生成給定約束條件的所有組合

如何生成4個塊的2個處理（A，B）的6個組合，以便在每個塊中有相同數量的A和B，使用R？生成給定約束條件的所有組合

"AABB","ABAB","ABBA","BBAA","BABA","BAAB"

P.S.組合數的計算方法如下：

如果

T = #treatments 

n = #treatments in each block = k*T,

組合的數量等於n!/[k!*k! (T times)]

謝謝

來源

2012-01-28 George Dontas

副查詢：你的意思是組合數是n！ /（（k！）^ T）？ – 2012-01-28 13:44:05

@Carl Witthoft是:-P – 2012-01-28 17:23:20

像這樣的東西應該工作：

library(gtools) 

t <- c('A','B') 
k <- 2 
n <- k * length(t) 

t2 <- rep(t, k) 
m <- permutations(n,n) 
res <- unique(apply(m,MARGIN=1,function(x) paste(t2[x],collapse=''))) 

-------------------------------------------------------------------- 
res 
[1] "ABAB" "ABBA" "AABB" "BAAB" "BABA" "BBAA"

來源

2012-01-28 12:55:03 digEmAll

multicool pac kage實現了一個算法，用於置換multisets ---完全是你想要執行的任務。這裏有什麼可以做，例如：

library(multicool) 

# Create a simple convenience function 
enumAllPartitions <- function(multiset) { 
    m1 <- initMC(multiset)  # Initialize the permutation object 
    N <- fact(length(multiset))/ # Calculate number of permutations 
     prod(fact(table(multiset))) 
    sapply(seq_len(N), function(X) paste(nextPerm(m1), collapse="")) 
} 

# Try it out with a few different multisets 
x <- c("A", "A", "B", "B") 
y <- c("G", "L", "L", "L") 
z <- c("X", "X", "Y", "Z", "Z") 

lapply(list(x,y,z), enumAllPartitions) 
[[1]] 
[1] "BBAA" "ABBA" "BABA" "ABAB" "AABB" "BAAB" 

[[2]] 
[1] "LLLG" "GLLL" "LGLL" "LLGL" 

[[3]] 
[1] "ZZYXX" "XZZYX" "ZXZYX" "ZZXYX" "XZZXY" "ZXZXY" "XZXZY" "XXZZY" "ZXXZY" 
[10] "ZZXXY" "YZZXX" "ZYZXX" "XZYZX" "ZXYZX" "YZXZX" "XYZZX" "YXZZX" "ZYXZX" 
[19] "XZYXZ" "ZXYXZ" "XZXYZ" "XXZYZ" "ZXXYZ" "YZXXZ" "XYZXZ" "YXZXZ" "XYXZZ" 
[28] "XXYZZ" "YXXZZ" "ZYXXZ"

來源

2012-01-29 04:56:28

預期的解決方案還可以使用新的iterpc包裝來實現。

I <- iterpc(c(2, 2), labels=c("A", "B"), ordered=TRUE) 
getall(I) 

#  [,1] [,2] [,3] [,4] 
# [1,] "A" "A" "B" "B" 
# [2,] "A" "B" "A" "B" 
# [3,] "A" "B" "B" "A" 
# [4,] "B" "A" "A" "B" 
# [5,] "B" "A" "B" "A" 
# [6,] "B" "B" "A" "A"

來源

2017-10-29 16:12:33 Prradep

生成給定約束條件的所有組合

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