给定你创建的矩阵tmp:

paste(tmp[1,], collapse = ",")

我假设有一些原因，为什么你创建一个矩阵使用cbind，而不是简单地:

tmp <- "GAD,AB"

> tmp = paste("GAD", "AB", sep = ",")
> tmp
[1] "GAD,AB"

我从谷歌中通过搜索R个串联字符串找到了这个:http://stat.ethz.ch/R-manual/R-patched/library/base/html/paste.html

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"

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)

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

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 = ", "))