Question

我有两个数据帧：

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

newstate = c("TX", "LA")
first =c("Jim","Mick")
df2 <- data.frame(first,newstate)
df2
  first newstate
1   Jim       TX
2  Mick       LA

我正在使用qdaptools查找功能：

library(qdaptools)
df1$match <- lookup(df1$first, df2[, 1:2])
> df1
  first state match
1   Jim    CA    TX
2  Mick    WA    LA
3  Paul    OR  <NA>
4   Ron    AZ  <NA>

有没有办法忽略nomatch或nomatch按原样返回现有变量？这将是期望结果的示例：

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

Answer 1

使用dplyr包，您可以使用coalesce()完成工作。在这里，我将因子转换为字符。如有必要，您希望将它们重新转换为因子。第一步是合并两个数据集并将因子转换为字符。然后，您使用newstate填写coalesce()中的NAs。

library(dplyr)

left_join(df1, df2, by = "first") %>%
mutate_all(funs(as.character)) %>%
mutate(newstate = coalesce(newstate, state))

#  first state newstate
#1   Jim    CA       TX
#2  Mick    WA       LA
#3  Paul    OR       OR
#4   Ron    AZ       AZ

Answer 2

一种天真的方式是用状态值替换NA！

> state = c("CA","WA","OR","AZ")
> first = c("Jim","Mick","Paul","Ron")
> df1 <- data.frame(first, state, stringsAsFactors = F)
> df1
  first state
1   Jim    CA
2  Mick    WA
3  Paul    OR
4   Ron    AZ
> 
> newstate = c("TX", "LA")
> first =c("Jim","Mick")
> df2 <- data.frame(first,newstate, stringsAsFactors = F)
> df2
  first newstate
1   Jim       TX
2  Mick       LA
> 
> df3 <- merge(df1,df2, by='first', all=TRUE)
> 
> #df3 <- as.character(df3)
> 
> df3$newstate[is.na(df3$newstate)] <- df3$state[is.na(df3$newstate)]
> 
> df3
  first state newstate
1   Jim    CA       TX
2  Mick    WA       LA
3  Paul    OR       OR
4   Ron    AZ       AZ
>

Answer 3

你也可以使用dplyer：

library(tidyverse)

df3 <- merge(x = df1, y = df2, all.x = TRUE) %>%
  mutate(state = as.character(state),
         newstate = as.character(newstate),
         newstate = if_else(is.na(newstate), state, newstate))

使用qdaptools（查找）将非匹配值返回到原样

3 个答案: