Question

我有一个看起来像的矩阵：

  1         2         3
1 part     of a      text1
2 part     of a      text2
3 part     of a      text3

我需要一个来自它的矢量：

c("part of a text1","part of a text 2","part of a text3"）

感谢您的帮助！

Answer 1

你可以试试这个

apply(df, 1, paste0, collapse=" ")
[1] "part of a text1" "part of a text2" "part of a text3"

Answer 2

我会在转换＆＃39;矩阵后使用do.call(paste。到data.frame，因为对于> 1e6行的大数据集来说会更快。话虽如此，并非每个数据集都是一个大数据集......

 do.call(paste, as.data.frame(m1))
 #[1] "part of a text1" "part of a text2" "part of a text3"

如果列数较少，例如3，则即使手动指定列并使用sprintf或paste，也会非常快。

  sprintf('%s %s %s', m1[,1] , m1[,2], m1[,3])
  #[1] "part of a text1" "part of a text2" "part of a text3"

基准

  m2 <- m1[rep(1:nrow(m1), 1e6),]
  system.time(do.call(paste, as.data.frame(m2)))
  #  user  system elapsed 
  # 2.873   0.000   1.816 

  system.time(apply(m2, 1, paste0, collapse=' '))
  #  user  system elapsed 
  #23.486   0.000  20.441 

  system.time(sprintf('%s %s %s', m2[,1], m2[,2], m2[,3]))
  #  user  system elapsed 
  #1.492   0.000   1.262 

  system.time(paste(m2[,1], m2[,2], m2[,3]))
  #   user  system elapsed 
  # 1.245   0.000   1.051

数据

 m1 <- structure(c("part", "part", "part", "of a", "of a", "of a", 
 "text1", 
 "text2", "text3"), .Dim = c(3L, 3L), .Dimnames = list(c("1", 
 "2", "3"), c("X1", "X2", "X3")))

在R中如何将基于字符的矩阵转换为向量？

2 个答案:

基准

数据