r - 使用R中的match函数按原样获得nomatch返回值

Question

我有一个更大的现有数据框。对于这个较小的示例，我想根据“第一”列用newstate(df2)替换某些变量(替换状态(df1))。我的问题是，由于新数据帧(df2)中只有某些名称匹配，因此值以NA形式返回。

现有数据框:

state = c("CA","WA","OR","AZ")
first = c("Jim","Mick","Paul","Ron")
df1 <- data.frame(first, state)

      first state
    1   Jim    CA
    2  Mick    WA
    3  Paul    OR
    4   Ron    AZ

新数据框与现有数据框匹配

state = c("CA","WA")
newstate = c("TX", "LA")
first =c("Jim","Mick")
df2 <- data.frame(first, state, newstate)

  first state newstate
1   Jim    CA       TX
2  Mick    WA       LA

尝试使用匹配，但在原始数据帧中未找到df2中匹配的“第一个”变量的情况下，为“状态”返回NA。

df1$state <- df2$newstate[match(df1$first, df2$first)]

  first state
1   Jim    TX
2  Mick    LA
3  Paul  <NA>
4   Ron  <NA>

有没有办法忽略不匹配或让不匹配按原样返回现有变量？这将是预期结果的示例:Jim/Mick的状态被更新，而Paul和Ron的状态没有改变。

      first state
    1   Jim    TX
    2  Mick    LA
    3  Paul    OR
    4   Ron    AZ

Answer 1

这是你想要的吗;顺便说一句，除非您真的想使用因子，否则请在data.frame调用中使用stringsAsFactors = FALSE。请注意在match调用中使用nomatch = 0。

> state = c("CA","WA","OR","AZ")
> first = c("Jim","Mick","Paul","Ron")
> df1 <- data.frame(first, state, stringsAsFactors = FALSE)
> state = c("CA","WA")
> newstate = c("TX", "LA")
> first =c("Jim","Mick")
> df2 <- data.frame(first, state, newstate, stringsAsFactors = FALSE)
> df1
  first state
1   Jim    CA
2  Mick    WA
3  Paul    OR
4   Ron    AZ
> df2
  first state newstate
1   Jim    CA       TX
2  Mick    WA       LA
> 
> # create an index for the matches
> indx <- match(df1$first, df2$first, nomatch = 0)
> df1$state[indx != 0] <- df2$newstate[indx]
> df1
  first state
1   Jim    TX
2  Mick    LA
3  Paul    OR
4   Ron    AZ