如何基于pos列匹配，用ifelse多选项比较R数据框列？

首先，先确保两个数据框按pos列精准匹配，避免因行顺序不一致导致的错误，我们可以先将df2的ref列合并到df1中：

# 合并df1和df2的ref列，按pos匹配
df_merged <- merge(df1, df2[, c("pos", "ref")], by = "pos", all.x = TRUE)

接下来分别实现两种需求输出：

形式一：将df1中的'-'替换为2 #

df3_form1 <- df_merged
# 遍历S01到S06列
for(i in 2:(ncol(df1))){
  df3_form1[[i]] <- ifelse(df_merged[[i]] == "-", 2, 
                           ifelse(df_merged[[i]] == df_merged$ref, 1, 0))
}
# 移除合并引入的ref列，还原原df1的列结构
df3_form1 <- df3_form1[, colnames(df1)]

# 查看结果
df3_form1

输出结果：

pos S01 S02 S03 S04 S05 S06
1    1   1   1   2   0   0   0
2    2   1   1   2   1   1   1
3    3   1   1   2   1   1   1
4    4   1   1   0   1   1   1
5    5   1   1   0   1   1   1
6    6   1   1   0   1   1   1
7    7   1   1   0   1   1   1
8    9   1   1   0   1   1   1
9   12   1   1   1   1   1   1
10  15   1   1   1   1   1   1

形式二：保留df1中的'-' #

注意：此形式下结果列会混合字符与数值，需将列转为字符类型：

df3_form2 <- df_merged
# 遍历S01到S06列
for(i in 2:(ncol(df1))){
  df3_form2[[i]] <- ifelse(df_merged[[i]] == "-", "-", 
                           ifelse(df_merged[[i]] == df_merged$ref, "1", "0"))
}
# 移除合并引入的ref列，还原原df1的列结构
df3_form2 <- df3_form2[, colnames(df1)]

# 查看结果
df3_form2

输出结果：

pos S01 S02 S03 S04 S05 S06
1    1   1   1   -   0   0   0
2    2   1   1   -   1   1   1
3    3   1   1   -   1   1   1
4    4   1   1   0   1   1   1
5    5   1   1   0   1   1   1
6    6   1   1   0   1   1   1
7    7   1   1   0   1   1   1
8    9   1   1   0   1   1   1
9   12   1   1   1   1   1   1
10  15   1   1   1   1   1   1

替代方案（用dplyr简化逻辑） #

如果已安装dplyr包，可用case_when让逻辑更清晰：

# 形式一的case_when版本
df3_form1[[i]] <- case_when(
  df_merged[[i]] == "-" ~ 2,
  df_merged[[i]] == df_merged$ref ~ 1,
  TRUE ~ 0
)

# 形式二的case_when版本
df3_form2[[i]] <- case_when(
  df_merged[[i]] == "-" ~ "-",
  df_merged[[i]] == df_merged$ref ~ "1",
  TRUE ~ "0"
)

内容的提问来源于stack exchange，提问作者abraham