英文:
Renaming columns if a specific character string doesn't exist
问题
我想将列名更改为包括字符 _16S_sed，如果它们尚未出现。例如，我有以下数据集：
> dput(dataframe1)
structure(list(GL11_DN_1_16S_sed = structure(1:3, .Label = c("val1",
"val2", "val3"), class = "factor"), GL12_UP_3_16S_sed = structure(1:3, .Label = c("val1",
"val2", "val3"), class = "factor"), GL13_1_DN = structure(1:3, .Label = c("val1",
"val2", "val3"), class = "factor")), class = "data.frame", row.names = c(NA,
-3L))
其中第三列需要添加字符 _16S_sed。谢谢。
英文:
I'm looking to rename colnames to include the characters _16S_sed if they're not present. For example I have the dataset
> dput(dataframe1)
structure(list(GL11_DN_1_16S_sed = structure(1:3, .Label = c("val1",
"val2", "val3"), class = "factor"), GL12_UP_3_16S_sed = structure(1:3, .Label = c("val1",
"val2", "val3"), class = "factor"), GL13_1_DN = structure(1:3, .Label = c("val1",
"val2", "val3"), class = "factor")), class = "data.frame", row.names = c(NA,
-3L))
where the third column needs the characters _16S_sed. TIA
答案1
得分: 3
以下是您要翻译的代码部分：
在基本的R中，一种方法是首先使用grep或grepl来识别缺少所需字符串的列，然后进行替换：
missing_colnames <- grep("_16S_sed", colnames(df), invert = TRUE) # 感谢 @rps1227
colnames(df)[missing_colnames] <- paste0(colnames(df[missing_colnames]), "_16S_sed")
# 或者
missing_colnames2 <- !grepl("_16S_sed", colnames(df))
colnames(df)[missing_colnames2] <- paste0(colnames(df)[missing_colnames2], "_16S_sed")
输出：
# GL11_DN_1_16S_sed GL12_UP_3_16S_sed GL13_1_DN_16S_sed
# 1 val1 val1 val1
# 2 val2 val2 val2
# 3 val3 val3 val3
希望这对您有所帮助。如果您需要进一步的帮助，请随时告诉我。
英文:
In base R, one approach would be to first identify the columns missing the desired string using grep or grepl, then replace:
missing_colnames <- grep("_16S_sed", colnames(df), invert = TRUE) # thanks @rps1227
colnames(df)[missing_colnames] <- paste0(colnames(df[missing_colnames]), "_16S_sed")
# or
missing_colnames2 <- !grepl("_16S_sed", colnames(df))
colnames(df)[missing_colnames2] <- paste0(colnames(df)[missing_colnames2], "_16S_sed")
Output:
# GL11_DN_1_16S_sed GL12_UP_3_16S_sed GL13_1_DN_16S_sed
# 1 val1 val1 val1
# 2 val2 val2 val2
# 3 val3 val3 val3
答案2
得分: 2
library(tidyverse)
df %>%
rename_with(~ if_else(str_detect(.x, "_16S_sed"), .x, str_c(.x, "_16S_sed")))
# A tibble: 3 × 3
GL11_DN_1_16S_sed GL12_UP_3_16S_sed GL13_1_DN_16S_sed
<fct> <fct> <fct>
1 val1 val1 val1
2 val2 val2 val2
3 val3 val3 val3
Glimpsed:
Rows: 3
Columns: 3
$ GL11_DN_1_16S_sed
$ GL12_UP_3_16S_sed
$ GL13_1_DN_16S_sed
<details>
<summary>英文:</summary>
library(tidyverse)
df %>%
rename_with(~ if_else(str_detect(.x, "_16S_sed"), .x, str_c(.x, "_16S_sed")))
# A tibble: 3 × 3
GL11_DN_1_16S_sed GL12_UP_3_16S_sed GL13_1_DN_16S_sed
<fct> <fct> <fct>
1 val1 val1 val1
2 val2 val2 val2
3 val3 val3 val3
Glimpsed:
Rows: 3
Columns: 3
$ GL11_DN_1_16S_sed <fct> val1, val2, val3
$ GL12_UP_3_16S_sed <fct> val1, val2, val3
$ GL13_1_DN_16S_sed <fct> val1, val2, val3
</details>
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