比较分组数据框中的行时返回列差异

Question

我想对每个组进行成对比较，并返回不匹配的行以及哪些列不同。下面是一个示例数据集，用于解释我的实际数据会有更多行和列的问题。

data=structure(list(ID = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 
13, 14, 15, 16, 17, 18, 19, 20), Common_1 = c("A", "A", "A", 
"A", "A", "A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B", 
"B", "B", "B", "B"), Common_2 = c("C", "C", "C", "C", "C", "D", 
"D", "D", "D", "D", "C", "C", "C", "C", "C", "D", "D", "D", "D", 
"D"), Common_3 = c("X", "X", "X", "X", "X", "X", "X", "X", "X", 
"X", "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y", "Y"), G = c(0, 
1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0), var_1 = c(1, 
3, 3, 3, 3, 1, 3, 2, 4, 3, 5, 5, 3, 4, 5, 1, 3, 5, 1, 4), var_2 = c("lev1", 
"lev1", "lev2", "lev2", "lev1", "lev2", "lev2", "lev1", "lev1", 
"lev2", "lev2", "lev2", "lev2", "lev1", "lev1", "lev1", "lev1", 
"lev1", "lev2", "lev2"), var_3 = c("on", "on", "on", "off", "off", 
"on", "on", "on", "off", "off", "on", "on", "on", "off", "off", 
"on", "on", "on", "off", "off"), var_4 = c("up", "up", "down", 
"down", "up", "down", "up", "down", "up", "up", "up", "up", "down", 
"down", "up", "up", "up", "up", "down", "down")), row.names = c(NA, 
-20L), class = c("tbl_df", "tbl", "data.frame"))

ID 是唯一标识符，Common_1,Common_2,Common_3 是分组变量，G 是我要比较的组，最后是其余列 var_1:var_4是确定差异的列。该过程是将G=0 中的每一行与G=1 进行比较，如果任何var 列存在差异，则返回不匹配的ID 组合以及哪些列不同。

这是Common_1=A、Common_2=C、Common_3=X 的所需结果，ID 用于行 G=0、所有分组变量、ID 用于G=1 不匹配和指示变量显示哪些列不同。

results=structure(list(ID = c(1, 1, 3, 3, 4, 4), Common_1 = c("A", "A", 
"A", "A", "A", "A"), Common_2 = c("C", "C", "C", "C", "C", "C"
), Common_3 = c("X", "X", "X", "X", "X", "X"), G = c(0, 0, 0, 
0, 0, 0), var_1 = c(1, 1, 0, 0, 0, 0), var_2 = c(0, 0, 1, 1, 
1, 1), var_3 = c(0, 1, 0, 1, 1, 0), var_4 = c(0, 0, 1, 1, 1, 
1), ID_diff = c(2, 5, 2, 5, 2, 5)), row.names = c(NA, -6L), class = c("tbl_df", 
"tbl", "data.frame"))

