从Python的奇数/偶数列表中删除偶数/奇数 [英] Remove an even/odd number from an odd/even Python list

问题描述

我试图更好地理解Python中的列表理解.我完成了一个关于代码战的在线挑战，提供了一个非常精致的解决方案，如下所示.

I am trying to better understand list comprehension in Python. I completed an online challenge on codewars with a rather inelegant solution, given below.

挑战是:

1. 给出一个偶数和一个奇数的列表，返回奇数
2. 给出一个奇数和一个偶数的列表，返回偶数

我对此的(不明确的)解决方案是:

My (inelegant) solution to this was:

def find_outlier(integers):
    o = []
    e = []
    for i in integers:
        if i % 2 == 0:
            e.append(i)
        else:
            o.append(i)
    # use sums to return int type
    if len(o) == 1:
        return sum(o)
    else:
        return sum(e)

哪个工作正常，但似乎蛮力的.我是否认为以(c0>和e等占位符列表开头的(大多数)函数很像菜鸟似的")错误?

Which works fine, but seems to be pretty brute force. Am I wrong in thinking that starting (most) functions with placeholder lists like o and e is pretty "noob-like"?

我想更好地理解为什么此解决方案适用于奇数列表，但在偶数列表上失败，以更好地理解列表理解:

I would love to better understand why this solution works for the odd list, but fails on the even list, in an effort to better understand list comprehension:

def find_outlier(integers):
    if [x for x in integers if x % 2 == 0]:
       return [x for x in integers if x % 2 == 0]
    elif [x for x in integers if x % 2 != 0]:
       return [x for x in integers if x % 2 != 0]
    else:
        print "wtf!"

o = [1,3,4,5]
e = [2,4,6,7]

In[1]: find_outlier(o)
Out[1]: [4]

In[2]: find_outlier(e)
Out[2]: [2, 4, 6]

Out[2]应该返回7的位置.

预先感谢您提供任何见解.

Thanks in advance for any insights.

推荐答案

您的尝试失败，因为第一个if是始终为真.您将始终拥有至少包含1个元素的列表；要么单数是奇数，并且您测试了一个带有所有偶数的列表，否则您的列表中有一个 one 偶数.只有 empty 列表为假.

Your attempt fails because the first if is always going to be true. You'll always have a list with at least 1 element; either the odd one out is odd and you tested a list with all even numbers, otherwise you have a list with the one even number in it. Only an empty list would be false.

列表理解并不是这里的最佳解决方案，不是.尝试用最少的已检查元素数来解决问题(前两个元素，如果它们的类型不同，则获得第3个打破平局，否则进行迭代，直到找到不适合尾部的元素):

List comprehensions are not the best solution here, no. Try to solve it instead with the minimum number of elements checked (the first 2 elements, if they differ in type get a 3rd to break the tie, otherwise iterate until you find the one that doesn't fit in the tail):

def find_outlier(iterable):
    it = iter(iterable)
    first = next(it)
    second = next(it)
    parity = first % 2
    if second % 2 != parity:
        # odd one out is first or second, 3rd will tell which
        return first if next(it) % 2 != parity else second
    else:
        # the odd one out is later on; iterate until we find the exception
        return next(i for i in it if i % 2 != parity)

如果输入可迭代输入中的元素少于3个，或者没有异常，则以上内容将引发StopIteration异常.如果出现多个异常(例如2个偶数后跟2个奇数；在这种情况下将返回第一个奇数)，它也将无法处理.

The above will throw a StopIteration exception if there are either fewer than 3 elements in the input iterable, or there is no exception to be found. It also won't handle the case where there is more than one exception (e.g. 2 even followed by 2 odd; the first odd value would be returned in that case).

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