多種組合和排列中的R

我想準備一個動態置換爲數據集設置，使得它看起來這個多種組合和排列中的R

ID  Val Perm 1 Perm 2 Perm3 Perm 4 Perm 5 Perm 6 Perm 7 Perm 8 Perm 9 Perm 10 Perm 11 Perm 12 Perm 13 Perm 14 Perm 15 Perm 16 Perm 17 Perm 18 Perm 19 Perm 20 Perm 21 Perm 22 Perm 23 Perm 24 Perm 25 Perm 26 Perm 27 Perm 28 Perm 29 Perm 30 Perm 31 Perm 32 Perm 33 Perm 34 Perm 35 Perm 36 
12,000  1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 
12,581  2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 
12,857  2 Range 3 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 
13,387  5 Range 3 Range 3 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 3 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 
15,846  2 Range 3 Range 3 Range 3 Range 2 Range 2 Range 2 Range 2 Range 2 Range 3 Range 3 Range 2 Range 2 Range 2 Range 2 Range 2 Range 3 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 
23,387  5 Range 3 Range 3 Range 3 Range 3 Range 2 Range 2 Range 2 Range 2 Range 3 Range 3 Range 3 Range 2 Range 2 Range 2 Range 2 Range 3 Range 3 Range 2 Range 2 Range 2 Range 2 Range 3 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 1 Range 1 Range 1 Range 1 Range 1 Range 1 
25,424  4 Range 3 Range 3 Range 3 Range 3 Range 3 Range 2 Range 2 Range 2 Range 3 Range 3 Range 3 Range 3 Range 2 Range 2 Range 2 Range 3 Range 3 Range 3 Range 2 Range 2 Range 2 Range 3 Range 3 Range 2 Range 2 Range 2 Range 3 Range 2 Range 2 Range 2 Range 2 Range 2 Range 2 Range 1 Range 1 Range 1 
25,424  2 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 2 Range 2 Range 3 Range 3 Range 3 Range 3 Range 3 Range 2 Range 2 Range 3 Range 3 Range 3 Range 3 Range 2 Range 2 Range 3 Range 3 Range 3 Range 2 Range 2 Range 3 Range 3 Range 2 Range 2 Range 3 Range 2 Range 2 Range 2 Range 2 Range 1 
25,932  6 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 2 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 2 Range 3 Range 3 Range 3 Range 3 Range 3 Range 2 Range 3 Range 3 Range 3 Range 3 Range 2 Range 3 Range 3 Range 3 Range 2 Range 3 Range 3 Range 2 Range 3 Range 2 Range 2 
25,932  1 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3 Range 3

從數據集

dput(df) 
structure(list(ID = c(12000, 12581, 12857, 13387, 15846, 23387, 
25424, 25424, 25932, 25932), Val = c(1L, 2L, 2L, 5L, 2L, 5L, 
4L, 2L, 6L, 1L)), .Names = c("ID", "Val"), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -10L), spec = structure(list(
    cols = structure(list(ID = structure(list(), class = c("collector_number", 
    "collector")), Val = structure(list(), class = c("collector_integer", 
    "collector"))), .Names = c("ID", "Val")), default = structure(list(), class = c("collector_guess", 
    "collector"))), .Names = c("cols", "default"), class = "col_spec"))

在這裏，我嘗試創建每個列通過保持「範圍1」不變並遞增「範圍2」並相應地減少「範圍3」。另外我試着它只有10行，但在未來的行數可以增加。雖然我曾與

apply(combinations(n=3,r=5,v = c("Range 1","Range 2","Range 3"),repeats.allowed = T),2,sort)

和

v <- c("Range 1","Range 2","Range 3");expand.grid(rep(list(v), 5))

但沒有成功，但嘗試。有沒有其他方法可以這樣做。任何想法將提前

DOMNICK

來源

2017-10-10 Domnick

【如何使一個偉大的[R重複的例子？（http://stackoverflow.com/questions/5963269） – Sotos

試圖使可再生的dataframe.kindly檢查更新 – Domnick

什麼'$ DF瓦爾'代表，以及它如何與排列相對應？什麼是從值生成列的邏輯？ – Gonzo

對於較短的解釋和更普遍的緣故appreciate.Thanks（如你所提到的行數可以擦菜板是10），我提議以下，你可以實現邏輯你自己：開始時，初始設置。

n <- 10 
k <- 3 
df <- as.data.frame(lapply(1:n, function(x) x <- k:1))

下面的代碼創建10個元素集合的所有唯一組合，其順序無關緊要。

all <- as.matrix(expand.grid(df)) 
all_sorted <- t(apply(all, 1, sort)) 
all_sorted_unique <- unique(all_sorted)

在10元，則所有的獨特元素必須出現

exclude_if_not_all <- 
    which( 
    apply(all_sorted_unique, 1, 
     function(x) length(unique(x))!=3) 
) 

t(all_sorted_unique[-exclude_if_not_all,])

結果符合市場預期。

 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26] [,27] [,28] [,29] [,30] [,31] [,32] [,33] [,34] [,35] [,36] 
[1,] 1 1 1 1 1 1 1 1 1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1 
[2,] 2 2 1 2 1 1 2 1 1  1  2  1  1  1  1  2  1  1  1  1  1  2  1  1  1  1  1  1  2  1  1  1  1  1  1  1 
[3,] 3 2 2 2 2 1 2 2 1  1  2  2  1  1  1  2  2  1  1  1  1  2  2  1  1  1  1  1  2  2  1  1  1  1  1  1 
[4,] 3 3 3 2 2 2 2 2 2  1  2  2  2  1  1  2  2  2  1  1  1  2  2  2  1  1  1  1  2  2  2  1  1  1  1  1 
[5,] 3 3 3 3 3 3 2 2 2  2  2  2  2  2  1  2  2  2  2  1  1  2  2  2  2  1  1  1  2  2  2  2  1  1  1  1 
[6,] 3 3 3 3 3 3 3 3 3  3  2  2  2  2  2  2  2  2  2  2  1  2  2  2  2  2  1  1  2  2  2  2  2  1  1  1 
[7,] 3 3 3 3 3 3 3 3 3  3  3  3  3  3  3  2  2  2  2  2  2  2  2  2  2  2  2  1  2  2  2  2  2  2  1  1 
[8,] 3 3 3 3 3 3 3 3 3  3  3  3  3  3  3  3  3  3  3  3  3  2  2  2  2  2  2  2  2  2  2  2  2  2  2  1 
[9,] 3 3 3 3 3 3 3 3 3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  2  2  2  2  2  2  2  2 
[10,] 3 3 3 3 3 3 3 3 3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3  3

你現在需要做的是隻將這個矩陣綁定到你的初始數據上。所有功能

customPermutation <- function(n, k){ 

    df <- as.data.frame(lapply(1:n, function(x) x <- 1:k)) 

    all <- as.matrix(expand.grid(df)) 
    all_sorted <- t(apply(all, 1, sort)) 
    all_sorted_unique <- unique(all_sorted) 

    exclude_if_not_all <- which(apply(all_sorted_unique, 1, function(x) length(unique(x))!=3)) 
    t(all_sorted_unique[-exclude_if_not_all,]) 

}

來源

2017-10-10 14:20:54 Gonzo

它按照我想要的方式工作。非常感謝，非常感謝。 – Domnick

多種組合和排列中的R

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