Question

我有一个数据集，其中不同的行具有不同的元素组合，我想拉出具有相同元素组合的行组。对于此示例数据集：

id <- c("A", "B", "C", "D")
X1 <- c(NA,NA,NA,"X1")
X2 <- c(NA,NA,"X2","X2")
X3 <- c("X3","X3","X3","X3")
X4 <- c("X4", "X4", "X4", "X4")
df <- data.frame(id,X1,X2,X3,X4)

> df
   id   X1   X2 X3 X4
   1  A <NA> <NA> X3 X4
   2  B <NA> <NA> X3 X4
   3  C <NA>   X2 X3 X4
   4  D   X1   X2 X3 X4

我希望能够退出

哪些ID有X1＆amp; X2＆amp; X3＆amp; X4（D）
哪个ids有！X1＆amp; X2＆amp; X3＆amp; X4（C）
哪个ids有！X1＆amp; ！X2＆amp; X3＆amp; X4（A和B）。

我尝试将数据框拆分为列表并删除空单元格，以便每个id在列表中获得自己的data.frame：

df.list <- split(df, seq(nrow(df)))
dfComplete.list <- lapply(df.list, function(remNA) remNA[,colSums(is.na(remNA)) < nrow(remNA)])

离开了我

> dfComplete.list
$`1`
  id X3 X4
1  1 X3 X4

$`2`
  id X3 X4
2  2 X3 X4

$`3`
  id X2 X3 X4
3  3 X2 X3 X4

$`4`
  id X1 X2 X3 X4
4  4 X1 X2 X3 X4

我很难过从这里出发去哪里。有没有办法根据它们共有的元素/列对列表中的数据帧进行分组？

我真正使用的真实数据集实际上有元素/列X7到X17，每个id有1到4个元素，所以理想的解决方案是能够识别我所存在的元素的所有组合数据

最后，在我将数据重新设置为上述格式之前，我的数据最初是以下格式，以防万一从原始格式中找到解决方案的方法更简单：

id <- c("A", "A", "B", "B", "C", "C", "C", "D", "D", "D", "D")
elements <- c("X3", "X4", "X3", "X4", "X2", "X3", "X4", "X1", "X2", "X3", "X4")
dataLong <- data.frame(id, elements)

> dataLong
  id elements
1   A       X3
2   A       X4
3   B       X3
4   B       X4
5   C       X2
6   C       X3
7   C       X4
8   D       X1
9   D       X2
10  D       X3
11  D       X4

提前感谢您的帮助！

Answer 1

reshape2::dcast函数可以帮助将数据从长格式转换为OP期望的格式。

#Data
id <- c("A", "A", "B", "B", "C", "C", "C", "D", "D", "D", "D")
elements <- c("X3", "X4", "X3", "X4", "X2", "X3", "X4", "X1", "X2", "X3", "X4")
dataLong <- data.frame(id, elements, stringsAsFactors = FALSE)

library(reshape2)

#Use dcast to get the result
dataLong %>% dcast(id~elements)
#   id   X1   X2 X3 X4
# 1  A <NA> <NA> X3 X4
# 2  B <NA> <NA> X3 X4
# 3  C <NA>   X2 X3 X4
# 4  D   X1   X2 X3 X4

Answer 2

我知道你想要计算独特的组合。我就是这样做的

library(dplyr)
library(tidyr)

dataLong %>% mutate(value=1) %>% 
  spread(elements, value) %>% 
  select(-id) %>% 
  group_by_all() %>% 
  summarise(count=n()) %>% ungroup()
#> # A tibble: 3 x 5
#>      X1    X2    X3    X4 count
#>   <dbl> <dbl> <dbl> <dbl> <int>
#> 1     1     1     1     1     1
#> 2    NA     1     1     1     1
#> 3    NA    NA     1     1     2

Answer 3

您可以使用tidyverse！使用arrange()有点多余，但我想向您展示该选项，因为它会安排您的数据框以反映您感兴趣的分组（您可以将其视为一种嵌套排序）。这可能就是您所需要的一切。

如果您想要实际计数，以及一列可以告诉您哪些ID对应于哪些组合，那么只需运行下面的完整代码即可。请注意，您必须在完整代码中添加所有变量（X7:X17）。在声明数据框时，您还希望使用stringsAsFactors = FALSE，这是一般的好习惯。

# Your example dataframe. Make sure to set stringsAsFactors = FALSE
id <- c("A", "B", "C", "D")
X1 <- c(NA,NA,NA,"X1")
X2 <- c(NA,NA,"X2","X2")
X3 <- c("X3","X3","X3","X3")
X4 <- c("X4", "X4", "X4", "X4")
df <- data.frame(id,X1,X2,X3,X4, stringsAsFactors = FALSE)

# We group rows by all unique combinations and then collapse those rows, 
# while recording which ids belong to which grouping, and how many there are 
# in each.
library(tidyverse)
ndf <- arrange(df, X1,X2,X3,X4) %>%
       group_by(X1,X2,X3,X4) %>%
       summarise(num = n(), id = paste(id, collapse=","))

# Output:
# A tibble: 3 x 6
# Groups:   X1, X2, X3 [?]
  X1    X2    X3    X4      num id   
  <chr> <chr> <chr> <chr> <int> <chr>
1 X1    X2    X3    X4        1 D    
2 <NA>  X2    X3    X4        1 C    
3 <NA>  <NA>  X3    X4        2 A,B

使用R按公共元素对数据帧中的行进行分组

3 个答案: