如何组合等序元素(功能编程)? [英] How to combine equal sequence elements (functional programming)?

问题描述

我想编写一个函数，该函数采用序列< 1、2、2、3>，并返回具有等元素的序列，这些元素的分组方式与<<< 1、1> ;、< 2,2> ，< 3 >>.

I want to write a function that takes in a sequence <1,1,2,2,3> and returns the sequence with equal elements grouped like <<1,1>, <2,2>, <3>>.

我使用的是序列，而不是列表，但是其中一些功能是相似的.我正在考虑使用的某些功能是map，reduce，制表，filter，append等.

I'm using sequences, not lists, but some of the functions are similar. Some of the functions I am thinking of using are map, reduce, tabulate, filter, append etc..

Reduce接受一个关联函数，并返回该运算符减少"的序列.因此，将op + 0降低为<1,2,3> = 6.

Reduce takes in an associative function and returns the sequence that is "reduced" by that operator. So, reduce op+ 0 <1,2,3> = 6.

我的第一个想法是使用map将序列提升一个级别. 因此，< 1,1,2,2,3> =><< 1>，< 1>，< 2>，< 2>，< 3 >>.

My first thought was to use map to raise the sequence by one level. So, <1,1,2,2,3> => <<1>,<1>,<2>,<2>,<3>>.

然后，我正在考虑使用reduce，在该方法中，我创建了一个函数，该函数接受像(x，y)这样的成对元素.如果x == y，那么我返回否则，我什么也不做.但是...这并不完全有效，因为函数在两种情况下都必须返回相同类型的东西.

Then, I was thinking of using reduce, in which I create a function that takes pairs of elements like (x,y). If x == y, then I return else, I do nothing. But...this doesn't exactly work since the function has to return something of the same type in both cases.

有人可以给我一些正确使用方法的提示，例如我可以使用哪些更高阶的函数?我正在使用SML，但我并没有要求任何人给我一个完整的答案，因此(使用任何一种功能语言，无论如何，)任何高级技巧都将不胜感激

Can someone give me some tips on the right path to take, like what higher order functions I could possibly use? I'm using SML but I'm not asking for anyone to give me a full blown answer so any high-level tips would be appreciated (in any functional language honestly)

推荐答案

我猜您所引用的reduce函数与F#中的fold函数相同:

I guess that the reduce function you are referring to is the same as the fold function in F#:

val fold : ('State -> 'Value -> 'State) -> 'State -> 'Value list -> 'State

这将获取一个值列表，以及一个初始状态和一个在迭代列表值的过程中转换状态的函数.

This takes a list of values, together with an initial state and a function that transforms the state while iterating through the values of the list.

您可以一次完成自己想做的事情.您需要保持一些状态.假设您在1,1,2,2,3的中间(例如，在第二个2上).现在您需要:

You can do what you want in a single fold. There are a couple of things that you'll need to keep in the state. Imagine you are somewhere in the middle of 1,1,2,2,3 (say, on the second 2). Now you'll need:

• 您当前正在收集的值-2
• 包含当前收集值的值列表-即[2](序列中的第一个2)
• 您先前收集的值列表-[ [1; 1] ].

• The value that you are currently collecting - that is 2
• A list of values containing the currently collected values - that is [2] (the first 2 from the sequence)
• A list of lists of values you collected previously - that is [ [1; 1] ].

您将从初始状态-1, [], []开始(使用-1作为一些不会出现在您的输入中的值).然后，您需要编写根据当前值转换状态的函数.这需要处理几种情况:

You would start with an initial state -1, [], [] (using -1 as some value that won't appear in your input). Then you need to write the function that transforms the state based on a current value. This needs to handle a couple of situations:

• 如果该值与您一直在收集的值不同，则需要将收集到的值列表添加到列表列表中(除非它为空)
• 当值相同时，您需要将其添加到现在收集的值列表中并继续

希望这能为您提供足够的信息，以找出如何执行此操作，而无需实际透露完整的源代码！

Hopefully, this gives you enough information to figure out how to do this, without actually revealing the full source code!

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