首页 > 解决方案 > 如何根据每行中的值对数据进行透视

问题描述

假设我们有一个这样的数据框:

> dput(data)
structure(list(Location = structure(1:18, .Label = c("a", "b", 
"c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", 
"p", "q", "r"), class = "factor"), C1 = c(7L, NA, 3L, 7L, NA, 
NA, 2L, 7L, NA, NA, NA, NA, 2L, NA, NA, NA, NA, NA), C2 = c(NA, 
8L, 1L, 1L, NA, 9L, 1L, 1L, NA, 1L, NA, 4L, 1L, NA, NA, NA, NA, 
1L), C3 = c(3L, 1L, 7L, NA, NA, NA, 7L, 2L, 5L, 4L, 9L, 10L, 
3L, 2L, 1L, 7L, NA, NA), C4 = c(NA, 2L, NA, 2L, 2L, 1L, 1L, 8L, 
8L, 5L, 6L, 15L, 15L, 5L, 5L, 2L, 15L, NA), C5 = c(NA, NA, NA, 
NA, 8L, NA, 2L, NA, 4L, 10L, 3L, 3L, 1L, NA, NA, 3L, NA, 8L)), class = "data.frame", row.names = c(NA, 
-18L))

记录数据的方式,我们有一个Location列,它代表一个已知的分组变量 levels a:r。然后我们有列C1:C5,它们本身代表 5 个集群,每个集群的样本Location根据某个任意变量进行分类。因此,每列的总和说明了每列有多少样本Location。例如,Location == a有 10 个样本,其中 7 个被分类为C1,3 个被分类为C3

我想创建一个列联表来执行卡方独立性检验,以查看Location集群分配是否独立。当数据以这种格式记录时,我们如何重塑数据来做到这一点?

更新:除非有一种更简单的方法可以根据每行中的值(可以直接对其执行卡方检验)从当前格式获取列联表,否则我希望我们必须将其转换为 tidy格式,其中每个原始样本有两列LocationCluster一个观察值,因此输出如下所示:

#there would be 10 observations for location a, 11 observations for b, and so on
Location   Cluster
a           C1
a           C1
a           C1
a           C1
a           C1
a           C1
a           C1
a           C3
a           C3
a           C3
b           C2
b           C2
b           C2
b           C2
b           C2
b           C2
b           C2
b           C2
b           C3
b           C4
b           C4
....

由此我们可以制作一个列联表并执行卡方检验

标签: rdplyrchi-squared

解决方案


我们可以重塑为“长”格式并用于uncount复制行

library(dplyr)
library(tidyr)
data %>%
   pivot_longer(cols = -Location, names_to = 'Cluster', values_drop_na = TRUE) %>%
   uncount(value)
# A tibble: 251 x 2
#   Location Cluster
#   <fct>    <chr>  
# 1 a        C1     
# 2 a        C1     
# 3 a        C1     
# 4 a        C1     
# 5 a        C1     
# 6 a        C1     
# 7 a        C1     
# 8 a        C3     
# 9 a        C3     
#10 a        C3     
# … with 241 more rows

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