匹配兩個data.frame（每個列名稱一個）並創建一個新的data.frame

Question

我有以下df1 ：

structure(list(rchX = c(0.562189054726368, 0.552238805970149, 
0.552238805970149, 0.54726368159204, 0.54726368159204, 0.54726368159204, 
0.54228855721393, 0.54228855721393, 0.537313432835821, 0.537313432835821
), frqX = c(0.925373134328358, 0.925373134328358, 0.915422885572139, 
0.965174129353234, 0.955223880597015, 0.875621890547264, 0.955223880597015, 
0.890547263681592, 0.900497512437811, 0.850746268656716), `1` = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), `2` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 
0), `3` = c(0, 1, 0, 1, 0, 0, 1, 0, 0, 0), `4` = c(0, 0, 0, 0, 
0, 0, 0, 0, 0, 0), `5` = c(0, 0, 0, 0, 0, 0, 0, 0, 1, 0), `6` = c(1, 
1, 1, 1, 1, 1, 1, 1, 1, 1), `7` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 
0), `8` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 1), `9` = c(0, 0, 0, 0, 
0, 0, 0, 0, 0, 0), `10` = c(0, 0, 0, 0, 0, 0, 0, 1, 0, 0), `11` = c(1, 
1, 1, 0, 0, 0, 0, 0, 0, 0), `12` = c(1, 0, 0, 1, 1, 1, 0, 1, 
1, 1), `13` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), `14` = c(0, 0, 
0, 0, 0, 0, 0, 0, 0, 0), `15` = c(0, 0, 0, 0, 0, 1, 0, 0, 0, 
0), `16` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), `17` = c(0, 0, 0, 
0, 0, 0, 0, 0, 0, 0), `18` = c(0, 0, 1, 0, 1, 0, 1, 0, 0, 0), 
    `19` = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), `20` = c(0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0)), class = "data.frame", row.names = c(NA, 
10L))

看起來像這樣：

        rchX      frqX 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
1  0.5621891 0.9253731 0 0 0 0 0 1 0 0 0  0  1  1  0  0  0  0  0  0  0  0
2  0.5522388 0.9253731 0 0 1 0 0 1 0 0 0  0  1  0  0  0  0  0  0  0  0  0
3  0.5522388 0.9154229 0 0 0 0 0 1 0 0 0  0  1  0  0  0  0  0  0  1  0  0
4  0.5472637 0.9651741 0 0 1 0 0 1 0 0 0  0  0  1  0  0  0  0  0  0  0  0
5  0.5472637 0.9552239 0 0 0 0 0 1 0 0 0  0  0  1  0  0  0  0  0  1  0  0
6  0.5472637 0.8756219 0 0 0 0 0 1 0 0 0  0  0  1  0  0  1  0  0  0  0  0
7  0.5422886 0.9552239 0 0 1 0 0 1 0 0 0  0  0  0  0  0  0  0  0  1  0  0
8  0.5422886 0.8905473 0 0 0 0 0 1 0 0 0  1  0  1  0  0  0  0  0  0  0  0
9  0.5373134 0.9004975 0 0 0 0 1 1 0 0 0  0  0  1  0  0  0  0  0  0  0  0
10 0.5373134 0.8507463 0 0 0 0 0 1 0 1 0  0  0  1  0  0  0  0  0  0  0  0

還有第二個具有相應名稱的data.frame：

df <- data.frame(
    a = seq(1:20),
    b = LETTERS[1:20]
)

    a b
1   1 A
2   2 B
3   3 C
4   4 D
5   5 E
6   6 F
7   7 G
8   8 H
9   9 I
10 10 J
11 11 K
12 12 L
13 13 M
14 14 N
15 15 O
16 16 P
17 17 Q
18 18 R
19 19 S
20 20 T

我想做的是檢查哪些列為1並與df的相應字母匹配。 第6列中的1表示“ F”，第11列中的1表示“ K”。 總有3個匹配項，因此新data.frame的前兩行將如下所示：

       rchX      frqX varA varB varC
1 0.5621891 0.9253731    F    K    L
2 0.5522388 0.9253731    C    F    K

誰能幫我？

Answer 1

如果我們需要基於apply的解決方案，我們可以做

cbind(df1[1:2],  t(apply(df1[-(1:2)], 1, function(x) 
 setNames(as.character(df$b), df$a)[names(x)[which(as.logical(x))]])))

---

或者可以使用tidyverse通過gather荷蘭國際集團為“長”格式，做一個left_join與鍵/ VAL數據集， summarise通過與行號，RCHX，frqX分組的輸出，並且separate成多列

library(tidyverse)
df1 %>% 
  mutate(rn = row_number()) %>% 
  gather(a, val, -rn, -rchX, -frqX) %>% 
  filter(val == 1) %>% 
  left_join(., df %>%
                  mutate(a = as.character(a))) %>% 
  select(-val)  %>% 
  group_by(rn, rchX, frqX) %>% 
  summarise(b = toString(b)) %>% 
  separate(b, into = str_c("Var", LETTERS[1:3])) %>%
  ungroup %>%
  select(-rn)
# A tibble: 10 x 5
#    rchX  frqX VarA  VarB  VarC 
#   <dbl> <dbl> <chr> <chr> <chr>
# 1 0.562 0.925 F     K     L    
# 2 0.552 0.925 C     F     K    
# 3 0.552 0.915 F     K     R    
# 4 0.547 0.965 C     F     L    
# 5 0.547 0.955 F     L     R    
# 6 0.547 0.876 F     L     O    
# 7 0.542 0.955 C     F     R    
# 8 0.542 0.891 F     J     L    
# 9 0.537 0.900 E     F     L    
#10 0.537 0.851 F     H     L    

---

我們還可以更有效地使用base R

m1 <- `dim<-`(setNames(as.character(df$b), 
   df$a)[names(df1)[-(1:2)][col(df1[-(1:2)])]], dim(df1[-(1:2)]))
out <- read.table(text= trimws(do.call(paste, 
 as.data.frame(replace(m1, df1[-(1:2)] == 0, "")))), header = FALSE)
cbind(df1[1:2], out)
#         rchX      frqX V1 V2 V3
#1  0.5621891 0.9253731  F  K  L
#2  0.5522388 0.9253731  C  F  K
#3  0.5522388 0.9154229  F  K  R
#4  0.5472637 0.9651741  C  F  L
#5  0.5472637 0.9552239  F  L  R
#6  0.5472637 0.8756219  F  L  O
#7  0.5422886 0.9552239  C  F  R
#8  0.5422886 0.8905473  F  J  L
#9  0.5373134 0.9004975  E  F  L
#10 0.5373134 0.8507463  F  H  L

Answer 2

在基R，一個方法是使用apply ，丟棄哪些是0值時，它們的名稱與比較a的柱df ，並得到相應的b值。

cbind(df1[1:2], t(apply(df1[-c(1:2)], 1, function(x) 
                df$b[match(names(x[x!=0]), df$a)])))

#           rchX         frqX 1 2 3
#1  0.5621890547 0.9253731343 F K L
#2  0.5522388060 0.9253731343 C F K
#3  0.5522388060 0.9154228856 F K R
#4  0.5472636816 0.9651741294 C F L
#5  0.5472636816 0.9552238806 F L R
#6  0.5472636816 0.8756218905 F L O
#7  0.5422885572 0.9552238806 C F R
#8  0.5422885572 0.8905472637 F J L
#9  0.5373134328 0.9004975124 E F L
#10 0.5373134328 0.8507462687 F H L