Python - 将用户输入整数拆分为列表,其中每个条目为2位数 [英] Python - Split user input integer into list, where each entry is 2 digits
问题描述
所以我试图将一个任意长的用户输入整数分成一个列表,其中每个条目是2位数,如果该数字有一个奇数的整数,则将唯一的一位数作为第一个数字。 (然后我将继续在它前面加零)
So I'm trying to split an arbitrarily long user input integer into a list where each entry is 2 digits, and if the number has an odd amount of integers put the only single digit as the first digit. (Where I will then proceed to put a zero in front of it)
我知道将用户整数输入放入列表中如下所示:
I know that putting user integer input into a list looks like:
userintegerlist = [int(i) for i in str(user_input)]
print userintegerlist
我的输入(比如 45346
)看起来像 [4, 5,3,4,6]
。但我希望它看起来像: [4,53,46]
。或者,如果输入为 68482238
,则输入为: [68,48,22,38]
。
And my input (say it's 45346
) will look like [4,5,3,4,6]
. But I want it to look like: [4,53,46]
. Or if input is 68482238
, it will be: [68,48,22,38]
.
这可能吗?顺便说一下,所有的代码都在Python中。
Is this possible? All the code is in Python by the way.
推荐答案
你可以很容易地使用字符串方法,因为其他答案已经所示。我将引导您访问相关的 itertools中的石斑鱼食谱。
You can do it with string methods fairly easily, as other answers have already shown. I direct you to the related grouper recipe in itertools.
我想提一下,用数学做这件事可能更有效率:
I want to mention that it may be more efficient to do it with maths:
>>> n = 45346
>>> output = []
>>> while n:
... output.append(n % 100)
... n //= 100
...
>>> output = output[::-1]
>>> print output
[4, 53, 46]
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