包装基础R重塑为便于使用的(Wrapping base R reshape for ease-of-

这是一个事实举世公认的是，R的基础重塑命令是迅速而有力，但有悲惨的语法。 因此，我写它周围的快速包装，我会扔在下一版本的taRifx包。 我这样做之前，但是，我想征求改进。

这里是我的版本，从@RichieCotton更新：

# reshapeasy: Version of reshape with way, way better syntax
 # Written with the help of the StackOverflow R community
 # x is a data.frame to be reshaped
 # direction is "wide" or "long"
 # vars are the names of the (stubs of) the variables to be reshaped (if omitted, defaults to everything not in id or vary)
 # id are the names of the variables that identify unique observations
 # vary is the variable that varies.  Going to wide this variable will cease to exist.  Going to long it will be created.
 # omit is a vector of characters which are to be omitted if found at the end of variable names (e.g. price_1 becomes price in long)
 # ... are options to be passed to stats::reshape
reshapeasy <- function( data, direction, id=(sapply(data,is.factor) | sapply(data,is.character)), vary=sapply(data,is.numeric), omit=c("_","."), vars=NULL, ... ) {
  if(direction=="wide") data <- stats::reshape( data=data, direction=direction, idvar=id, timevar=vary, ... )
  if(direction=="long") {
    varying <- which(!(colnames(data) %in% id))
    data <- stats::reshape( data=data, direction=direction, idvar=id, varying=varying, timevar=vary, ... )
  }
  colnames(data) <- gsub( paste("[",paste(omit,collapse="",sep=""),"]$",sep=""), "", colnames(data) )
  return(data)
}

请注意，您可以从广角移动到长而不改变比其他方向的选项。 对我来说，这是关键的可用性。

我很高兴，如果你聊天或E-mail给我你的信息给在功能帮助文件确认任何实质性的改善。

改进可能落在在以下几个方面：

• 命名的功能和它的参数
• 更一般化的（目前处理一个相当具体的情况下，我认为是目前最常见的，但它尚未用尽统计的能力::重塑）
• 代码改进

例子

样本数据

x.wide <- structure(list(surveyNum = 1:6, pio_1 = structure(c(2L, 2L, 1L, 
2L, 1L, 1L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2"), class = "factor"), pio_2 = structure(c(2L, 1L, 2L, 1L, 
2L, 2L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2"), class = "factor"), pio_3 = structure(c(2L, 2L, 1L, 1L, 
2L, 1L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2"), class = "factor"), caremgmt_1 = structure(c(2L, 1L, 1L, 
2L, 1L, 2L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2"), class = "factor"), caremgmt_2 = structure(c(1L, 2L, 2L, 
2L, 2L, 1L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2"), class = "factor"), caremgmt_3 = structure(c(1L, 2L, 1L, 
2L, 1L, 1L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2"), class = "factor"), prev_1 = structure(c(1L, 2L, 2L, 1L, 
1L, 2L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2"), class = "factor"), prev_2 = structure(c(2L, 2L, 1L, 2L, 
1L, 1L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2"), class = "factor"), prev_3 = structure(c(2L, 1L, 2L, 2L, 
1L, 1L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2"), class = "factor"), price_1 = structure(c(2L, 1L, 2L, 5L, 
3L, 4L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2", "3", "4", "5", "6"), class = "factor"), price_2 = structure(c(6L, 
5L, 5L, 4L, 4L, 2L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2", "3", "4", "5", "6"), class = "factor"), price_3 = structure(c(3L, 
5L, 2L, 5L, 4L, 5L), .Names = c("1", "2", "3", "4", "5", "6"), .Label = c("1", 
"2", "3", "4", "5", "6"), class = "factor")), .Names = c("surveyNum", 
"pio_1", "pio_2", "pio_3", "caremgmt_1", "caremgmt_2", "caremgmt_3", 
"prev_1", "prev_2", "prev_3", "price_1", "price_2", "price_3"
), idvars = "surveyNum", rdimnames = list(structure(list(surveyNum = 1:24), .Names = "surveyNum", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
"14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24"
), class = "data.frame"), structure(list(variable = structure(c(1L, 
1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L), .Label = c("pio", 
"caremgmt", "prev", "price"), class = "factor"), .id = c(1L, 
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L)), .Names = c("variable", 
".id"), row.names = c("pio_1", "pio_2", "pio_3", "caremgmt_1", 
"caremgmt_2", "caremgmt_3", "prev_1", "prev_2", "prev_3", "price_1", 
"price_2", "price_3"), class = "data.frame")), row.names = c(NA, 
6L), class = c("cast_df", "data.frame"))

