ループベクトル値

Question

例データと出力式のRの文字列内：

> DF 
    A B C 
1 11 22 88 
2 11 22 47 
3 2 30 21 
4 3 30 21 

> r 
[1] "A==A[i] & B==B[i] "   "A==A[i] & C==C[i] "   
[3] "B==B[i] & C==C[i] "   "A==A[i] & B==B[i] & C==C[i] " 

実行されるコード：

> output=list() 
> for (j in r){ 
+ for (i in 1:nrow(DF)){ 
+  
+  output[[j]][i]=j 
+ } 
+ } 
> output 
$`A==A[i] & B==B[i] ` 
[1] "A==A[i] & B==B[i] " "A==A[i] & B==B[i] " "A==A[i] & B==B[i] " 
[4] "A==A[i] & B==B[i] " 

$`A==A[i] & C==C[i] ` 
[1] "A==A[i] & C==C[i] " "A==A[i] & C==C[i] " "A==A[i] & C==C[i] " 
[4] "A==A[i] & C==C[i] " 

$`B==B[i] & C==C[i] ` 
[1] "B==B[i] & C==C[i] " "B==B[i] & C==C[i] " "B==B[i] & C==C[i] " 
[4] "B==B[i] & C==C[i] " 

$`A==A[i] & B==B[i] & C==C[i] ` 
[1] "A==A[i] & B==B[i] & C==C[i] " "A==A[i] & B==B[i] & C==C[i] " 
[3] "A==A[i] & B==B[i] & C==C[i] " "A==A[i] & B==B[i] & C==C[i] " 

> output=purrr::flatten_chr(output) 
> output 
[1] "A==A[i] & B==B[i] "   "A==A[i] & B==B[i] "   
[3] "A==A[i] & B==B[i] "   "A==A[i] & B==B[i] "   
[5] "A==A[i] & C==C[i] "   "A==A[i] & C==C[i] "   
[7] "A==A[i] & C==C[i] "   "A==A[i] & C==C[i] "   
[9] "B==B[i] & C==C[i] "   "B==B[i] & C==C[i] "   
[11] "B==B[i] & C==C[i] "   "B==B[i] & C==C[i] "   
[13] "A==A[i] & B==B[i] & C==C[i] " "A==A[i] & B==B[i] & C==C[i] " 
[15] "A==A[i] & B==B[i] & C==C[i] " "A==A[i] & B==B[i] & C==C[i] " 

私の目標は、特定の値[I]と同じ出力を得ることでした、B [i]とC [i]はDFから、すなわち、最終的な出力は次のようになります。

> output 
[1] "A==11 & B==22 "   "A==11 & B==22 "   
[3] "A==2 & B==30 "    "A==3 & B==30 "   
[5] "A==11 & C==88 "   "A==11 & C==47 "   
[7] "A==2 & C==21 "    "A==3 & C==21 "   
[9] "B==22 & C==88 "   "B==22 & C==47 "   
[11] "B==30 & C==21 "   "B==30 & C==21 "   
[13] "A==11 & B==22 & C==88 " "A==11 & B==22 & C==47 " 
[15] "A==2 & B==30 & C==21 " "A==3 & B==30 & C==21 " 

誰かがHELことができれば、私は感謝されますこの事でp me。

Answer 1

難しさは、ここでの問題は、rベクターで指定されている方法から来て、より良い期待される結果と一致するように、ミラノの答えを変更することは可能であろう。その代わりの

r <- c(
    "A==A[i] & B==B[i]", 
    "A==A[i] & C==C[i]", 
    "B==B[i] & C==C[i]", 
    "A==A[i] & B==B[i] & C==C[i]" 
) 

、私はrの各要素には、あなただけの列の名前を特定する必要があることを考えると、別のルートを取りますよ。たとえば、最初のケースでは、AとBが必要です。そもそも名を取得するための整頓にevalを使用する関数にそれをすべてアップをラッピング

map(names, ~{ 
    glue("{name} == {values}", name = ., values = DF[[.]]) 
}) %>% 
    reduce(paste, sep = " & ") 
#> [1] "A == 11 & B == 22" "A == 11 & B == 22" "A == 2 & B == 30" 
#> [4] "A == 3 & B == 30" 

：

library(purrr) 
library(glue) 
library(rlang) 
names <- c("A", "B") 
DF <- data.frame(
    A = c(11, 11, 2, 3), 
    B = c(22, 22, 30, 30), 
    C = c(88, 47, 21, 21) 
) 

map(names, ~{ 
    glue("{name} == {values}", name = ., values = DF[[.]]) 
}) 
#> [[1]] 
#> A == 11 
#> A == 11 
#> A == 2 
#> A == 3 
#> 
#> [[2]] 
#> B == 22 
#> B == 22 
#> B == 30 
#> B == 30 

あなたはその後、reduceそれだけで、あなたのデータフレームの1行に1つの文字列を取得することができます。

tests <- function(data, ...){ 
    names <- map_chr(quos(...), f_name) 

    map(names, ~{ 
    glue("{name} == {values}", name = ., values = data[[.]]) 
    }) %>% 
    reduce(paste, sep = " & ") 

} 

あなたはtests 4回呼び出すことによって結果を得ることができるように：

を

c(
    tests(DF, A, B), 
    tests(DF, A, C), 
    tests(DF, B, C), 
    tests(DF, A, B, C) 
) 
#> [1] "A == 11 & B == 22"   "A == 11 & B == 22"   
#> [3] "A == 2 & B == 30"   "A == 3 & B == 30"   
#> [5] "A == 11 & C == 88"   "A == 11 & C == 47"   
#> [7] "A == 2 & C == 21"   "A == 3 & C == 21"   
#> [9] "B == 22 & C == 88"   "B == 22 & C == 47"   
#> [11] "B == 30 & C == 21"   "B == 30 & C == 21"   
#> [13] "A == 11 & B == 22 & C == 88" "A == 11 & B == 22 & C == 47" 
#> [15] "A == 2 & B == 30 & C == 21" "A == 3 & B == 30 & C == 21" 

Answer 2

この出力は後で表示されますか？

DF <- data.frame(A = c(11, 11, 2, 3), 
       B = c(22, 22, 30, 30), 
       C = c(88, 47, 21, 21)) 

r <- c("A==A[i] & B==B[i]", "A==A[i] & C==C[i]", 
     "B==B[i] & C==C[i]", "A==A[i] & B==B[i] & C==C[i]") 

output=list() 
for (j in r){ 
    for (i in 1:nrow(DF)) 
    output[[j]][[i]] <- DF[with(DF, eval(parse(text = j))), ] 
} 

output 

$`A==A[i] & B==B[i]` 
$`A==A[i] & B==B[i]`[[1]] 
    A B C 
1 11 22 88 
2 11 22 47 

$`A==A[i] & B==B[i]`[[2]] 
    A B C 
1 11 22 88 
2 11 22 47 

$`A==A[i] & B==B[i]`[[3]] 
    A B C 
3 2 30 21 

$`A==A[i] & B==B[i]`[[4]] 
    A B C 
4 3 30 21 


$`A==A[i] & C==C[i]` 
$`A==A[i] & C==C[i]`[[1]] 
    A B C 
1 11 22 88 
...