69

我喜欢reshape2包，因为它让生活变得如此轻松。通常，Hadley 会在他以前的包中进行改进，以实现流线型、更快运行的代码。我想我会给tidyr一个旋转，从我读到的内容我认为与reshape2gather非常相似。但是在阅读了文档之后，我无法完成相同的任务。meltgathermelt

数据视图

dput这是数据的视图（帖子末尾的实际数据）：

  teacher yr1.baseline     pd yr1.lesson1 yr1.lesson2 yr2.lesson1 yr2.lesson2 yr2.lesson3
1       3      1/13/09 2/5/09      3/6/09     4/27/09     10/7/09    11/18/09      3/4/10
2       7      1/15/09 2/5/09      3/3/09      5/5/09    10/16/09    11/18/09      3/4/10
3       8      1/27/09 2/5/09      3/3/09     4/27/09     10/7/09    11/18/09      3/5/10

代码

这是melt时尚的代码，我在gather. 我怎样才能gather做同样的事情melt？ 

library(reshape2); library(dplyr); library(tidyr)

dat %>% 
   melt(id=c("teacher", "pd"), value.name="date") 

dat %>% 
   gather(key=c(teacher, pd), value=date, -c(teacher, pd))

期望的输出

   teacher     pd     variable     date
1        3 2/5/09 yr1.baseline  1/13/09
2        7 2/5/09 yr1.baseline  1/15/09
3        8 2/5/09 yr1.baseline  1/27/09
4        3 2/5/09  yr1.lesson1   3/6/09
5        7 2/5/09  yr1.lesson1   3/3/09
6        8 2/5/09  yr1.lesson1   3/3/09
7        3 2/5/09  yr1.lesson2  4/27/09
8        7 2/5/09  yr1.lesson2   5/5/09
9        8 2/5/09  yr1.lesson2  4/27/09
10       3 2/5/09  yr2.lesson1  10/7/09
11       7 2/5/09  yr2.lesson1 10/16/09
12       8 2/5/09  yr2.lesson1  10/7/09
13       3 2/5/09  yr2.lesson2 11/18/09
14       7 2/5/09  yr2.lesson2 11/18/09
15       8 2/5/09  yr2.lesson2 11/18/09
16       3 2/5/09  yr2.lesson3   3/4/10
17       7 2/5/09  yr2.lesson3   3/4/10
18       8 2/5/09  yr2.lesson3   3/5/10

数据

dat <- structure(list(teacher = structure(1:3, .Label = c("3", "7", 
    "8"), class = "factor"), yr1.baseline = structure(1:3, .Label = c("1/13/09", 
    "1/15/09", "1/27/09"), class = "factor"), pd = structure(c(1L, 
    1L, 1L), .Label = "2/5/09", class = "factor"), yr1.lesson1 = structure(c(2L, 
    1L, 1L), .Label = c("3/3/09", "3/6/09"), class = "factor"), yr1.lesson2 = structure(c(1L, 
    2L, 1L), .Label = c("4/27/09", "5/5/09"), class = "factor"), 
        yr2.lesson1 = structure(c(2L, 1L, 2L), .Label = c("10/16/09", 
        "10/7/09"), class = "factor"), yr2.lesson2 = structure(c(1L, 
        1L, 1L), .Label = "11/18/09", class = "factor"), yr2.lesson3 = structure(c(1L, 
        1L, 2L), .Label = c("3/4/10", "3/5/10"), class = "factor")), .Names = c("teacher", 
    "yr1.baseline", "pd", "yr1.lesson1", "yr1.lesson2", "yr2.lesson1", 
    "yr2.lesson2", "yr2.lesson3"), row.names = c(NA, -3L), class = "data.frame")
4

3 回答 3

89

您的gather行应如下所示：

dat %>% gather(variable, date, -teacher, -pd)

This says "Gather all variables except teacher and pd, calling the new key column 'variable' and the new value column 'date'."

As an explanation, note the following from the help(gather) page:

 ...: Specification of columns to gather. Use bare variable names.
      Select all variables between x and z with ‘x:z’, exclude y
      with ‘-y’. For more options, see the select documentation.

Since this is an ellipsis, the specification of columns to gather is given as separate (bare name) arguments. We wish to gather all columns except teacher and pd, so we use -.

于 2014-10-23T19:54:31.560 回答
8

In tidyr 1.0.0 this task is accomplished with the more flexible pivot_longer().

The equivalent syntax would be

library(tidyr)
dat %>% pivot_longer(cols = -c(teacher, pd), names_to = "variable", values_to = "date")

which says, correspondingly, "pivot everything longer except teacher and pd, calling the new variable column "variable" and the new value column "date".

Note that the long data comes back in order firstly of the columns of the previous data frame that were pivoted, unlike from gather, which came back in the order of the new variable column. To rearrange the resultant tibble, use dplyr::arrange().

于 2019-09-24T13:40:18.597 回答
0

My solution

    dat%>%
    gather(!c(teacher,pd),key=variable,value=date)
于 2022-01-01T07:52:37.660 回答