x.long <- structure(list(.id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), pio = structure(c(2L, 
2L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 
2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 2L, 
1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 
1L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 
1L, 2L, 2L, 1L, 2L, 1L, 1L), .Label = c("1", "2"), class = "factor"), 
    caremgmt = structure(c(2L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 1L, 
    2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 
    1L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 
    1L, 2L, 1L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 
    1L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 1L, 
    1L, 2L, 2L), .Label = c("1", "2"), class = "factor"), prev = structure(c(1L, 
    2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 
    1L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 
    2L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 
    2L, 2L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 
    1L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 2L, 2L, 2L), .Label = c("1", 
    "2"), class = "factor"), price = structure(c(2L, 1L, 2L, 
    5L, 3L, 4L, 1L, 5L, 4L, 3L, 1L, 2L, 6L, 6L, 5L, 4L, 6L, 3L, 
    5L, 6L, 3L, 1L, 2L, 4L, 3L, 5L, 2L, 5L, 4L, 5L, 6L, 6L, 4L, 
    6L, 4L, 1L, 2L, 3L, 1L, 2L, 2L, 5L, 1L, 6L, 1L, 3L, 4L, 3L, 
    6L, 5L, 5L, 4L, 4L, 2L, 2L, 2L, 6L, 3L, 1L, 4L, 4L, 5L, 1L, 
    3L, 6L, 1L, 3L, 5L, 1L, 3L, 6L, 2L), .Label = c("1", "2", 
    "3", "4", "5", "6"), class = "factor"), surveyNum = c(1L, 
    2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 
    15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 1L, 2L, 
    3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 
    16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 1L, 2L, 3L, 
    4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 
    17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L)), .Names = c(".id", 
"pio", "caremgmt", "prev", "price", "surveyNum"), row.names = c(NA, 
-72L), class = "data.frame")

例子

> x.wide
  surveyNum pio_1 pio_2 pio_3 caremgmt_1 caremgmt_2 caremgmt_3 prev_1 prev_2 prev_3 price_1 price_2 price_3
1         1     2     2     2          2          1          1      1      2      2       2       6       3
2         2     2     1     2          1          2          2      2      2      1       1       5       5
3         3     1     2     1          1          2          1      2      1      2       2       5       2
4         4     2     1     1          2          2          2      1      2      2       5       4       5
5         5     1     2     2          1          2          1      1      1      1       3       4       4
6         6     1     2     1          2          1          1      2      1      1       4       2       5
> reshapeasy( x.wide, "long", NULL, id="surveyNum", vary="id", sep="_" )
    surveyNum id pio caremgmt prev price
1.1         1  1   2        2    1     2
2.1         2  1   2        1    2     1
3.1         3  1   1        1    2     2
4.1         4  1   2        2    1     5
5.1         5  1   1        1    1     3
6.1         6  1   1        2    2     4
1.2         1  2   2        1    2     6
2.2         2  2   1        2    2     5
3.2         3  2   2        2    1     5
4.2         4  2   1        2    2     4
5.2         5  2   2        2    1     4
6.2         6  2   2        1    1     2
1.3         1  3   2        1    2     3
2.3         2  3   2        2    1     5
3.3         3  3   1        1    2     2
4.3         4  3   1        2    2     5
5.3         5  3   2        1    1     4
6.3         6  3   1        1    1     5

> head(x.long)
  .id pio caremgmt prev price surveyNum
1   1   2        2    1     2         1
2   1   2        1    2     1         2
3   1   1        1    2     2         3
4   1   2        2    1     5         4
5   1   1        1    1     3         5
6   1   1        2    2     4         6

> head(reshapeasy( x.long, direction="wide", id="surveyNum", vary=".id" ))
  surveyNum pio.1 caremgmt.1 prev.1 price.1 pio.3 caremgmt.3 prev.3 price.3 pio.2 caremgmt.2 prev.2 price.2
1         1     2          2      1       2     2          1      2       3     2          1      2       6
2         2     2          1      2       1     2          2      1       5     1          2      2       5
3         3     1          1      2       2     1          1      2       2     2          2      1       5
4         4     2          2      1       5     1          2      2       5     1          2      2       4
5         5     1          1      1       3     2          1      1       4     2          2      1       4
6         6     1          2      2       4     1          1      1       5     2          1      1       2

Answer 1:

我也想看到一个选项，命令的输出，因为这是我不喜欢在基地R.约重塑作为例子的一件事，让我们使用Stata的学习模块：宽重塑数据长 ，你是已经熟悉了。 我看的例子是“孩子的身高和体重在1岁和2岁”的例子。

这是我通常与做reshape()

# library(foreign)
kidshtwt = read.dta("http://www.ats.ucla.edu/stat/stata/modules/kidshtwt.dta")
kidshtwt.l = reshape(kidshtwt, direction="long", idvar=1:2, 
                     varying=3:6, sep="", timevar="age")
# The reshaped data is correct, just not in the order I want it
# so I always have to do another step like this
kidshtwt.l = kidshtwt.l[order(kidshtwt.l$famid, kidshtwt.l$birth),]

