如何仅用数据框 2 匹配的值替换数据场 1 中特定列的缺失值

Question

我有两个数据框，如下所示：

structure(list(ID = c("a", "a", "c", "d", "a", "f"), a1 = c("2001", 
"2000", "2000", "1998", "2006", "2002"), value = c("100", NA, 
"200", NA, "400", "500")), class = "data.frame", row.names = c(NA, 
-6L))

ID2 <- c("x","a","c","d","n","f","g","m")
a2 <- c(2000,2000,2000,2000,2000,2000,2000,2000)
value2 <- c(100, 1000, 20000, 1000, 400,500,1,1)

  ID   a1 value
1  a 2001   100
2  a 2000  <NA>
3  c 2000   200
4  d 1998  <NA>
5  a 2006   400
6  f 2002   500

  ID2   a2 value2
1   x 2000    100
2   a 2000   1000
3   c 2000  20000
4   d 2000   1000
5   n 2000    400
6   f 2000    500
7   g 2000      1
8   m 2000      1

我想做什么：我想替换仅缺少的值列值数据框 1 的相应值来自数据框 2 的 value2 列。年份和 ID 应该匹配。我使用 left_join 尝试了以下代码；然而，这给了我错误的结果

r <- left_join(data, data2, by=c("ID"="ID2","a1"="a2"))

有什么建议吗？

预期结果

  ID   a1 value
1  a 2001   100
2  a 2000  **1000**
3  c 2000   200
4  d 1998  <NA>
5  a 2006   400
6  f 2002   500

Answer 1

这是一个 tidyverse 方法。

请注意，因为数据帧 1 中的 a1 和 value 是 character，我还将数据帧 2 的相应列更改为 character。

你需要做的是left_join他们在一起，coalescevalue和value2。

library(tidyverse)

ID2 <- c("x","a","c","d","n","f","g","m")
a2 <- c("2000","2000","2000","2000","2000","2000","2000","2000")
value2 <- c("100", "1000", "20000", "1000", "400","500","1","1")
df2 <- data.frame(ID2, a2, value2)

left_join(df1, df2, by = c("ID" = "ID2", "a1" = "a2")) %>% 
  transmute(ID = ID, a1 = a1, value = coalesce(value, value2))

  ID   a1 value
1  a 2001   100
2  a 2000  1000
3  c 2000   200
4  d 1998  <NA>
5  a 2006   400
6  f 2002   500

Answer 2

仅使用基础 R。

merge(df1, df2, by.x=c('ID', 'a1'), by.y=c('ID2', 'a2'), all.x=TRUE) |>
  transform(value=ifelse(is.na(value), value2, value), value2=NULL)
#   ID   a1 value
# 1  a 2000  1000
# 2  a 2001   100
# 3  a 2006   400
# 4  c 2000   200
# 5  d 1998    NA
# 6  f 2002   500

数据：

df1 <- structure(list(ID = c("a", "a", "c", "d", "a", "f"), a1 = c(2001L, 
2000L, 2000L, 1998L, 2006L, 2002L), value = c(100L, NA, 200L, 
NA, 400L, 500L)), class = "data.frame", row.names = c("1", "2", 
"3", "4", "5", "6"))
df2 <- structure(list(ID2 = c("x", "a", "c", "d", "n", "f", "g", "m"
), a2 = c(2000L, 2000L, 2000L, 2000L, 2000L, 2000L, 2000L, 2000L
), value2 = c(100L, 1000L, 20000L, 1000L, 400L, 500L, 1L, 1L)), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8"))

如何仅用数据框 2 匹配的值替换数据场 1 中特定列的缺失值

How to replace only missing value of specific column from datafarm 1 by value matched from data frame 2

replace

join

r

na