用tidyselect和正则表达式重命名R数据的列

Question

我有一个dataframe，它的列名是编号和一些复杂文本的组合：

1. A1.你好，

，

1. A1a.祝您愉快，

，

.

1. Z7d.其他一些标题

现在我只想保留"A1.“、"A1a.”、"Z7d."，删除前面的数字和结尾的文本。你知道如何用tidyselect和regex做这件事吗？

Answer 1

你可以用这个正则表达式-

names(df) <- sub('\\d+\\.\\s+([A-Za-z0-9]+).*', '\\1', names(df))
names(df)
#[1] "A1"  "A1a" "Z7d"

如果您想要一个rename_with答案，也可以在tidyverse中使用相同的正则表达式。

library(dplyr)
df %>% rename_with(~sub('\\d+\\.\\s+([A-Za-z0-9]+).*', '\\1', .))

#          A1        A1a        Z7d
#1  0.5755992  0.4147519 -0.1474461
#2  0.1347792 -0.6277678  0.3263348
#3  1.6884930  1.3931306  0.8809109
#4 -0.4269351 -1.2922231 -0.3362182
#5 -2.0032113  0.2619571  0.4496466

数据

df <- structure(list(`1. A1. Good day` = c(0.575599213383783, 0.134779160673435, 
1.68849296209512, -0.426935114884432, -2.00321125417319), `2. A1a. Have a nice day` = c(0.414751904860513, 
-0.627767775889949, 1.39313055331098, -1.29222310608057, 0.261957078465535
), `99. Z7d. Some other titles` = c(-0.147446140558093, 0.326334824433201, 
0.880910933597998, -0.336218174873965, 0.449646567320979)), 
class = "data.frame", row.names = c(NA, -5L))

Answer 2

我们可以使用str_extract

library(stringr)
names(df) <- str_extract(names(df), "(?<=\\.\\s)[^.]+")
names(df)
[1] "A1"  "A1a" "Z7d"

数据

df <- structure(list(`1. A1. Good day` = c(0.575599213383783, 0.134779160673435, 
1.68849296209512, -0.426935114884432, -2.00321125417319), `2. A1a. Have a nice day` = c(0.414751904860513, 
-0.627767775889949, 1.39313055331098, -1.29222310608057, 0.261957078465535
), `99. Z7d. Some other titles` = c(-0.147446140558093, 0.326334824433201, 
0.880910933597998, -0.336218174873965, 0.449646567320979)), 
class = "data.frame", row.names = c(NA, -5L))