更新：添加结果说明

我正在对G=0 和G=1 进行成对比较。前两行结果的派生如下： 同组Common_1=A,Common_2=C,Common_3=X

现在比较 ID=1 和 ID=2

var_1 不同，因此 1 放在 var_1 列中，其余为 0。 ID_diff=2 因为那是与ID=1 不同的ID

比较 ID=1 和 ID=5

var_1 和 var_3 是不同的，所以每列都放置一个 1，其余为 0。 ID_diff=5 因为那是不同于 ID=1 的 ID

我尝试编写一个函数来循环使用 G=0 遍历每个案例，并与 G=1 但在提取不匹配信息时遇到困难的每个案例进行比较，感谢任何帮助。

Ronak Shah 解决方案的结果有效，但我无法正确显示结果。

> var_col <- grep('^var', names(data))
> 
> apply_fun <- function(tmp) {
+     df1 <- subset(tmp, G == 0)
+     df2 <- subset(tmp, G == 1)
+     lapply(seq(nrow(df1)), function(x) {
+         df3 <- df1[rep(x, nrow(df2)), ]
+         df3$ID_diff <- df2$ID
+         df3[var_col] <- +(df1[rep(x, nrow(df2)), var_col] != df2[var_col])
+         df3
+     })
+ }
> 
> 
> library(dplyr)
> data %>%
+     group_by(across(starts_with('Common'))) %>%
+     summarise(data = apply_fun(cur_data_all())) %>%
+     ungroup %>%
+     select(data) %>%
+     tidyr::unnest(data)
`summarise()` regrouping output by 'Common_1', 'Common_2', 'Common_3' (override with `.groups` argument)
# A tibble: 22 x 10
      ID Common_1 Common_2 Common_3     G var_1[,1]  [,2]  [,3]  [,4] var_2[,1]  [,2]  [,3]  [,4] var_3[,1]  [,2]  [,3]  [,4] var_4[,1]  [,2]
   <dbl> <chr>    <chr>    <chr>    <dbl>     <int> <int> <int> <int>     <int> <int> <int> <int>     <int> <int> <int> <int>     <int> <int>
 1     1 A        C        X            0         1     0     0     0         1     0     0     0         1     0     0     0         1     0
 2     1 A        C        X            0         1     0     1     0         1     0     1     0         1     0     1     0         1     0
 3     3 A        C        X            0         0     1     0     1         0     1     0     1         0     1     0     1         0     1
 4     3 A        C        X            0         0     1     1     1         0     1     1     1         0     1     1     1         0     1
 5     4 A        C        X            0         0     1     1     1         0     1     1     1         0     1     1     1         0     1
 6     4 A        C        X            0         0     1     0     1         0     1     0     1         0     1     0     1         0     1
 7     7 A        D        X            0         1     0     0     1         1     0     0     1         1     0     0     1         1     0
 8     8 A        D        X            0         1     1     0     0         1     1     0     0         1     1     0     0         1     1
 9     9 A        D        X            0         1     1     1     1         1     1     1     1         1     1     1     1         1     1
10    10 A        D        X            0         1     0     1     1         1     0     1     1         1     0     1     1         1     0
# ... with 12 more rows, and 3 more variables: [,3] <int>, [,4] <int>, ID_diff <dbl>

Answer 1

试试这个：

我们创建一个函数，将G == 0 的每一行与G == 1 的每一行进行比较，并根据比较将var_col 中的值更改为1/0。

var_col <- grep('^var', names(data))

apply_fun <- function(tmp) {
  df1 <- subset(tmp, G == 0)
  df2 <- subset(tmp, G == 1)
  lapply(seq(nrow(df1)), function(x) {
    df3 <- df1[rep(x, nrow(df2)), ]
    df3$ID_diff <- df2$ID
    df3[var_col] <- +(df1[rep(x, nrow(df2)), var_col] != df2[var_col])
    df3
  })
}

按组应用数据：

library(dplyr)
data %>%
  group_by(across(starts_with('Common'))) %>%
  summarise(data = apply_fun(cur_data_all())) %>%
  ungroup %>%
  select(data) %>%
  tidyr::unnest(data)

# A tibble: 22 x 10
#      ID Common_1 Common_2 Common_3     G var_1 var_2 var_3 var_4 ID_diff
#   <dbl> <chr>    <chr>    <chr>    <dbl> <int> <int> <int> <int>   <dbl>
# 1     1 A        C        X            0     1     0     0     0       2
# 2     1 A        C        X            0     1     0     1     0       5
# 3     3 A        C        X            0     0     1     0     1       2
# 4     3 A        C        X            0     0     1     1     1       5
# 5     4 A        C        X            0     0     1     1     1       2
# 6     4 A        C        X            0     0     1     0     1       5
# 7     7 A        D        X            0     1     0     0     1       6
# 8     8 A        D        X            0     1     1     0     0       6
# 9     9 A        D        X            0     1     1     1     1       6
#10    10 A        D        X            0     1     0     1     1       6
# … with 12 more rows

cur_data_all() 和across 需要最新安装的dplyr。我的packageVersion('dplyr') 是1.0.1’。