使用其他数据填充缺失值？

Question

A <- data.frame(Item_A = c("00EF", "00EF", "00EF", "00EF", "00EF", "00FR", "00FR"),  
                Item_B = c(NA, NA, NA, NA, "JAMES RIVER", NA, NA))

B <- data.frame(Item_A = c("00EF", "00EF", "00EF", "00FR", "00FR"), 
                Item_B = c("JAMES RIVER", NA, "JAMES RIVER",
                           "RICE MIDSTREAM", "RICE MIDSTREAM"))

预计：

A <- data.frame(Item_A = c("00EF", "00EF", "00EF", "00EF", "00EF", "00FR", "00FR"),  
                Item_B = c("JAMES RIVER", "JAMES RIVER", "JAMES RIVER", 
                         "JAMES RIVER", "JAMES RIVER", "RICE MIDSTREAM", "RICE MIDSTREAM"))

B <- data.frame(Item_A = c("00EF", "00EF", "00EF", "00FR", "00FR"), 
                Item_B = c("JAMES RIVER", "JAMES RIVER", "JAMES RIVER", 
                           "RICE MIDSTREAM", "RICE MIDSTREAM"))

我必须根据 Item_A 相同的其他行的 Item_B 填写第 Item_B 项。例如，数据集A中Item_B的第一个到第四个观测需要变成"JAMES RIVER"。

能否请您提出一种方法来填充 R 中的缺失值？我尝试了很多技术，但无法得到我想要的。

Answer 1

您可以试试 tidyr 库助手 fill

library(tidyr)
A %>% 
  tidyr::fill(Item_B, .direction = "down") %>% 
  tidyr::fill(Item_B, .direction = "up")

  Item_A      Item_B
1   00FF JAMES RIVER
2   00FF JAMES RIVER
3   00FF JAMES RIVER
4   00FF JAMES RIVER
5   00FF JAMES RIVER
6   00FR JAMES RIVER
7   00FR JAMES RIVER

Answer 2

据我了解这个问题，这不是只是一个简单地在每个 data.frame 的一列中填充缺失值的练习。我相信这需要在查找或映射 table:

的帮助下填写属于 Item_A 的 Item_B 的值

library(data.table)
# create mapping table from both data.frames
map <- unique(rbindlist(list(A, B)))[!is.na(Item_B)]
# or, in case there are additional columns besides Item_A and Item_B
map <- unique(rbindlist(list(A, B))[!is.na(Item_B), .(Item_A, Item_B)])
map

   Item_A         Item_B
1:   00FF    JAMES RIVER
2:   00EF    JAMES RIVER
3:   00FR RICE MIDSTREAM

# join and replace
setDT(A)[map, on = c("Item_A"), Item_B := i.Item_B][]

   Item_A         Item_B
1:   00FF    JAMES RIVER
2:   00FF    JAMES RIVER
3:   00FF    JAMES RIVER
4:   00FF    JAMES RIVER
5:   00FF    JAMES RIVER
6:   00FR RICE MIDSTREAM
7:   00FR RICE MIDSTREAM

setDT(B)[map, on = c("Item_A"), Item_B := i.Item_B][]

   Item_A         Item_B
1:   00EF    JAMES RIVER
2:   00EF    JAMES RIVER
3:   00EF    JAMES RIVER
4:   00FR RICE MIDSTREAM
5:   00FR RICE MIDSTREAM

在连接期间，有两列名为 Item_B，一个来自第一个数据 table，A（或 B，resp.），另一个来自第二个数据tablemap。为了区分它们，i.前缀表示i.Item_B应该取自map。

Answer 3

您可以尝试创建一个字典数据框。

library(dplyr)
dictionnary <- bind_rows(A,B) %>% 
           filter(!is.na(Item_B)) %>% 
           distinct
find_name <- function(id){
  name <- dictionnary[["Item_B"]][which(dictionnary[["Item_A"]]==id)]
  return(name)
}
test_id <- c("00EF","00EF","00EF","00FR","00FR")
new_names <- sapply(test_id ,find_name )

然后您可以声明您的数据框：

New_A <- data.frame(Item_A=c("00FF","00FF","00FF","00FF","00FF","00FR","00FR"),
                Item_B=sapply(c("00FF","00FF","00FF","00FF","00FF","00FR","00FR"),find_name))

New_B <- data.frame(Item_A=c("00EF","00EF","00EF","00FR","00FR"), 
                Item_B=sapply(c("00EF","00EF","00EF","00FR","00FR"),find_name))

Answer 4

@YXCHEN 根据您的输入更新

lookup_df <- unique(rbindlist(list(A, B)))[!is.na(Item_B)] 

left_join(A %>% select(Item_A), lookup_df)
left_join(B %>% select(Item_A), lookup_df)

使用其他数据填充缺失值？

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