Question

我正在尝试构建一个函数F，用于替换stings'df'数据框中的目标字符串'str'，逐列，逐行，根据列名作为要替换的子字符串，和列值作为替换。 result是替换字符串的字符串向量长度'rownum' 将'colnum'替换为每个字符串作为输出。

一个例子可以说明最佳：

str <- "Hi, I am name and I am age years old! - said name "

df <- data.frame(name = c('John', 'Richard','Edward'), age =c('10','26','12'))

F(str,df)

"Hi, I am John and I am 10 years old! - said John "

"Hi, I am Richard and I am 26 years old! - said Richard "

"Hi, I am Edward and I am 12 years old! - said Edward "

我为这份工作写了一个函数：

F <- function(str,df)
{
  x <- str
  for(i in names(df)){
    x <- unname(mapply(gsub,i,df[[i]],x))
  }
  return(x)
}

它似乎有效，但我的印象是它不高效也不优雅。

有没有办法避免循环？
是必要的吗？
可以在'str'是多行文本时使用，而不仅仅是a 单行？

感谢您的帮助

Answer 1

也许是另一个“隐藏”for循环的选项：

library(stringi)
f <- function(str, df) 
  apply(df, 1, stri_replace_all, str=str, fixed=names(df), merge=T, vec=F)  
f("Hi, I am name and I am age years old! - said name ", df)
# [1] "Hi, I am John and I am 10 years old! - said John "      
# [2] "Hi, I am Richard and I am 26 years old! - said Richard "
# [3] "Hi, I am Edward and I am 12 years old! - said Edward "

str <- "Hi, I am name and I am age years old! - said name\n
Hi, I am name and I am age years old! - said name"
f(str, df)
# [1] "Hi, I am John and I am 10 years old! - said John\n\nHi, I am John and I am 10 years old! - said John"            
# [2] "Hi, I am Richard and I am 26 years old! - said Richard\n\nHi, I am Richard and I am 26 years old! - said Richard"
# [3] "Hi, I am Edward and I am 12 years old! - said Edward\n\nHi, I am Edward and I am 12 years old! - said Edward"

Answer 2

Mustache是通过模板进行此类字符串操作的绝佳解决方案。对于简单的字符串/模板，我也会使用sprintf。对于更复杂的模板，我肯定会使用Mustache。

Mustache的R实现是whisker - 包

在你的情况下，可以这样做，例如通过：

#install.packages("whisker")
library(whisker)
template <- 
"Hi, I am {{name}} and I am {{age}} years old! - 
said {{name}}"

df <- data.frame(name = c('John', 'Richard','Edward'), age =c('10','26','12'))

out <- apply(df, 1, function(x) whisker.render(template, x))

给你：

[1] "Hi, I am John and I am 10 years old! -\nsaid John"      
[2] "Hi, I am Richard and I am 26 years old! -\nsaid Richard"
[3] "Hi, I am Edward and I am 12 years old! -\nsaid Edward"

存在换行符（\n）是输出。

您还可以使用readLines初步阅读模板，而不是在代码中对其进行硬编码。

Answer 3

最简单的方法（由评论中的@RomanLustrik提供）：

str <- "Hi, I am %s and I am %s years old! - said %s "
sprintf(str, df$name, df$age, df$name)

结果：

[1] "Hi, I am John and I am 10 years old! - said John "      
[2] "Hi, I am Richard and I am 26 years old! - said Richard "
[3] "Hi, I am Edward and I am 12 years old! - said Edward "

Answer 4

我们可以通过编程方式完成此任务（灵感来自@ RomanLustrik的想法

do.call(sprintf, c(cbind(df, name2=df$name), fmt = gsub("name|age", "%s", str)))
#[1] "Hi, I am John and I am 10 years old! - said John "    
#[2] "Hi, I am Richard and I am 26 years old! - said Richard "
#[3] "Hi, I am Edward and I am 12 years old! - said Edward "

R - 使用字符串数据帧顺序替换字符串

4 个答案: