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为什么这个递归映射函数使用所有可用内存?

[英]Why is this recursive map function using all available memory?

I am trying to implement the map function in Python using recursion, running it on Google Colab.我正在尝试使用递归在 Python 中实现 map 函数,并在 Google Colab 上运行它。

def randfun(var):
    return var * 2

def mapr(f, itr):
    if len(itr) == 1:
        return f(itr[0])
    else:
        amendedlist = list(mapr(randfun, f(itr[1:])))
        #print(amendedlist)
        return amendedlist

li = [1,2,3,4,5]
result = mapr(randfun, li)
print(result)

The error I'm getting is this: Your session crashed after using all available RAM .我得到的错误是: Your session crashed after using all available RAM It produces a bunch of logs, which I can't decrypt, but am happy to reproduce.它产生了一堆日志,我无法解密,但很高兴重现。

Looks like the code is failing to halt the recursion and it goes on infinitely, but again I'm not sure.看起来代码无法停止递归并且它无限地继续,但我又不确定。 What am I doing wrong?我究竟做错了什么?

This worked for me这对我有用

def randfun(var):
    return var * 2

def mapr(f, itr):
    if len(itr) == 1:
        return [f(itr[0])]
    amendedlist = [f(itr[0])] + mapr(f, itr[1:])
    return amendedlist

li = [1,2,3,4,5]
result = mapr(randfun, li)
print(result) #[2, 4, 6, 8, 10]

When you recurse, you need to pass a shorter list to mapr() , eg itr[1:] .当您递归时,您需要将较短的列表传递给itr[1:] mapr() ,例如itr[1:] But you're passing f(itr[1:]) instead.但是您正在传递f(itr[1:]) When itr == [1, 2, 3, 4, 5] , f(itr[1:]) is f([2, 3, 4, 5]) which is [2, 3, 4, 5, 2, 3, 4, 5] .itr == [1, 2, 3, 4, 5]f(itr[1:])f([2, 3, 4, 5])[2, 3, 4, 5, 2, 3, 4, 5] So the argument is getting shorter, not longer, and you never reach the base case.所以争论越来越短,而不是越来越长,你永远不会达到基本情况。

The recursion should just pass itr[1:] , not f(itr[1:]) since the arguments to f() should be list elements, not the list itself.递归应该只传递itr[1:] ,而不是f(itr[1:])因为f()的参数应该是列表元素,而不是列表本身。

Your base case is also wrong.你的基本情况也是错误的。 mapr() must always return a list, but in the base case you're just returning the result of the function without wrapping it into a list. mapr()必须始终返回一个列表,但在基本情况下,您只是返回函数的结果而不将其包装到列表中。

It would be best to use an empty list as the base case, so the function works for empty lists.最好使用空列表作为基本情况,因此该函数适用于空列表。

def mapr(f, itr):
    if len(itr) == 0:
        return []
    else:
        amendedlist = [f(itr[0])] + mapr(f, itr[1:])
        #print(amendedlist)
        return amendedlist

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