Question

为什么reshape2 melt会为我返回value = NA？

它适用于重塑，但不适用于reshape2：

以下是一个示例数据文件：

"","station_id","year","month","day","h1","h2","h3","h4","h5","h6","h7","h8","h9","h10","h11","h12","h13","h14","h15","h16","h17","h18","h19","h20","h21","h22","h23","h24"
"1",1,2004,1,1,46,46,45,41,39,35,33,33,36,47,53,54,55,55,55,55,52,46,40,40,39,38,40,41
"2",1,2004,1,2,43,44,46,46,47,47,47,47,47,47,47,49,52,56,54,56,57,53,50,47,46,45,45,45
"3",1,2004,1,3,45,46,46,44,43,46,46,47,51,55,56,59,65,68,69,68,68,65,64,63,62,63,63,62
"4",1,2004,1,4,63,62,62,62,60,60,60,62,60,64,64,66,71,70,71,72,71,68,67,67,65,64,65,64
"5",1,2004,1,5,64,63,65,64,64,64,64,64,65,66,66,67,68,68,66,66,66,66,63,54,52,49,47,47
"6",1,2004,1,6,47,46,45,43,41,41,39,39,40,43,45,44,45,46,46,46,45,39,39,39,38,36,32,32

我们说它保存为/tmp/foo.csv，然后：

使用重塑：

$ R
...
Type 'q()' to quit R.

> library("reshape")
Loading required package: plyr

Attaching package: ‘reshape’

The following object(s) are masked from ‘package:plyr’:

    rename, round_any

> hlist <- NULL; for(z in 1:24) { hlist <- cbind(hlist, sprintf("h%d",z)) }
> 
> thh <- read.csv('/tmp/foo.csv')
> thm <- melt(thh,measure.vars=hlist,variable="hour")
> head(thm)
  station_id year month day hour value
1          1 2004     1   1   h1    46
2          1 2004     1   2   h1    43
3          1 2004     1   3   h1    45
4          1 2004     1   4   h1    63
5          1 2004     1   5   h1    64
6          1 2004     1   6   h1    47
> q()

使用reshape2：

$ R
...
Type 'q()' to quit R.

> library("reshape2")
> hlist <- NULL; for(z in 1:24) { hlist <- cbind(hlist, sprintf("h%d",z)) }
> 
> thh <- read.csv('/tmp/foo.csv')
> thm <- melt(thh,measure.vars=hlist,variable="hour")
> head(thm)
  station_id year month day hour value
1          1 2004     1   1   h1    NA
2          1 2004     1   2   h1    NA
3          1 2004     1   3   h1    NA
4          1 2004     1   4   h1    NA
5          1 2004     1   5   h1    NA
6          1 2004     1   6   h1    NA
> q()

您可以看到使用library("reshape")，value列包含数字，但对于libary("reshape2")，它有NA，用于相同的数据。

Answer 1

有更好的方法可以做你想做的事。

以下所有内容适用于melt()的{{1}}：

reshape2

似乎是来自＆＃34;重塑＆＃34;的# Not using hlist melt(th, measure.vars=5:ncol(th), variable="hour") melt(th, id.vars=1:4, variable="hour") # Using your hlist hlist <- NULL; for(z in 1:24) { hlist <- cbind(hlist, sprintf("h%d",z)) } melt(th, measure.vars=as.vector(hlist), variable="hour") # Using an alternative hlist hlist <- paste0("h", 1:24) melt(th, measure.vars=hlist, variable="hour")接受矩阵作为melt()的输入，但来自＆＃34; reshape2＆＃34;的measure.vars。没有（我觉得更合理）。

更新：示例可重现的问题

仅供参考，以下是一种从头到尾的方式，您可以通过方便其他Stack Overflow用户复制和粘贴的方式分享此问题：

melt()

现在，展示你的问题。您可以使用# Use set.seed when you want to use random numbers # but want others to have the same data as you. set.seed(1) # Make up some data that mimics your actual dataset # Does not have to be your exact dataset th <- cbind( data.frame(station = rep(LETTERS[1:3], each = 3), year = 2004, month = rep(1:3, times = 3)), setNames(data.frame( matrix(sample(100, 45, replace = TRUE), nrow = 9)), paste0("h", 1:5))) hlist <- NULL; for(z in 1:5) { hlist <- cbind(hlist, sprintf("h%d",z)) } # Cleanup any unnecessary stuff that your code leaves behind in the workspace rm(z)而不必退出并重新启动R.

detach(package:package_name)

希望这有帮助！

为什么reshape2熔化返回值= NA？

1 个答案:

更新：示例可重现的问题