Question

我要订购以下数据

col.to_order <- paste(rep(c("a",'b','c'),times=100), 1:100,sep = "")

[1] "a1"   "b2"   "c3"   "a4"   "b5"   "c6"   "a7"   "b8"   "c9"   "a10"  "b11"  "c12"  "a13"  "b14"  "c15"  "a16"  "b17"  "c18" 
 [19] "a19"  "b20"  "c21"  "a22"  "b23"  "c24"  "a25"  "b26"  "c27"  "a28"  "b29"  "c30"  "a31"  "b32"  "c33"  "a34"  "b35"  "c36" 
 [37] "a37"  "b38"  "c39"  "a40"  "b41"  "c42"  "a43"  "b44"  "c45"  "a46"  "b47"  "c48"  "a49"  "b50"  "c51"  "a52"  "b53"  "c54" 
 [55] "a55"  "b56"  "c57"  "a58"  "b59"  "c60"  "a61"  "b62"  "c63"  "a64"  "b65"  "c66"  "a67"  "b68"  "c69"  "a70"  "b71"  "c72" 
 [73] "a73"  "b74"  "c75"  "a76"  "b77"  "c78"  "a79"  "b80"  "c81"  "a82"  "b83"  "c84"  "a85"  "b86"  "c87"  "a88"  "b89"  "c90"

我想通过以下方式订购数据

"a1" "b1" "c1" "a2" "b2" "c2" "a3" "b3" "c3" ... "a100" "b100" "c100".

我该怎么做？

Answer 1

当您将其称为列时，我认为您具有df。然后，dplyr的可能性可能是：

df %>%
 arrange(as.numeric(gsub("\\D", "", col.to_order)))

    col.to_order
1             a1
2             b1
3             c1
4             b2
5             c2
6             a2
7             c3
8             a3
9             b3
10            a4

使用base R（这只是一个向量）：

col.to_order[order(as.numeric(gsub("\\D", "", col.to_order)))]

  [1] "a1"   "b1"   "c1"   "b2"   "c2"   "a2"   "c3"   "a3"   "b3"   "a4"   "b4"

Answer 2

使用tidyverse的选项是使用parse_number提取数字部分

library(readr)
library(dplyr)
df %>% 
    arrange(parse_number(col.to_order))
#     col.to_order
#1             a1
#2             b1
#3             c1
#4             b2
#5             c2
#6             a2
#7             c3
#8             a3
#9             b3
#10            a4
#...

数据

df <- data.frame(col.to_order, stringsAsFactors = FALSE)

按因子水平排序数据

2 个答案:

数据