6

作为试点调查的一部分，我向每个 Turker 提供了四个选项中的一组选项。数据如下所示：

> so
  WorkerId pio_1_1 pio_1_2 pio_1_3 pio_1_4 pio_2_1 pio_2_2 pio_2_3 pio_2_4
1        1     Yes      No      No      No      No      No     Yes      No
2        2      No     Yes      No      No     Yes      No     Yes      No
3        3     Yes     Yes      No      No     Yes      No     Yes      No

我希望它看起来像这样：

WorkerId set pio1 pio2 pio3 pio4
       1   1  Yes   No   No   No
       1   2   No   No  Yes   No
...

我可以通过多种方式来解决这个问题，但都不是很优雅：

  • 用正则表达式和反向引用交换数字的顺序，然后使用 reshape()
  • 编写我自己的小函数来解析下划线之间的第一个数字，然后将其重新整形
  • 拆分然后堆叠列（依赖于正确的排序）

但在我看来，所有这些都忽略了你可能称之为“双宽”格式的数据有自己的结构的想法。我很想为此使用 reshape2 包，但是尽管使用 cast() 生成了数据，但我看不到任何可以帮助我真正融化这个 data.frame 的选项。

欢迎提出建议。

so <- structure(list(WorkerId = 1:3, pio_1_1 = structure(c(2L, 1L, 
2L), .Label = c("No", "Yes"), class = "factor"), pio_1_2 = structure(c(1L, 
2L, 2L), .Label = c("No", "Yes"), class = "factor"), pio_1_3 = structure(c(1L, 
1L, 1L), .Label = c("No", "Yes"), class = "factor"), pio_1_4 = structure(c(1L, 
1L, 1L), .Label = "No", class = "factor"), pio_2_1 = structure(c(1L, 
2L, 2L), .Label = c("No", "Yes"), class = "factor"), pio_2_2 = structure(c(1L, 
1L, 1L), .Label = c("No", "Yes"), class = "factor"), pio_2_3 = structure(c(2L, 
2L, 2L), .Label = c("No", "Yes"), class = "factor"), pio_2_4 = structure(c(1L, 
1L, 1L), .Label = "No", class = "factor")), .Names = c("WorkerId", 
"pio_1_1", "pio_1_2", "pio_1_3", "pio_1_4", "pio_2_1", "pio_2_2", 
"pio_2_3", "pio_2_4"), row.names = c(NA, 3L), class = "data.frame")
4

5 回答 5

4

如果我们调用您的原始数据集 dat，这将使用 base：

dat2 <- reshape(dat, 
    varying=list(pio_1= c(2, 6), pio_2= c(3,7), pio_3= c(4,8), pio_4= c(5,9) ),
    v.names=c(paste0("pio_",1:4)), 
    idvar = "WorkerId",
    direction="long", 
    timevar="set") 
row.names(dat2) <- NULL
dat2[order(dat2$WorkerId, dat2$set), ]

产生：

  WorkerId set pio_1 pio_2 pio_3 pio_4
1        1   1   Yes    No    No    No
2        1   2    No    No   Yes    No
3        2   1    No   Yes    No    No
4        2   2   Yes    No   Yes    No
5        3   1   Yes   Yes    No    No
6        3   2   Yes    No   Yes    No

编辑：（好吧，我忍不住想尝试让它更容易自动化）

y <- do.call('rbind', strsplit(names(dat)[-1], "_"))[, c(1, 3, 2)]
names(dat) <- c(names(dat)[1], paste0(y[, 1], "_", y[, 2], ".", y[, 3]))

dat2 <- reshape(dat, 
    varying=2:9, 
    idvar = "WorkerId",
    direction="long", 
    timevar="set")
row.names(dat2) <- NULL
dat2[order(dat2$WorkerId, dat2$set), ]
于 2012-04-15T12:40:20.543 回答
4

我建议对你的名字做一些 gsub 以使它们变成 R 更喜欢的形式，换句话说，时间变量是最后一项，而不是中间项，并带有“。” 作为默认分隔符。

试试这个：

names(so) = gsub("([a-z])_([0-9])_([0-9])", "\\1_\\3\\.\\2", names(so))
so.l = reshape(so, direction="long", varying=2:9, timevar="set", idvar=1)

然后，如果要按 WorkerId 排序：

so.l = so.l[order(so.l$WorkerId), ]
于 2012-04-15T13:15:23.970 回答
4

Here's another solution using reshape2 and stringr

melt.wide = function(data, id.vars, new.names) {
  require(reshape2)
  require(stringr)
  data.melt = melt(data, id.vars=id.vars)
  new.vars = data.frame(do.call(
    rbind, str_extract_all(data.melt$variable, "[0-9]+")))
  names(new.vars) = new.names
  cbind(data.melt, new.vars)
}

Then, you use it like this:

> so.long = melt.wide(so, id.vars=1, new.names=c("set", "option"))
> dcast(so.long, WorkerId + set ~ option)
  WorkerId set   1   2   3  4
1        1   1 Yes  No  No No
2        1   2  No  No Yes No
3        2   1  No Yes  No No
4        2   2 Yes  No Yes No
5        3   1 Yes Yes  No No
6        3   2 Yes  No Yes No

I think that using stringr might prove to be a more simple solution than those in the functions that have been suggested so far.