由于这是一个恼人的步骤，我总是要经过整形的数据时，我认为这将是添加到您的功能非常有用。

我也建议至少具有与最后一列为了重塑从做同样的事情的选项long到wide 。

实施例功能列顺序

我不知道对这个融入你的函数的最佳方式，但我把这个共同基础上的变量名称基本模式进行排序的数据帧。

col.name.sort = function(data, patterns) {
  a = names(data)
  b = length(patterns)

  subs = vector("list", b)

  for (i in 1:b) {
    subs[[i]] = sort(grep(patterns[i], a, value=T))
    }
  x = unlist(subs)
  data[ , x ]
}

它可以以下面的方式被使用。 试想一下，我们救了你的输出reshapeasy long到wide例如为名为数据帧a ，我们希望它是“surveyNum”，“caremgmt”（1-3），“上一页”（1-3），“PIO有序“（1-3），以及 ”价格“（1-3），我们可以使用：

col.name.sort(a, c("sur", "car", "pre", "pio", "pri"))

Answer 2:

一些初步的想法：

我一直认为，方向指令“宽”和“长”是一个有点模糊。 难道他们的意思是你想将数据转换成该格式，或该数据已经是这种格式？ 这是你需要学习或查找的东西。 您可以通过具有分离功能避免问题reshapeToWide和reshapeToLong 。 作为奖励，每个函数的签名少了一个参数。

---

我不认为你的意思是包括行

varying <- which(!(colnames(x.wide) %in% "surveyNum"))

因为它是指一个特定的数据集。

---

我喜欢data到x为第一个参数，因为它清楚地表明，该输入应该是一个数据帧。

---

它通常是更好的形式有没有默认参数的第一。 因此， vars应该跟从id和vary 。

---

您可以为默认id和vary ？ reshape::melt默认为因子和字符列ID和数字列而有所不同。

Answer 3:

我觉得可能是在你的例子是错误的。 对于从广角要长，我得到以下错误：

> reshapeasy( x.wide, "long", NULL, id="surveyNum", vary="id", sep="_" )
Error in gsub(paste("[", paste(omit, collapse = "", sep = ""), "]$", sep = ""),  : 
  invalid regular expression '[]$', reason 'Missing ']''

卸下NULL可以解决该问题。 这使我问，什么是预期目的NULL ？

我还认为，该功能会，如果它产生改进的time由默认的变量，如果没有明确由用户指定（如在完成reshape()

参见，例如，从基部以下reshpae()

> head(reshape(x.wide, direction="long", idvar=1, varying=2:13, sep="_"))
    surveyNum time pio caremgmt prev price
1.1         1    1   2        2    1     2
2.1         2    1   2        1    2     1
3.1         3    1   1        1    2     2
4.1         4    1   2        2    1     5
5.1         5    1   1        1    1     3
6.1         6    1   1        2    2     4

如果我熟悉这一点，我看你的函数负责“变”对我来说，我可能会尝试：

> head(reshapeasy( x.wide, "long", id="surveyNum", sep="_" ))
Error in `row.names<-.data.frame`(`*tmp*`, value = paste(d[, idvar], times[1L],  : 
  duplicate 'row.names' are not allowed
In addition: Warning message:
non-unique value when setting 'row.names': ‘1.1’

但是，这并不是一个非常有用的错误。 也许包括自定义错误消息可能是你最后的功能非常有用。

允许用户设置而异NULL ，因为你在你目前的版本功能都做了，也似乎并不高明给我。 这会产生这样的输出：

> head(reshapeasy( x.wide, "long", id="surveyNum", NULL, sep="_" ))
    surveyNum pio caremgmt prev price
1.1         1   2        2    1     2
2.1         2   2        1    2     1
3.1         3   1        1    2     2
4.1         4   2        2    1     5
5.1         5   1        1    1     3
6.1         6   1        2    2     4

这种输出的问题是，如果我需要重塑回广，我不能轻易地做到这一点。 因此，我认为，产生的是固定重塑的默认选项time变量，但让这可能是一个非常有用的功能，用户覆盖。

Answer 4:

也许对于那些谁是懒惰，不喜欢输入变量名，你可以添加以下到您的函数的开始：

  if (is.numeric(id) == 1) {
    id = colnames(data)[id]
  } else if (is.numeric(id) == 0) {
    id = id
  }

  if (is.numeric(vary) == 1) {
    vary = colnames(data)[vary]
  } else if (is.numeric(vary) == 0) {
    vary = vary
  }

然后，用你的例子之后，你可以使用以下简写：

reshapeasy(x.wide, direction="long", id=1, sep="_", vary="id")
reshapeasy(x.long, direction="wide", id=6, vary=1)

（我知道，它可能不是很好的做法，因为代码可能会不易阅读以下易于理解被人以后，但它频繁发生。）

文章来源: Wrapping base R reshape for ease-of-use