统计两个嵌套列表对应元素的公共子元素数量

需求说明 #

现有两个嵌套列表list1和list2，二者的顶级元素名称均为ENSG编号。需要针对每个ENSG编号，统计两个列表中对应子元素（格式为chrXX_XXXX_X_X_b38的列名）的公共数量，最终输出包含两列的数据框：第一列为ENSG编号，第二列为公共子元素的数量。

输入列表 #

list1

list1 <- list(ENSG00000040608 = structure(list(chr22_20230714_G_A_b38 = 0.0000953181301665087, 
                             chr22_20230737_G_A_b38 = -0.00124036704551427, 
                             chr22_20231229_T_A_b38 = 0.000808061558738542, 
                             chr22_20231474_G_A_b38 = 0.000387528601423933, 
                             chr22_20231667_C_G_b38 = -0.000120624028990859), 
                                          row.names = c(NA, -1L), 
                                          class = c("tbl_df", "tbl", "data.frame")), 
              ENSG00000020766 = structure(list(chr22_47157062_G_A_b38 = 0.00000909931572319958, 
                             chr22_47157212_G_A_b38 = -0.000124084106569373, 
                             chr22_47157394_C_G_b38 = -0.0000752774417069946, 
                             chr22_47157559_G_A_b38 = 0.0000808446315377557, 
                             chr22_47157607_T_C_b38 = 0.000237979025556899), 
                                          row.names = c(NA, -1L), 
                                          class = c("tbl_df", "tbl", "data.frame")))

list2

list2 <- list(ENSG00000040608 = structure(list(chr22_20230714_G_A_b38 = c(1L, 1L, 1L, 2L, 1L), 
                             chr22_20230737_G_A_b38 = c(0L, 0L, 0L, 0L, 0L), 
                             chr22_20231229_T_A_b38 = c(1L, 0L, 1L, 0L, 1L), 
                             chr22_20231555_A_T_b38 = c(0L,1L, 0L, 0L, 0L), 
                             chr22_20231667_C_G_b38 = c(1L, 1L, 1L, 2L, 1L)), 
                                          row.names = c(NA, -5L), 
                                          class = c("tbl_df", "tbl", "data.frame")), 
              ENSG00000020766 = structure(list(chr22_47157062_G_A_b38 = c(0L,1L, 0L, 0L, 0L), 
                            chr22_47157212_G_A_b38 = c(0L, 0L, 1L, 1L, 2L), 
                            chr22_47157394_C_G_b38 = c(0L, 1L, 1L, 1L, 2L), 
                            chr22_47207559_T_C_b38 = c(0L, 1L, 0L, 0L, 0L)), 
                                          row.names = c(NA, -4L), 
                                          class = c("tbl_df", "tbl", "data.frame")))

解决方案 #

方法1：基础R实现 #

# 提取每个ENSG对应的子元素列名
list1_cols <- lapply(list1, colnames)
list2_cols <- lapply(list2, colnames)

# 计算每个ENSG的公共子元素数量
common_counts <- mapply(function(x, y) length(intersect(x, y)), list1_cols, list2_cols)

# 转换为目标数据框
result_df <- data.frame(
  ENSGs = names(common_counts),
  Common_Count = as.numeric(common_counts),
  row.names = NULL
)

# 输出结果
print(result_df)

方法2：tidyverse实现 #

library(tidyverse)

result_df <- tibble(ENSGs = names(list1)) %>%
  mutate(
    # 获取list1中每个ENSG的子元素列名
    list1_subs = map(list1, colnames),
    # 获取list2中对应ENSG的子元素列名
    list2_subs = map(ENSGs, ~colnames(list2[[.x]])),
    # 统计公共子元素数量
    Common_Count = map2_int(list1_subs, list2_subs, ~length(intersect(.x, .y)))
  ) %>%
  select(ENSGs, Common_Count)

# 输出结果
print(result_df)

运行结果 #

两种方法都会得到如下数据框：

ENSGs Common_Count
1 ENSG00000040608            4
2 ENSG00000020766            3

逻辑说明 #

1. 每个ENSG对应的子元素是嵌套列表中数据框的列名，因此先提取每个数据框的列名；
2. 对每一组ENSG的列名集合取交集，用length()统计交集元素的数量；
3. 将结果整理成包含ENSG编号和公共数量的两列数据框。

内容的提问来源于stack exchange，提问作者Maya_Cent