使用str_replace_all()更改data.frame()中的多个列名()
我读了这篇文章并练习了匹配模式,但我仍然没有弄清楚 我有一个同样尺寸的面板,每年数次。现在,我想以合乎逻辑的方式重命名它们。我的原始数据看起来有点像这样使用str_replace_all()更改data.frame()中的多个列名(),r,pattern-matching,rename,stringr,R,Pattern Matching,Rename,Stringr,我读了这篇文章并练习了匹配模式,但我仍然没有弄清楚 我有一个同样尺寸的面板,每年数次。现在,我想以合乎逻辑的方式重命名它们。我的原始数据看起来有点像这样 set.seed(667) dta <- data.frame(id = 1:6, R1213 = runif(6), R1224 = runif(6, 1, 2), R1255 = runif(6, 2, 3),
set.seed(667)
dta <- data.frame(id = 1:6,
R1213 = runif(6),
R1224 = runif(6, 1, 2),
R1255 = runif(6, 2, 3),
R1235 = runif(6, 3, 4))
# install.packages(c("tidyverse"), dependencies = TRUE)
require(tidyverse)
(tbl <- dta %>% as_tibble())
#> # A tibble: 6 x 5
#> id R1213 R1224 R1255 R1235
#> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 0.488 1.60 2.07 3.07
#> 2 2 0.692 1.42 2.76 3.19
#> 3 3 0.262 1.34 2.33 3.82
#> 4 4 0.330 1.77 2.61 3.93
#> 5 5 0.582 1.92 2.15 3.86
#> 6 6 0.930 1.88 2.56 3.59
每个调用A
实际上都来自同一年,比如2017年,但是需要附加后缀.1
,.2
,等等。我一遍又一遍地使用paste0('A.2017',1:3)
,但这一次有三个就足够了
tbl <- dta %>% as_tibble()
names(tbl) <- tbl %>% names() %>%
str_replace_all('^R1.[125].$', paste0('A.2017.', 1:3)) %>%
str_replace_all('^R1.[7].$', paste0('A.2018.', 1))
tbl
#> Warning message:
#> In stri_replace_all_regex(string, pattern, fix_replacement(replacement), :
#> longer object length is not a multiple of shorter object length
#> > tbl
#> # A tibble: 6 x 5
#> id A.2017.2 A.2017.3 A.2017.1 R1235
#> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 0.488 1.60 2.07 3.07
#> 2 2 0.692 1.42 2.76 3.19
#> 3 3 0.262 1.34 2.33 3.82
#> 4 4 0.330 1.77 2.61 3.93
#> 5 5 0.582 1.92 2.15 3.86
#> 6 6 0.930 1.88 2.56 3.59
tbl%为不兼容()
名称(待定)%names()%>%
str_replace_all(“^R1[125].$”,paste0('A.2017',1:3))%>%
str_replace_all(“^R1[7].$”,粘贴0('A.2018',1))
tbl
#>警告信息:
#>在stri_replace_all_regex(字符串、模式、修复_replacement(replacement))中:
#>较长的对象长度不是较短对象长度的倍数
#>>待定
#>#tibble:6 x 5
#>id A.2017.2 A.2017.3 A.2017.1 R1235
#>
#> 1 1 0.488 1.60 2.07 3.07
#> 2 2 0.692 1.42 2.76 3.19
#> 3 3 0.262 1.34 2.33 3.82
#> 4 4 0.330 1.77 2.61 3.93
#> 5 5 0.582 1.92 2.15 3.86
#> 6 6 0.930 1.88 2.56 3.59
这确实出现了,但顺序颠倒了,我被告知较长的对象长度不是较短对象长度的倍数,但我不是
3
正确的长度吗?我希望以一种更干净、更简单的方式来做这件事。而且,我真的不喜欢名称(tbl)基于David的建议-使用dplyr::rename_at
进行如下操作怎么样
library(dplyr)
## Get data
set.seed(667)
dta <- data.frame(id = 1:6,
R1213 = runif(6),
R1224 = runif(6, 1, 2),
R1255 = runif(6, 2, 3),
R1235 = runif(6, 3, 4)) %>%
as_tibble()
## Rename
dta <- dta %>%
rename_at(.vars = grep('^R1.[125].$', names(.)),
.funs = ~paste0("A.2017.", 1:length(.)))
dta
#> # A tibble: 6 x 5
#> id A.2017.1 A.2017.2 A.2017.3 R1235
#> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 0.196 1.74 2.51 3.49
#> 2 2 0.478 1.85 2.06 3.69
#> 3 3 0.780 1.32 2.21 3.26
#> 4 4 0.705 1.49 2.49 3.33
#> 5 5 0.942 1.59 2.66 3.58
#> 6 6 0.906 1.90 2.87 3.93
朝着整洁的数据方向发展
现在变量已经被重命名,您可能会发现以整洁的格式保存数据很有用
library(tidyr)
library(dplyr)
#Use tidy dataframe gather all variables, split by "." and drop A column (or keep if a measurement id)
renamed_dta %>%
gather(key = "measure", value = "value", -id) %>%
separate(measure, c("A", "year", "measure"), "[[.]]") %>%
select(-A)
#> # A tibble: 24 x 4
#> id year measure value
#> <int> <chr> <chr> <dbl>
#> 1 1 2017 1 0.196
#> 2 2 2017 1 0.478
#> 3 3 2017 1 0.780
#> 4 4 2017 1 0.705
#> 5 5 2017 1 0.942
#> 6 6 2017 1 0.906
#> 7 1 2017 2 1.74
#> 8 2 2017 2 1.85
#> 9 3 2017 2 1.32
#> 10 4 2017 2 1.49
#> # ... with 14 more rows
library(tidyr)
图书馆(dplyr)
#使用tidy dataframe收集所有变量,按“.”拆分并删除一列(如果是度量id,则保留)
已重命名\u dta%>%
聚集(key=“measure”、value=“value”、-id)%>%
单独(计量单位,c(“A”、“年”、“计量单位”),“[.]]”%>%
选择(-A)
#>#tibble:24 x 4
#>id年份度量值
#>
#> 1 1 2017 1 0.196
#> 2 2 2017 1 0.478
#> 3 3 2017 1 0.780
#> 4 4 2017 1 0.705
#> 5 5 2017 1 0.942
#> 6 6 2017 1 0.906
#> 7 1 2017 2 1.74
#> 8 2 2017 2 1.85
#> 9 3 2017 2 1.32
#> 10 4 2017 2 1.49
#>#…还有14行
你试过在
处重命名吗?这是一个很好的答案。非常感谢!你的函数对我特别有用。谢谢。太好了,很乐意帮忙。你能接受它作为答案吗?
library(dplyr)
library(purrr)
## Get data
set.seed(667)
dta <- data.frame(id = 1:6,
R1213 = runif(6),
R1224 = runif(6, 1, 2),
R1255 = runif(6, 2, 3),
R1235 = runif(6, 3, 4)) %>%
as_tibble()
## Define a function to keep a hold out data set, then rename iteratively for each pattern and replacement.
rename_multiple_years <- function(df, patterns,
replacements,
hold_out_var = "id") {
hold_out_df <- df %>%
select_at(.vars = hold_out_var)
rename_df <- map2_dfc(patterns, replacements, function(pattern, replacement) {
df %>%
rename_at(.vars = grep(pattern, names(.)),
.funs = ~paste0(replacement, 1:length(.))) %>%
select_at(.vars = grep(replacement, names(.)))
})
final_df <- bind_cols(hold_out_df, rename_df)
return(final_df)
}
## Call function on specified patterns and replacements
renamed_dta <- dta %>%
rename_multiple_years(patterns = c("^R1.[125].$", "^R1.[3].$"),
replacements = c("A.2017.", "A.2018."))
renamed_dta
#> # A tibble: 6 x 5
#> id A.2017.1 A.2017.2 A.2017.3 A.2018.1
#> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 0.196 1.74 2.51 3.49
#> 2 2 0.478 1.85 2.06 3.69
#> 3 3 0.780 1.32 2.21 3.26
#> 4 4 0.705 1.49 2.49 3.33
#> 5 5 0.942 1.59 2.66 3.58
#> 6 6 0.906 1.90 2.87 3.93
library(tidyr)
library(dplyr)
#Use tidy dataframe gather all variables, split by "." and drop A column (or keep if a measurement id)
renamed_dta %>%
gather(key = "measure", value = "value", -id) %>%
separate(measure, c("A", "year", "measure"), "[[.]]") %>%
select(-A)
#> # A tibble: 24 x 4
#> id year measure value
#> <int> <chr> <chr> <dbl>
#> 1 1 2017 1 0.196
#> 2 2 2017 1 0.478
#> 3 3 2017 1 0.780
#> 4 4 2017 1 0.705
#> 5 5 2017 1 0.942
#> 6 6 2017 1 0.906
#> 7 1 2017 2 1.74
#> 8 2 2017 2 1.85
#> 9 3 2017 2 1.32
#> 10 4 2017 2 1.49
#> # ... with 14 more rows