0,0),
P3 = c(0,0,1,0,0,0,0,0,0,0,0,0,0),
P4 = c(0,0,0,1,0,0,0,0,0,0,0,0,0),
P5 = c(0,0,0,0,1,0,0,0,0,0,0,0,0),
P6 = c(0,0,0,0,0,1,0,0,0,0,0,0,0),
P7 = c(0,0,0,0,0,0,1,0,0,0,0,0,0),
P8 = c(0,0,0,0,0,0,0,1,0,0,0,0,0),
P9 = c(0,0,0,0,0,0,0,0,1,0,0,0,0),
)
print(df)
所以我有这个:
# A tibble: 13 x 15
period sales P1 P10 P11 P12 P13 P2 P3 P4 P5 P6
<dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 201501 92 1 0 0 0 0 0 0 0 0 0
2 201502 60 0 0 0 0 0 1 0 0 0 0
3 201503 31 0 0 0 0 0 0 1 0 0 0
4 201504 74 0 0 0 0 0 0 0 1 0 0
5 201505 82 0 0 0 0 0 0 0 0 1 0
6 201506 86 0 0 0 0 0 0 0 0 0 1
7 201507 19 0 0 0 0 0 0 0 0 0 0
8 201508 32 0 0 0 0 0 0 0 0 0 0
9 201509 99 0 0 0 0 0 0 0 0 0 0
10 201510 47 0 1 0 0 0 0 0 0 0 0
11 201511 21 0 0 1 0 0 0 0 0 0 0
12 201512 77 0 0 0 1 0 0 0 0 0 0
13 201513 25 0 0 0 0 1 0 0 0 0 0
# ... with 3 more variables: P7 <dbl>, P8 <dbl>, P9 <dbl>
是否有一种自动的方法来获得相同的tibble，但P+数字列的正确顺序......。P1,P2,P3,P4等等等等...。
我们可以用 mixedsortmixedorder 从 gtools :
library(dplyr)
df %>% select(period, sales, gtools::mixedsort(names(.)))
# period sales P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11 P12 P13
# <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 201501 34 1 0 0 0 0 0 0 0 0 0 0 0 0
# 2 201502 22 0 1 0 0 0 0 0 0 0 0 0 0 0
# 3 201503 17 0 0 1 0 0 0 0 0 0 0 0 0 0
# 4 201504 91 0 0 0 1 0 0 0 0 0 0 0 0 0
# 5 201505 27 0 0 0 0 1 0 0 0 0 0 0 0 0
# 6 201506 58 0 0 0 0 0 1 0 0 0 0 0 0 0
# 7 201507 57 0 0 0 0 0 0 1 0 0 0 0 0 0
# 8 201508 2 0 0 0 0 0 0 0 1 0 0 0 0 0
# 9 201509 24 0 0 0 0 0 0 0 0 1 0 0 0 0
#10 201510 92 0 0 0 0 0 0 0 0 0 1 0 0 0
#11 201511 21 0 0 0 0 0 0 0 0 0 0 1 0 0
#12 201512 59 0 0 0 0 0 0 0 0 0 0 0 1 0
#13 201513 4 0 0 0 0 0 0 0 0 0 0 0 0 1
这是一个基本的R解决方案，它依赖于你在开始时有额外的两列。
i1 <- order(as.numeric(gsub('\\D+', '', names(df[-c(1:2)]))))
df <- df[c(1, 2, i1 + 2)]
现在调查这些名字，我们得到
names(df)
#[1] "period" "sales" "P1" "P2" "P3" "P4" "P5" "P6" "P7" "P8" "P9" "P10" "P11" "P12" "P13"