我如何连接(合并,组合)两个值? 例如,我有:

tmp = cbind("GAD", "AB")
tmp
#      [,1]  [,2]
# [1,] "GAD" "AB"

我的目标是将“tmp”中的两个值连接到一个字符串:

tmp_new = "GAD,AB"

哪个函数可以帮我做这个?


当前回答

Glue是作为tidyverse的一部分开发的一个新函数、数据类和包,具有许多扩展功能。它结合了paste、sprintf和前面其他答案的特性。

tmp <- tibble::tibble(firststring = "GAD", secondstring = "AB")
(tmp_new <- glue::glue_data(tmp, "{firststring},{secondstring}"))
#> GAD,AB

由reprex包于2019-03-06创建(v0.2.1)

是的,对于这个问题中的简单例子来说,它有点过头了,但在许多情况下都很强大。(参见https://glue.tidyverse.org/)

快速的例子比较粘贴与下面。胶水代码更容易输入,看起来也更容易阅读。

tmp <- tibble::tibble(firststring = c("GAD", "GAD2", "GAD3"), secondstring = c("AB1", "AB2", "AB3"))
(tmp_new <- glue::glue_data(tmp, "{firststring} and {secondstring} went to the park for a walk. {firststring} forgot his keys."))
#> GAD and AB1 went to the park for a walk. GAD forgot his keys.
#> GAD2 and AB2 went to the park for a walk. GAD2 forgot his keys.
#> GAD3 and AB3 went to the park for a walk. GAD3 forgot his keys.
(with(tmp, paste(firststring, "and", secondstring, "went to the park for a walk.", firststring, "forgot his keys.")))
#> [1] "GAD and AB1 went to the park for a walk. GAD forgot his keys."  
#> [2] "GAD2 and AB2 went to the park for a walk. GAD2 forgot his keys."
#> [3] "GAD3 and AB3 went to the park for a walk. GAD3 forgot his keys."

由reprex包于2019-03-06创建(v0.2.1)

其他回答

正如其他人所指出的,paste()是正确的方法。但是每次需要使用非默认分隔符时,都必须输入paste(str1, str2, str3, sep= "),这很烦人。

您可以非常容易地创建包装器函数,使工作变得更加简单。例如,如果你发现自己经常连接没有分隔符的字符串,你可以这样做:

p <- function(..., sep='') {
    paste(..., sep=sep, collapse=sep)
}

或者如果你经常想从一个向量连接字符串(比如PHP中的implode()):

implode <- function(..., sep='') {
     paste(..., collapse=sep)
}

允许你这样做:

p('a', 'b', 'c')
#[1] "abc"
vec <- c('a', 'b', 'c')
implode(vec)
#[1] "abc"
implode(vec, sep=', ')
#[1] "a, b, c"

此外,还有内置的paste0,它做的事情与我的implode相同,但不允许自定义分隔符。它比paste()稍微高效一些。

考虑这样一种情况,字符串是列,结果应该是一个新列:

df <- data.frame(a = letters[1:5], b = LETTERS[1:5], c = 1:5)

df$new_col <- do.call(paste, c(df[c("a", "b")], sep = ", ")) 
df
#  a b c new_col
#1 a A 1    a, A
#2 b B 2    b, B
#3 c C 3    c, C
#4 d D 4    d, D
#5 e E 5    e, E

如果需要粘贴所有列,可以跳过[c("a", "b")]子设置。

# you can also try str_c from stringr package as mentioned by other users too!
do.call(str_c, c(df[c("a", "b")], sep = ", ")) 

另一种方法:

sprintf("%s you can add other static strings here %s",string1,string2)

它有时比paste()函数有用。%s表示将包含主观字符串的位置。

注意,这将在你试图构建路径时派上用场:

sprintf("/%s", paste("this", "is", "a", "path", sep="/"))

输出

/this/is/a/path
paste()

才是正确的选择。正如之前的海报所指出的,粘贴可以做两件事:

将值串联成一个“字符串”,例如:

> paste("Hello", "world", sep=" ")
[1] "Hello world"

其中参数sep指定在参数之间用于连接的字符, 或者折叠字符向量

> x <- c("Hello", "World")
> x
[1] "Hello" "World"
> paste(x, collapse="--")
[1] "Hello--World"

其中参数collapse指定要折叠的vector元素之间使用的字符。

你甚至可以将两者结合起来:

> paste(x, "and some more", sep="|-|", collapse="--")
[1] "Hello|-|and some more--World|-|and some more"

Help.search()是一个方便的函数,例如:

> help.search("concatenate")

将引导您粘贴()。