在R中是否有一种方法可以创建m的二进制集,其中包含n个列的所有组合,而按行只有 1的值可以是&#34; 1&#34;?< / p>
例如,对于n = 2和m = 2,我们将分别具有m的以下组合:
.affix{
top:60px;
}
.affix-bottom {
bottom:270px;
position: absolute;
}
但是,例如,这些不允许: (11,00),(01,11),(00,11),(11,10),(11,11)
答案 0 :(得分:2)
这与您的其他question非常相似。在这个问题的my answer中,我们看到重新解释这个问题,使攻击更加容易。因此,对于这个问题,我们可以将其缩小为:&#34;如何生成2 的所有成对排列的重复?&#34;
我们可以使用几乎完全相同的设置,只有这次我们设置参数repeats.allowed = TRUE
。
library(gtools)
bitPairwise2 <- function(numBits, groupSize) {
t(sapply(t(permutations(numBits + 1, groupSize,
c(0, 2^(0:(numBits-1))), repeats.allowed = TRUE)),
function(x) {as.integer(intToBits(x))})[1:numBits, ])
}
bitPairwise2(2,2)
[,1] [,2]
[1,] 0 0 ## (00,00)
[2,] 0 0
[3,] 0 0 ## (00,10)
[4,] 1 0
[5,] 0 0 ## (00,01)
[6,] 0 1
[7,] 1 0 ## (10,00)
[8,] 0 0
[9,] 1 0 ## (10,10)
[10,] 1 0
[11,] 1 0 ## (10,01)
[12,] 0 1
此函数可以推广到任意数量的位以及任意数量的组。例如,所有可能的3位三元组由下式给出:
## first 9 groups
bitPairwise2(3, 3)[1:27, ]
[,1] [,2] [,3]
[1,] 0 0 0 ## (000,000,000)
[2,] 0 0 0
[3,] 0 0 0
[4,] 0 0 0 ## (000,000,100)
[5,] 0 0 0
[6,] 1 0 0
[7,] 0 0 0 ## (000,000,010)
[8,] 0 0 0
[9,] 0 1 0
[10,] 0 0 0 ## (000,000,001)
[11,] 0 0 0
[12,] 0 0 1
[13,] 0 0 0 ## (000,100,000)
[14,] 1 0 0
[15,] 0 0 0
[16,] 0 0 0 ## (000,100,100)
[17,] 1 0 0
[18,] 1 0 0
[19,] 0 0 0 ## (000,100,010)
[20,] 1 0 0
[21,] 0 1 0
[22,] 0 0 0 ## (000,100,001)
[23,] 1 0 0
[24,] 0 0 1
[25,] 0 0 0 ## (000,010,000)
[26,] 0 1 0
[27,] 0 0 0
以下是最后9组:
a <- bitPairwise2(3, 3)[166:192, ]
row.names(a) <- 166:192
a
[,1] [,2] [,3]
166 0 0 1 ## (001,100,001)
167 1 0 0
168 0 0 1
169 0 0 1 ## (001,010,000)
170 0 1 0
171 0 0 0
172 0 0 1 ## (001,010,100)
173 0 1 0
174 1 0 0
175 0 0 1 ## (001,010,010)
176 0 1 0
177 0 1 0
178 0 0 1 ## (001,010,001)
179 0 1 0
180 0 0 1
181 0 0 1 ## (001,001,000)
182 0 0 1
183 0 0 0
184 0 0 1 ## (001,001,100)
185 0 0 1
186 1 0 0
187 0 0 1 ## (001,001,010)
188 0 0 1
189 0 1 0
190 0 0 1 ## (001,001,001)
191 0 0 1
192 0 0 1
如果您需要列表中的输出,请尝试以下操作:
test <- bitPairwise2(4, 3)
numGroups <- nrow(test)/3
makeGroupList <- function(mat, nG, groupSize) {
lapply(1:nG, function(x) {
s <- groupSize*(x-1) + 1
e <- s + (groupSize - 1)
mat[s:e, ]
})
}
makeGroupList(test, numGroups, 3)
[[1]]
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 0 0
[3,] 0 0 0 0
[[2]]
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 0 0
[3,] 1 0 0 0
[[3]]
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 0 0
[3,] 0 1 0 0
. . . . .
. . . . .
. . . . .