Haskell 中的列表理解
我一直在使用以下代码来获取预定数量的数字的所有组合：
getList x = [ [a,b,c] | a <- [1..x], b <- [1..x], c <- [1..x]]
一开始这很好，但我希望扩展该程序以处理非常大的列表，并且我一直认为必须有一个更好的方法来做到这一点。我将如何创建一个函数，它采用与此处相同的参数 x ，以及子列表有多少项的另一个参数。对于四个项目，我会去修改代码：
getList x = [ [a,b,c,d] | a <- [1..x], b <- [1..x], c <- [1..x], d <- [1..x]]
它不需要是列表理解。感谢您的帮助。
I've been using the following code to get all combinations of a pre-determined amount of numbers:
getList x = [ [a,b,c] | a <- [1..x], b <- [1..x], c <- [1..x]]
This was fine to begin with, but I'm looking to expand the program to handle very large lists, and I keep thinking there must be a better way to do this. How would I create a function that takes the same parameter x as here, and also another parameter for how many items the sublists have. For four items I would go and modify the code:
getList x = [ [a,b,c,d] | a <- [1..x], b <- [1..x], c <- [1..x], d <- [1..x]]
It doesn't need to be a list comprehension. Thank you for any help.
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我相信您想要的是
Control.Monad中的replicateM函数。列表 monad 基于“尝试所有可能的组合”，并且普通的
replicate通过重复某个项目一定次数来创建列表。因此，在给定一些可能值的列表的情况下，replicateM 的结果是从该列表中多次选择项目的所有可能方法的列表。例如：
因此，要将您的函数扩展到任意重复，您可以使用类似以下内容的内容：
...其中您的原始
getList将相当于getListN 3。I believe what you want would be the
replicateMfunction inControl.Monad.The list monad is based on "trying all possible combinations", and plain
replicatecreates a list by repeating an item some number of times. So the result ofreplicateMis, given some list of possible values, a list of all possible ways to select an item from that list some number of times.For example:
So to extend your function to arbitrary repetitions, you'd use something like:
...where your original
getListwould be equivalent togetListN 3.万一，任何人都喜欢非 Monadic 解决方案来理解内部工作原理（尽管通过
replicateM的解决方案很棒！）：本质上，通过
foldl的实现完全可以工作与replacatM 解决方案的方式相同。In case, anyone likes a non-Monadic-solution to understand to inner workings (although, the soliution via
replicateMis great!):Essentially, this implementation via
foldlworks exactly in the same way as thereplacatM-solution does.