R-查找值的唯一排列 [英] R- Find Unique Permutations of Values

问题描述

我希望创建一个包含两个不同值的向量的所有可能排列，其中我控制每个值的比例.

例如，如果我有一个长度为3的向量，并且我希望所有可能的组合都包含一个单一的1，那么我想要的输出就是一个看起来像这样的列表:

list.1 <- list(c(1,0,0), c(0,1,0), c(0,0,1))

相反，如果我希望所有可能的组合包含三个1，那么我想要的输出是一个看起来像这样的列表:

list.3 <- list(c(1,1,1))

换句话说，1和0值的模式很重要，但是所有1都应被视为与所有其他1相同.

基于此处和其他位置的搜索，我尝试了几种方法:

expand.grid(0:1, 0:1, 0:1)  # this includes all possible combinations of 1, 2, or 3 ones
permn(c(0,1,1))             # this does not treat the ones as identical (e.g. it produces (0,1,1) twice)
unique(permn(c(0,1,1)))     # this does the job!

因此，使用软件包combinat中的函数permn似乎很有希望.但是，在我将其扩展到我的实际问题(长度为20，具有50％1s和50％0s的向量)时，我遇到了问题:

unique(permn(c(rep(1,10), rep(0, 10))))

# returns the error:
Error in vector("list", gamma(n + 1)) : 
  vector size specified is too large

我的理解是，发生这种情况是因为，在对permn的调用中，它创建了一个包含所有可能排列的列表，即使其中许多排列是相同的，并且该列表太大，R无法处理.

有人对如何解决这个问题有建议吗?

很抱歉，如果以前已经回答过-SO问题很多，包含相似的语言，但问题不同，我无法找到满足我需要的解决方案！

解决方案

expand.grid包括所有排列不应该成为一个破坏者.只需在以下位置添加一个子集:

combinations <- function(size, choose) {

  d <- do.call("expand.grid", rep(list(0:1), size))
  d[rowSums(d) == choose,]

}

combinations(size=10, choose=3)
#    Var1 Var2 Var3 Var4 Var5 Var6 Var7 Var8 Var9 Var10
# 8     1    1    1    0    0    0    0    0    0     0
# 12    1    1    0    1    0    0    0    0    0     0
# 14    1    0    1    1    0    0    0    0    0     0
# 15    0    1    1    1    0    0    0    0    0     0
# 20    1    1    0    0    1    0    0    0    0     0
# 22    1    0    1    0    1    0    0    0    0     0
...

I am hoping to create all possible permutations of a vector containing two different values, in which I control the proportion of each of the values.

For example, if I have a vector of length three and I want all possible combinations containing a single 1, my desired output is a list looking like this:

list.1 <- list(c(1,0,0), c(0,1,0), c(0,0,1))

In contrast, if I want all possible combinations containing three 1s, my desired output is a list looking like this:

list.3 <- list(c(1,1,1))

To put it another way, the pattern of the 1 and 0 values matter, but all 1s should be treated as identical to all other 1s.

Based on searching here and elsewhere, I've tried several approaches:

expand.grid(0:1, 0:1, 0:1)  # this includes all possible combinations of 1, 2, or 3 ones
permn(c(0,1,1))             # this does not treat the ones as identical (e.g. it produces (0,1,1) twice)
unique(permn(c(0,1,1)))     # this does the job!

So, using the function permn from the package combinat seems promising. However, where I scale this up to my actual problem (a vector of length 20, with 50% 1s and 50% 0s, I run into problems:

unique(permn(c(rep(1,10), rep(0, 10))))

# returns the error:
Error in vector("list", gamma(n + 1)) : 
  vector size specified is too large

My understanding is that this is happening because, in the call to permn, it makes a list containing all possible permutations, even though many of them are identical, and this list is too large for R to handle.

Does anyone have a suggestion for how to work around this?

Sorry if this has been answered previously - there are many, many SO questions containing similar language but different problems and I have not bene able to find a solution which meets my needs!

解决方案

It should not be a dealbreaker that expand.grid includes all permutations. Just add a subset after:

combinations <- function(size, choose) {

  d <- do.call("expand.grid", rep(list(0:1), size))
  d[rowSums(d) == choose,]

}

combinations(size=10, choose=3)
#    Var1 Var2 Var3 Var4 Var5 Var6 Var7 Var8 Var9 Var10
# 8     1    1    1    0    0    0    0    0    0     0
# 12    1    1    0    1    0    0    0    0    0     0
# 14    1    0    1    1    0    0    0    0    0     0
# 15    0    1    1    1    0    0    0    0    0     0
# 20    1    1    0    0    1    0    0    0    0     0
# 22    1    0    1    0    1    0    0    0    0     0
...

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