Question

我有一组患者数据df，我试图在R中取消标识。

structure(list(name = structure(c(2L, 5L, 1L, 6L, 4L, 3L), .Label = c("Andrew", 
                                                                      "Jim", "Kurt", "Lester", "Mickey", "Taylor"), class = "factor"), 
               heart_rate = c(78L, 82L, 67L, 105L, 85L, 94L), age = c(35L, 
                                                                      23L, 43L, 52L, 33L, 45L), partner = structure(c(5L, 2L, 6L, 
                                                                                                                      1L, 3L, 4L), .Label = c("Andrew", "Jim ", "Kurt ", "Lester ", 
                                                                                                                                              "Mickey ", "Taylor "), class = "factor")), class = "data.frame", row.names = c(NA, 
                                                                                                                                                                                                                             -6L))

我想基于名为name的对象的partner列替换id和key列的名称

structure(list(name = structure(c(2L, 5L, 1L, 6L, 4L, 3L), .Label = c("Andrew", 
"Jim", "Kurt", "Lester", "Mickey", "Taylor"), class = "factor"), 
    id = structure(c(2L, 5L, 1L, 6L, 4L, 3L), .Label = c("A3", 
    "J9", "K5", "L4", "M4", "T7"), class = "factor")), class = "data.frame", row.names = c(NA, 
-6L))

我可以使用此代码取消标识name列

df[["name"]] <- key[ match(df[['name']], key[['name']] ) , 'id']

但是，当我尝试使用此代码取消标识partner列

df[["partner"]] <- key[ match(df[['partner']], key[['name']] ) , 'id']

我的数据框看起来像这样

structure(list(name = structure(c(2L, 5L, 1L, 6L, 4L, 3L), .Label = c("A3", 
"J9", "K5", "L4", "M4", "T7"), class = "factor"), heart_rate = c(78L, 
82L, 67L, 105L, 85L, 94L), age = c(35L, 23L, 43L, 52L, 33L, 45L
), partner = structure(c(NA, NA, NA, 1L, NA, NA), .Label = c("A3", 
"J9", "K5", "L4", "M4", "T7"), class = "factor")), row.names = c(NA, 
-6L), class = "data.frame")

有人有什么建议吗？可能只适用于一行中的数据集的所有列的方法的加分点和代码的说明，将不胜感激。

Answer 1

问题在于，在partner的{{1}}列中，大多数单词后面都有一个空格：

df

这意味着.Label = c("Andrew", "Jim ", "Kurt ", "Lester ", "Mickey ", "Taylor ")除了名称“ Andrew”（它会正确返回该索引）之外，找不到完全匹配的内容。

解决此问题的方法是使用

从match()列中删除空格

partner

然后您的代码可以正常工作：

df$partner = trimws(df$partner)

R用来自另一个对象的ID将多列中的名称替换为ID

1 个答案: