如何在 R 中迭代地找到集合的所有可能子集? [英] How do I find all possible subsets of a set iteratively in R?

问题描述

所以我知道以下命令会将所需长度 y 的所有可能组合存储在列表中，其中 y <j:

So I know that the following command would store all possible combinations of a desired length y in a list, where y < j:

 lapply(y, function(x) combn(j,x))

但我不希望将它们全部存储在列表中，因为稍后我将只访问它们一次，因此将它们存储在内存中效率不高.有没有一种方法可以让我在某种循环或其他东西中生成每个组合，然后在我完成计算之后，它只会给我下一个组合?所以基本上我想迭代地产生组合而不是先存储它们.

But I don't want them all to be stored in a list because later on I will be accessing them only once so it's not efficient to store them in memory. Is there a way where I can just produce each combination in some sort of a loop or something, and then after I'm done performing a calculation, it would just give me the next combination? So basically I want to produce the combinations iteratively instead of storing them first.

所以在伪代码中，我想要的是:

So in pseudo code, what i'd like to have is:

#loop that will generate each possible combination one by one
loop{
  operation that uses combination
}

推荐答案

不需要循环(lapply 或其他):

No need for loops (lapply or otherwise):

combn(1:4,2)
#      [,1] [,2] [,3] [,4] [,5] [,6]
# [1,]    1    1    1    2    2    3
# [2,]    2    3    4    3    4    4

计算组合总和的示例:

combn(1:4,2,FUN=sum)
# [1] 3 4 5 5 6 7

一个用户定义函数的例子:

An example with a user defined function:

x <- 11:14
combn(1:4,2,FUN=function(i,a) sum(a[i]),a=x)
#[1] 23 24 25 25 26 27

这里(在匿名函数中)i 是用作索引和参数的组合 a 是我将 x 传递给的向量.

Here (in the anonymous function) i is the combination used as index and argument a is a vector to which I pass x.

用户定义的命名函数也是如此:

And the same with a user-defined named function:

fun <- function(i,a) sum(a[i])
combn(1:4,2,FUN=fun,a=x)

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