从另一个数据帧

Question

替换数据帧一定的价值，我有两个数据框：

df1 <- data.frame(id = c("LABEL1", "LABEL2", "LABEL3", "LABEL4", "LABEL5", "LABEL6"),matrix(1:60,6,10)) 
df1[c(4:6), c(2:4)] = NA 

df2 = data.frame(id = c("LABEL3", "LABEL4", "LABEL5", "LABEL6"),matrix(seq(100,10000, length.out = 32),4,8)) 

我想用一个密钥值=“ID”来查找DF2仅从DF1缺失的数值。这里是所需的输出： enter image description here

这里是我尝试的方法： 1. merge：但我得到X1：X3的重复列。 2.匹配：

df1[,2]= df2[,2][match(df1$id, df2$id)] 

但我会在DF1覆盖的标签3。从qdap包 3.查找：

library(qdap) 
apply(df1, 2, lookup, df2) 

相同的结果的方法2.

谢谢！

Answer 1

你可以使用tidyr整齐数据的形式工作，然后dplyr表一步的解释

结合

library(dplyr) 
library(tidyr) 

在某种程度上与管

df1 %>% 
    mutate(id = as.character(id)) %>% 
    gather(key = "col", value = "val", -id) %>% 
    left_join(df2 %>% 
       mutate(id = as.character(id)) %>% 
       gather(key = "col", value = "val", -id), 
      by =c("id", "col")) %>% 
    transmute(id, col, val = ifelse(is.na(val.x), val.y, val.x)) %>% 
    spread(col, val) %>% 
    select(id, num_range("X", 1:10)) 
#>  id  X1  X2  X3 X4 X5 X6 X7 X8 X9 X10 
#> 1 LABEL1 1.0000 7.000 13.000 19 25 31 37 43 49 55 
#> 2 LABEL2 2.0000 8.000 14.000 20 26 32 38 44 50 56 
#> 3 LABEL3 3.0000 9.000 15.000 21 27 33 39 45 51 57 
#> 4 LABEL4 419.3548 1696.774 2974.194 22 28 34 40 46 52 58 
#> 5 LABEL5 738.7097 2016.129 3293.548 23 29 35 41 47 53 59 
#> 6 LABEL6 1058.0645 2335.484 3612.903 24 30 36 42 48 54 60 

步骤

# id as character instead of factor df1 <- df1 %>% mutate(id = as.character(id)) # tidy data df1 <- df1 %>% gather(key = "col", value = "val", -id) # print result as dplyr tbl df1 %>% as.tbl() #> Source: local data frame [60 x 3] #> #> id col val #> (chr) (chr) (int) #> 1 LABEL1 X1 1 #> 2 LABEL2 X1 2 #> 3 LABEL3 X1 3 #> 4 LABEL4 X1 NA #> 5 LABEL5 X1 NA #> 6 LABEL6 X1 NA #> 7 LABEL1 X2 7 #> 8 LABEL2 X2 8 #> 9 LABEL3 X2 9 #> 10 LABEL4 X2 NA #> .. ... ... ... # idem on df2 df2 <- df2 %>% mutate(id = as.character(id)) %>% tidyr::gather(key = "col", value = "val", -id) # print result as dplyr tbl df2 %>% as.tbl() #> Source: local data frame [32 x 3] #> #> id col val #> (chr) (chr) (dbl) #> 1 LABEL3 X1 100.0000 #> 2 LABEL4 X1 419.3548 #> 3 LABEL5 X1 738.7097 #> 4 LABEL6 X1 1058.0645 #> 5 LABEL3 X2 1377.4194 #> 6 LABEL4 X2 1696.7742 #> 7 LABEL5 X2 2016.1290 #> 8 LABEL6 X2 2335.4839 #> 9 LABEL3 X3 2654.8387 #> 10 LABEL4 X3 2974.1935 #> .. ... ... ... # join only id and col level of df1 with df2 new.df <- left_join(df1, df2, by = c("id", "col")) # print result as dplyr tbl new.df %>% as.tbl() #> Source: local data frame [60 x 4] #> #> id col val.x val.y #> (chr) (chr) (int) (dbl) #> 1 LABEL1 X1 1 NA #> 2 LABEL2 X1 2 NA #> 3 LABEL3 X1 3 100.0000 #> 4 LABEL4 X1 NA 419.3548 #> 5 LABEL5 X1 NA 738.7097 #> 6 LABEL6 X1 NA 1058.0645 #> 7 LABEL1 X2 7 NA #> 8 LABEL2 X2 8 NA #> 9 LABEL3 X2 9 1377.4194 #> 10 LABEL4 X2 NA 1696.7742 #> .. ... ... ... ... #replace NA in col val.x from df1 by value val.y of df2 # and only keep id, col and new column val new.df <- new.df %>% transmute(id, col, val = ifelse(is.na(val.x), val.y, val.x)) new.df %>% as.tbl() #> Source: local data frame [60 x 3] #> #> id col val #> (chr) (chr) (dbl) #> 1 LABEL1 X1 1.0000 #> 2 LABEL2 X1 2.0000 #> 3 LABEL3 X1 3.0000 #> 4 LABEL4 X1 419.3548 #> 5 LABEL5 X1 738.7097 #> 6 LABEL6 X1 1058.0645 #> 7 LABEL1 X2 7.0000 #> 8 LABEL2 X2 8.0000 #> 9 LABEL3 X2 9.0000 #> 10 LABEL4 X2 1696.7742 #> .. ... ... ... # put back data in wide format new.df %>% spread(col, val) %>% select(id, num_range("X", 1:10)) # put column in same order as df1 #> id X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 #> 1 LABEL1 1.0000 7.000 13.000 19 25 31 37 43 49 55 #> 2 LABEL2 2.0000 8.000 14.000 20 26 32 38 44 50 56 #> 3 LABEL3 3.0000 9.000 15.000 21 27 33 39 45 51 57 #> 4 LABEL4 419.3548 1696.774 2974.194 22 28 34 40 46 52 58 #> 5 LABEL5 738.7097 2016.129 3293.548 23 29 35 41 47 53 59 #> 6 LABEL6 1058.0645 2335.484 3612.903 24 30 36 42 48 54 60 

Answer 2

可能有人有比我更好的方法，但这应该起作用。这确实假定列的顺序是相同的，所以要小心。

row.matches = match(df1$id, df2$id) 
nas = which(is.na(df1), arr.ind = TRUE) 
replacements = nas 
replacements[ ,1] = row.matches[nas[ ,1]] 
df1[nas] = df2[replacements] 

从本质上讲，所有我做的是找到匹配的行并在港定居发生在df1指数。使用匹配的行向量替换这些NA索引的行索引，并将df1中的那些值替换为df2中的相应值。