对比两个数据集的edta_complete列，求完整条目数及差异subject_id

需求说明 #

• 现有两个数据集df和df1，其中edta_complete列的0代表数据不完整，1代表数据完整
• 需要完成两项操作：
  1. 分别统计两个数据集中edta_complete=1的subject_id数量
  2. 对比两个数据集，找出完整条目更多的那个数据集中，与另一个数据集存在差异的subject_id

示例数据集 #

df

df <- structure (list(subject_id = c("191-5467", "191-6784", "191-3457", "191-0987", "191-1245", "191-2365"), edta_complete = c("1","0","1","1","1","0")), class = "data.frame", row.names = c (NA, -6L))

df1

df1 <- structure (list(subject_id = c("191-5467", "191-6784", "191-3457", "191-0987", "191-1245", "191-2365"), edta_complete = c("1","1","1","1","1","1")), class = "data.frame", row.names = c (NA, -6L))

1. 统计完整条目数量 #

# 统计df中edta_complete=1的条目数
df %>% filter(edta_complete == 1) %>% nrow()
# 输出：[1] 4

# 统计df1中edta_complete=1的条目数
df1 %>% filter(edta_complete == 1) %>% nrow()
# 输出：[1] 6

2. 找出差异的subject_id #

通过合并数据集对比edta_complete列的取值，筛选出df1中标记为完整但df中标记为不完整的subject_id：

# 合并两个数据集，保留subject_id和对应edta_complete值
merged_df <- merge(df, df1, by = "subject_id", suffixes = c("_df", "_df1"))

# 筛选差异条目
diff_subjects <- merged_df %>%
  filter(edta_complete_df == "0" & edta_complete_df1 == "1") %>%
  pull(subject_id)

# 输出结果
diff_subjects
# 输出：[1] "191-6784" "191-2365"

内容的提问来源于stack exchange，提问作者Thandi