r 重命名列名以包含 0
r rename column names to include 0s
这是我的数据集
Id Date Col_a_1 Col_a_2 Col_a_3 Col_a_12 Col_a_65
1 02/19/2020 0 1 2 0 4
2 02/10/2020 1 2 0 1 3
1 03/11/2020 2 1 3 1 0
4 10/29/2020 1 0 2 1 0
我喜欢给那些以 1,2,3 结尾的列名加 0
最终预期的数据集是具有这些列名的数据集
Id Date col_b_01 col_b_02 col_b_03 col_b_12 col_a_65
1 02/19/2020 0 1 2 0 4
2 02/10/2020 1 2 0 1 3
1 03/11/2020 2 1 3 1 0
4 10/29/2020 1 0 2 1 0
虽然我可以单独重命名它们，但我喜欢更有效地这样做，但不确定如何。非常感谢任何建议。提前致谢。
names(dat)[3:7]
# [1] "Col_a_1" "Col_a_2" "Col_a_3" "Col_a_12" "Col_a_65"
gsub("_([0-9])$", "_0\1", names(dat)[3:7])
# [1] "Col_a_01" "Col_a_02" "Col_a_03" "Col_a_12" "Col_a_65"
names(dat)[3:7] <- gsub("_([0-9])$", "_0\1", names(dat)[3:7])
dat
# Id Date Col_a_01 Col_a_02 Col_a_03 Col_a_12 Col_a_65
# 1 1 02/19/2020 0 1 2 0 4
# 2 2 02/10/2020 1 2 0 1 3
# 3 1 03/11/2020 2 1 3 1 0
# 4 4 10/29/2020 1 0 2 1 0
您可以在 rename_with -
中使用 str_replace
library(dplyr)
library(stringr)
df %>%
rename_with(~str_replace(., '\d+', function(m) sprintf('%02s', m)),
starts_with('Col'))
# Id Date Col_a_01 Col_a_02 Col_a_03 Col_a_12 Col_a_65
#1 1 02/19/2020 0 1 2 0 4
#2 2 02/10/2020 1 2 0 1 3
#3 1 03/11/2020 2 1 3 1 0
#4 4 10/29/2020 1 0 2 1 0
数据
df <- structure(list(Id = c(1L, 2L, 1L, 4L), Date = c("02/19/2020",
"02/10/2020", "03/11/2020", "10/29/2020"), Col_a_1 = c(0L, 1L,
2L, 1L), Col_a_2 = c(1L, 2L, 1L, 0L), Col_a_3 = c(2L, 0L, 3L,
2L), Col_a_12 = c(0L, 1L, 1L, 1L), Col_a_65 = c(4L, 3L, 0L, 0L
)), class = "data.frame", row.names = c(NA, -4L))
这是我的数据集
Id Date Col_a_1 Col_a_2 Col_a_3 Col_a_12 Col_a_65
1 02/19/2020 0 1 2 0 4
2 02/10/2020 1 2 0 1 3
1 03/11/2020 2 1 3 1 0
4 10/29/2020 1 0 2 1 0
我喜欢给那些以 1,2,3 结尾的列名加 0
最终预期的数据集是具有这些列名的数据集
Id Date col_b_01 col_b_02 col_b_03 col_b_12 col_a_65
1 02/19/2020 0 1 2 0 4
2 02/10/2020 1 2 0 1 3
1 03/11/2020 2 1 3 1 0
4 10/29/2020 1 0 2 1 0
虽然我可以单独重命名它们，但我喜欢更有效地这样做，但不确定如何。非常感谢任何建议。提前致谢。
names(dat)[3:7]
# [1] "Col_a_1" "Col_a_2" "Col_a_3" "Col_a_12" "Col_a_65"
gsub("_([0-9])$", "_0\1", names(dat)[3:7])
# [1] "Col_a_01" "Col_a_02" "Col_a_03" "Col_a_12" "Col_a_65"
names(dat)[3:7] <- gsub("_([0-9])$", "_0\1", names(dat)[3:7])
dat
# Id Date Col_a_01 Col_a_02 Col_a_03 Col_a_12 Col_a_65
# 1 1 02/19/2020 0 1 2 0 4
# 2 2 02/10/2020 1 2 0 1 3
# 3 1 03/11/2020 2 1 3 1 0
# 4 4 10/29/2020 1 0 2 1 0
您可以在 rename_with -
str_replace
library(dplyr)
library(stringr)
df %>%
rename_with(~str_replace(., '\d+', function(m) sprintf('%02s', m)),
starts_with('Col'))
# Id Date Col_a_01 Col_a_02 Col_a_03 Col_a_12 Col_a_65
#1 1 02/19/2020 0 1 2 0 4
#2 2 02/10/2020 1 2 0 1 3
#3 1 03/11/2020 2 1 3 1 0
#4 4 10/29/2020 1 0 2 1 0
数据
df <- structure(list(Id = c(1L, 2L, 1L, 4L), Date = c("02/19/2020",
"02/10/2020", "03/11/2020", "10/29/2020"), Col_a_1 = c(0L, 1L,
2L, 1L), Col_a_2 = c(1L, 2L, 1L, 0L), Col_a_3 = c(2L, 0L, 3L,
2L), Col_a_12 = c(0L, 1L, 1L, 1L), Col_a_65 = c(4L, 3L, 0L, 0L
)), class = "data.frame", row.names = c(NA, -4L))