sapply与lapply在读取文件和rbind时的对比';我在嘲笑他们
我遵循哈德利的思路:读取多个sapply与lapply在读取文件和rbind时的对比';我在嘲笑他们,r,sapply,rbind,read.csv,R,Sapply,Rbind,Read.csv,我遵循哈德利的思路:读取多个CSV文件,然后将它们转换为一个数据帧。我还试验了lappy与sapply的对比,如上所述 这是我的第一个CSV文件: dput(File1) structure(list(First.Name = structure(c(1L, 2L, 1L, 1L, 1L), .Label = c("A", "C"), class = "factor"), Last.Name = structure(c(1L, 2L, 2L, 2L, 2L), .Label = c("B",
CSV
文件,然后将它们转换为一个数据帧。我还试验了lappy
与sapply
的对比,如上所述
这是我的第一个CSV文件:
dput(File1)
structure(list(First.Name = structure(c(1L, 2L, 1L, 1L, 1L), .Label = c("A",
"C"), class = "factor"), Last.Name = structure(c(1L, 2L, 2L,
2L, 2L), .Label = c("B", "D"), class = "factor"), Income = c(55L,
23L, 34L, 45L, 44L), Tax = c(23L, 21L, 22L, 24L, 25L), Location = structure(c(3L,
3L, 1L, 4L, 2L), .Label = c("Americas", "AP", "EMEA", "LATAM"
), class = "factor")), .Names = c("First.Name", "Last.Name",
"Income", "Tax", "Location"), class = "data.frame", row.names = c(NA,
-5L))
dput(File2)
structure(list(First.Name = structure(c(1L, 2L, 1L, 1L, 1L), .Label = c("A",
"C"), class = "factor"), Last.Name = structure(c(1L, 2L, 2L,
2L, 2L), .Label = c("B", "D"), class = "factor"), Income = c(55L,
55L, 55L, 55L, 55L), Tax = c(24L, 24L, 24L, 24L, 24L), Location = structure(c(3L,
3L, 1L, 4L, 2L), .Label = c("Americas", "AP", "EMEA", "LATAM"
), class = "factor")), .Names = c("First.Name", "Last.Name",
"Income", "Tax", "Location"), class = "data.frame", row.names = c(NA,
-5L))
这是我的第二个CSV文件:
dput(File1)
structure(list(First.Name = structure(c(1L, 2L, 1L, 1L, 1L), .Label = c("A",
"C"), class = "factor"), Last.Name = structure(c(1L, 2L, 2L,
2L, 2L), .Label = c("B", "D"), class = "factor"), Income = c(55L,
23L, 34L, 45L, 44L), Tax = c(23L, 21L, 22L, 24L, 25L), Location = structure(c(3L,
3L, 1L, 4L, 2L), .Label = c("Americas", "AP", "EMEA", "LATAM"
), class = "factor")), .Names = c("First.Name", "Last.Name",
"Income", "Tax", "Location"), class = "data.frame", row.names = c(NA,
-5L))
dput(File2)
structure(list(First.Name = structure(c(1L, 2L, 1L, 1L, 1L), .Label = c("A",
"C"), class = "factor"), Last.Name = structure(c(1L, 2L, 2L,
2L, 2L), .Label = c("B", "D"), class = "factor"), Income = c(55L,
55L, 55L, 55L, 55L), Tax = c(24L, 24L, 24L, 24L, 24L), Location = structure(c(3L,
3L, 1L, 4L, 2L), .Label = c("Americas", "AP", "EMEA", "LATAM"
), class = "factor")), .Names = c("First.Name", "Last.Name",
"Income", "Tax", "Location"), class = "data.frame", row.names = c(NA,
-5L))
这是我的密码:
dat1 <-",First.Name,Last.Name,Income,Tax,Location\n1,A,B,55,23,EMEA\n2,C,D,23,21,EMEA\n3,A,D,34,22,Americas\n4,A,D,45,24,LATAM\n5,A,D,44,25,AP"
dat2 <-",First.Name,Last.Name,Income,Tax,Location\n1,A,B,55,24,EMEA\n2,C,D,55,24,EMEA\n3,A,D,55,24,Americas\n4,A,D,55,24,LATAM\n5,A,D,55,24,AP"
tc1 <- textConnection(dat1)
tc2 <- textConnection(dat2)
merged_file <- do.call(rbind, lapply(list(tc1,tc2), read.csv))
以下是输出:
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 1 1 1
[2,] 1 2 2 2 2
[3,] 55 23 34 45 44
[4,] 23 21 22 24 25
[5,] 3 3 1 4 2
[6,] 1 2 1 1 1
[7,] 1 2 2 2 2
[8,] 55 55 55 55 55
[9,] 24 24 24 24 24
[10,] 3 3 1 4 2
我非常感谢你的帮助。我对R相当陌生,不知道发生了什么 这个问题与因素无关,它是通用的
sapply
vslappy
。
为什么sapply
的理解如此错误,而lappy
的理解却正确记住,在R中,数据帧是列的列表。并且每列可以有不同的类型
将列列表返回到lappy
,从而正确进行连接。它将相应的列保持在一起。因此,你的因素出现正确rbind
但是。。。sapply
- 返回数值矩阵。。。(与数据帧不同,矩阵只能有一种类型)
- …更糟糕的是
- 所以
将两个5x6输入数据帧转换为转置的6x5矩阵(列现在对应于行)sapply
- 所有数据都强制为数字(垃圾!)
- 然后,
row-“将这两个数字的垃圾6x5矩阵串联成一个垃圾12x5矩阵。由于列已被转换为行,因此将矩阵连接在一起的行组合了数据类型,显然您的因子被弄乱了rbind
小结:只需使用
lappy
为什么要将lappy
更改为sapply
lappy
是这里合适的函数,而且效率更高。顺便说一句,paste
是矢量化的。@RichScriven-我只是在尝试理解为什么当我使用sapply
而不是lappy
时,输出是不同的。“虽然它工作得很好”,但作为一个可复制的例子,它甚至根本不工作。我们没有你的路径,所以它会失败。从textConnection()
而不是文件中读取数据帧最简单。我编辑了你的代码。这个问题与因素无关,它是通用的sapply vs lapply。副本