替换/翻译字符串中的字符

发布于 2024-11-28 10:56:51 字数 552 浏览 3 评论 0原文

我有一个带有字符列的数据框:

df <- data.frame(var1 = c("aabbcdefg", "aabbcdefg"))
df
#        var1
# 1 aabbcdefg
# 2 aabbcdefg

我想替换几个不同的单个字符,例如从“a”到“h”,从“b”到“i”等等。目前,我多次调用 gsub

df$var1 <- gsub("a", "h", df$var1)
df$var1 <- gsub("b", "i", df$var1)
df$var1 <- gsub("c", "j", df$var1)
df$var1 <- gsub("d", "k", df$var1)
df$var1 <- gsub("e", "l", df$var1)
df$var1 <- gsub("f", "m", df$var1)
df
#        var1
# 1 hhiijklmg
# 2 hhiijklmg

但是,我确信还有更优雅的解决方案。有什么想法可以继续吗?

I have a data frame with a character column:

df <- data.frame(var1 = c("aabbcdefg", "aabbcdefg"))
df
#        var1
# 1 aabbcdefg
# 2 aabbcdefg

I want to replace several different individual characters, e.g. from "a" to "h", from "b" to "i" and so on. Currently I use several calls to gsub:

df$var1 <- gsub("a", "h", df$var1)
df$var1 <- gsub("b", "i", df$var1)
df$var1 <- gsub("c", "j", df$var1)
df$var1 <- gsub("d", "k", df$var1)
df$var1 <- gsub("e", "l", df$var1)
df$var1 <- gsub("f", "m", df$var1)
df
#        var1
# 1 hhiijklmg
# 2 hhiijklmg

However, I'm sure there are more elegant solutions. Any ideas ho to proceed?

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折戟 2024-12-05 10:56:51

您想要 chartr

df$var1 <- chartr("abcdef", "hijklm", df$var1)
df
#        var1
# 1 hhiijklmg
# 2 hhiijklmg

You want chartr:

df$var1 <- chartr("abcdef", "hijklm", df$var1)
df
#        var1
# 1 hhiijklmg
# 2 hhiijklmg
兮颜 2024-12-05 10:56:51

您可以创建 fromto 向量:

from <- c('a','b','c','d','e','f')
to <- c('h','i','j','k','l','m')

然后对 gsub 函数进行向量化:

gsub2 <- function(pattern, replacement, x, ...) {
for(i in 1:length(pattern))
x <- gsub(pattern[i], replacement[i], x, ...)
x
}

这给出:

> df <- data.frame(var1 = c("aabbcdefg", "aabbcdefg"))
> df$var1 <- gsub2(from, to, df$var1)
> df
       var1
1 hhiijklmg
2 hhiijklmg

You can create from and to vectors:

from <- c('a','b','c','d','e','f')
to <- c('h','i','j','k','l','m')

and then vectorialize the gsub function:

gsub2 <- function(pattern, replacement, x, ...) {
for(i in 1:length(pattern))
x <- gsub(pattern[i], replacement[i], x, ...)
x
}

Which gives:

> df <- data.frame(var1 = c("aabbcdefg", "aabbcdefg"))
> df$var1 <- gsub2(from, to, df$var1)
> df
       var1
1 hhiijklmg
2 hhiijklmg
又怨 2024-12-05 10:56:51

如果您不想使用 Chartr 因为替换可能不止一个字符,那么另一种选择是使用 gsubfn 包中的 gsubfn (我知道这不是 gsub,而是 gsub 的扩展)。下面是一个示例:

> library(gsubfn)
> tmp <- list(a='apple',b='banana',c='cherry')
> gsubfn('.', tmp, 'a.b.c.d')
[1] "apple.banana.cherry.d"

替换也可以是一个函数,它接受匹配并返回该匹配的替换值。

If you don't want to use chartr because the substitutions may be more than one character, then another option is to use gsubfn from the gsubfn package (I know this is not gsub, but is an expansion on gsub). Here is one example:

> library(gsubfn)
> tmp <- list(a='apple',b='banana',c='cherry')
> gsubfn('.', tmp, 'a.b.c.d')
[1] "apple.banana.cherry.d"

The replacement can also be a function that would take the match and return the replacement value for that match.

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