如何在R语言中替换dataframe指定行的数值？

原始数据 #

df = data.frame(row.names = c("IL6", "AS-1", "KPMG", "DS55", "CD44", "Type"), 
                Class1 = c(1, 19, 18, 11, 2, "ACC"), 
                Class2 = c(3, 12, 20, 9, 12, "ACC"), 
                Class3 = c(6, 5, 9, 16, 9, "BRCA"), 
                Class4 = c(3, 9, 18, 15, 14, "BRCA"), 
                Class5 = c(4, 20, 19, 19, 16, "BRCA"), 
                Class6 = c(3, 12, 20, 9, 12, "SRCA"), 
                Class7 = c(6, 5, 9, 16, 9, "SRCA"), 
                Class8 = c(3, 9, 18, 15, 14, "GBM"))

需求 #

根据映射规则 ACC=1, BRCA=2, SRCA=3, GBM=4，替换行名为Type的所有列值。

报错情况 #

尝试以下代码时出现错误：

# mapping values
mapping <- c("ACC" = "1", "BRCA" = "2", "SRCA" = "3", "GBM" = "4")

# Replace the values in the 'Type' row
df["Type", ] <- mapping[df["Type", ]]

错误信息：

Error in mapping[df["Type", ]]: invalid subscript type 'list'

问题原因 #

df["Type", ]返回的是1行的DataFrame，属于列表（list）类型，而向量mapping的索引只能接受原子向量（如字符向量），因此触发类型不匹配的错误。

解决方案 #

将df["Type", ]转换为字符向量后再进行映射替换，有两种常用方式：

方法1：使用as.character()转换

mapping <- c("ACC" = "1", "BRCA" = "2", "SRCA" = "3", "GBM" = "4")
df["Type", ] <- mapping[as.character(df["Type", ])]

方法2：使用unlist()转换

mapping <- c("ACC" = "1", "BRCA" = "2", "SRCA" = "3", "GBM" = "4")
df["Type", ] <- mapping[unlist(df["Type", ])]

验证结果 #

执行后得到期望的输出：

output <- data.frame(row.names = c("IL6", "AS-1", "KPMG", "DS55", "CD44", "Type"), 
                     Class1 = c(1, 19, 18, 11, 2, "1"), 
                     Class2 = c(3, 12, 20, 9, 12, "1"), 
                     Class3 = c(6, 5, 9, 16, 9, "2"), 
                     Class4 = c(3, 9, 18, 15, 14, "2"), 
                     Class5 = c(4, 20, 19, 19, 16, "2"), 
                     Class6 = c(3, 12, 20, 9, 12, "3"), 
                     Class7 = c(6, 5, 9, 16, 9, "3"), 
                     Class8 = c(3, 9, 18, 15, 14, "4"))

内容的提问来源于stack exchange，提问作者nicholaspooran