我喜欢reshape2包,因为它让生活变得如此简单。通常情况下,Hadley在他以前的包中进行了改进,使代码能够简化,运行速度更快。我想我应该试一试tidyr,从我读到的内容来看,我认为gather
与reshape2中的melt
非常相似。但在阅读文档后,我无法让gather
执行与melt
相同的任务。
数据视图
下面是数据视图(post结束时dput
格式的实际数据):
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
这是我在gather
中尝试的melt
形式的代码。怎样才能让gather
做和melt
一样的事情呢?
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))
所需的输出
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
Data
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")
发布于 2014-10-24 03:54:32
您的gather
行应该如下所示:
dat %>% gather(variable, date, -teacher, -pd)
这表示“收集除teacher
和pd
之外的所有变量,将新键列称为'variable‘,将新值列称为’date‘。”
作为解释,请注意help(gather)
页面中的以下内容:
...: 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.
因为这是一个省略号,所以要收集的列的规范是作为单独的(裸名称)参数给出的。我们希望收集除teacher
和pd
之外的所有列,因此我们使用-
。
发布于 2019-09-24 21:40:19
在tidyr 1.0.0中,这项任务是通过更灵活的pivot_longer()
完成的。
等价的语法是
library(tidyr)
dat %>% pivot_longer(cols = -c(teacher, pd), names_to = "variable", values_to = "date")
相应地,它表示“透视除teacher
和pd
之外的所有内容,将新变量列称为" variable”,将新值列称为"date“。
请注意,长数据首先按透视后的前一个数据帧的列的顺序返回,而不像gather
那样按新变量列的顺序返回。要重新排列生成的tibble,请使用dplyr::arrange()
。
发布于 2022-01-01 07:52:38
我的解决方案
dat%>%
gather(!c(teacher,pd),key=variable,value=date)
https://stackoverflow.com/questions/26536251
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