A "triple wide" example

Here's why I like this solution: it also works if your data is, say, triple wide. Here's an example, (with data modified from here):

triplewide = structure(list(ID = 1:4, w1d1t1 = c(4L, 3L, 2L, 2L), w1d1t2 = c(5L, 
4L, 3L, 3L), w1d2t1 = c(6L, 5L, 5L, 4L), w1d2t2 = c(5L, 4L, 5L, 
2L), w2d1t1 = c(6L, 5L, 4L, 3L), w2d1t2 = c(5L, 4L, 5L, 5L), 
    w2d2t1 = c(6L, 3L, 6L, 3L), w2d2t2 = c(7L, 4L, 3L, 2L)), .Names = c("ID", 
"w1d1t1", "w1d1t2", "w1d2t1", "w1d2t2", "w2d1t1", "w2d1t2", "w2d2t1", 
"w2d2t2"), class = "data.frame", row.names = c(NA, -4L))

This is what it looks like to start with:

> triplewide
  ID w1d1t1 w1d1t2 w1d2t1 w1d2t2 w2d1t1 w2d1t2 w2d2t1 w2d2t2
1  1      4      5      6      5      6      5      6      7
2  2      3      4      5      4      5      4      3      4
3  3      2      3      5      5      4      5      6      3
4  4      2      3      4      2      3      5      3      2

A variable name like w1d1t1 means "week 1, day 1, test 1". Let's say that your expected "tidy data" should be a dataset with the columns "ID", "week", "day", "trial 1", and "trial 2", then you can use the function as follows:

> triplewide.long = melt.wide(triplewide, id.vars="ID",
+                             new.names=c("week", "day", "trial"))
> dcast(triplewide.long, ID + week + day ~ trial)
   ID week day 1 2
1   1    1   1 4 5
2   1    1   2 6 5
3   1    2   1 6 5
4   1    2   2 6 7
5   2    1   1 3 4
6   2    1   2 5 4
7   2    2   1 5 4
8   2    2   2 3 4
9   3    1   1 2 3
10  3    1   2 5 5
11  3    2   1 4 5
12  3    2   2 6 3
13  4    1   1 2 3
14  4    1   2 4 2
15  4    2   1 3 5
16  4    2   2 3 2
于 2012-04-16T08:10:41.673 回答
3

I am not sure if this is too obvious, but here goes. It should be self-explanatory. Pass in your so dataframe and it returns the reshaped data. 

library("reshape2")

reshape.middle <- function(dat) {
    dat <- melt(so, id="WorkerId")
    dat$set <- substr(dat$variable, 5,5)
    dat$name <- paste(substr(dat$variable, 1, 4),
                      substr(dat$variable, 7, 7),
                      sep="")
    dat$variable <- NULL

    dat <- melt(dat, id=c("WorkerId", "set", "name"))
    dat$variable <- NULL

    return(dcast(dat, WorkerId + set ~ name))
}

so # initial form
so <- reshape.middle(so)
so # as needed

Hope this helps.

于 2012-04-15T13:40:59.330 回答
1

Here's what I eventually went with, largely based off @gauden 's approach. In re-reading Hadley's tidy data pdf, it turns out he advises a similar course.

melt.wide <- function(data, id.vars, new.names, sep=".", variable.name="variable", ... ) {
  # Guess number of variables currently wide
  colnames(data) <- sub( paste0(sep,"$"), "",  colnames(data) )
  wide.vars <- colnames(data)[grep( sep, colnames(data) )]
  n.wide <- str_count( wide.vars, sep )
  stopifnot(length(new.names)==unique(n.wide))
  # Melt
  data.melt <- melt(data,id.vars=id.vars,measure.vars=wide.vars,...)
  new <- stack.list(str_split(data.melt$variable,sep))
  colnames(new) <- c(variable.name,new.names)
  data.melt <- subset(data.melt,select=c(-variable))
  cbind(data.melt,new)
}


# Stacks lists of data.frames (e.g. from replicate() )
stack.list <- function( x, label=FALSE, ... ) {
  ret <- x[[1]]
  if(label) { ret$from <- 1 }
  if(length(x)==1) return(ret)
  for( i in seq(2,length(x)) ) {
    new <- x[[i]]
    if(label) { new$from <- i }
    ret <- rbind(ret,new)
  }
  return(ret)
}

> dat<-melt.wide(so,id.vars="WorkerId",new.names=c("set","option"),sep="_")
> dcast(dat, WorkerId + set ~ option)
  WorkerId set   1   2   3  4
1        1   1 Yes  No  No No
2        1   2  No  No Yes No
3        2   1  No Yes  No No
4        2   2 Yes  No Yes No
5        3   1 Yes Yes  No No
6        3   2 Yes  No Yes No
于 2012-04-16T01:39:44.570 回答