递归函数和列表追加/扩展 [英] Recursive functions and lists appending/extending

问题描述

这是一个非常简单的代码，代替了一个更大的问题，但我希望我可以分块处理它。我将从我的第一个问题开始。

This is a very simple code in place of a bigger problem, but I'm hoping I can tackle it in chunks. I'll start with my first problem.

def testrecurse(z,target):
    x=[]
    if z<target:
        z*=2
        x.append(z)
        x.extend(testrecurse(z,target))
    return x

这是一个测试函数，可以帮助我的大脑退出递归。它需要一个数字，然后显示两个数字的所有乘法，直到达到目标数字。所以如果我输入：

This is a test function to help my brain out with recursion. It takes a number, then shows all the multiplications of two until it hits the target number. so if I enter:

testrecurse(1,1000)

我收到：

I receive:

[2, 4, 8, 16, 32, 64, 128, 256, 512, 1024]

很棒！输出看起来很好，很干净。但这是我的问题，我很难在我的输出中追加或添加第一个值。这是我想要的输出看起来像。

which is great! Output looks good and clean. But here's my problem, I'm having a hard time appending or adding that very first value, in my output. Here's what I want the output to look like.

[1,2, 4, 8, 16, 32, 64, 128, 256, 512, 1024]

我试过改变$ / b

I've tried changing

x=[] to x=[z]

$ b $然后我收到：

but then I receive:

[1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64, 128, 128, 256, 256, 512, 512, 1024, 1024]

任何帮助将不胜感激，我是新的递归，它使我的头疼。

any help would be appreciated, I'm new to recursion and it makes my head hurt.

推荐答案

How about this?

def testrecurse(z, target):
    if z >= target:
        return []
    return [z] + testrecurse(2 * z, target)

：

Example:

>>> testrecurse(1, 1000)
[1, 2, 4, 8, 16, 32, 64, 128, 256, 512]

请注意，它不再包含 1024 。如果你想这样做，把第三行改为

Note that it does not include 1024 any more. If you want this, change the third line to

        return [z]

当然，你通常不会递归地写这个，而是使用作为循环或 itertools.takewhile（）。

Of course you wouldn't normally write this recursively, but rather use a for loop or itertools.takewhile().

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