Question

>>> odd,even=[ ],[ ]
>>> [even.append(x) if x%2==0 else odd.append(x) for x in range(51)]
[None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None]
>>> odd
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49]

我的问题：即使没有在开头声明odd, even = [], []，是否可以将列表理解语句中的奇数和偶数分开？

Answer 1

最好只循环一次。这是6行，但它们是快速行

odd, even=[ ], [ ]
for x in range(51):
    if x%2:
        odd.append(x)
    else:
        even.append(x)

Answer 2

even,odd = [],[]
for x in range(51): 
    (odd if x%2 else even).append(x)

Answer 3

考虑一下

evens = [i for i in xrange(1,1000) if i % 2]
odds = [i for i in xrange(1,1000) if i % 2 != 0]

Answer 4

>>> even, odd = [[x for x in range(51) if x%2 == 0], [x for x in range(51) if x%2 == 1]]
>>> even
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50]
>>> odd
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49]

我不推荐这种方法，但是我想这是一种“只有一个循环”的方法：

>>> even, odd = zip(*[(x, x + 1) for x in range(0, 51, 2)])
>>> even
(0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50)
>>> odd
(1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51)

Answer 5

我想提请注意itertools，特别是Python帮助中的itertools recipes。在这里，我们可以使用partition函数（向后移植到Python 2.7）

from itertools import tee, ifilter, ifilterfalse

def partition(pred, iterable):
    'Use a predicate to partition entries into false entries and true entries'
    # partition(is_odd, range(10)) --> 0 2 4 6 8   and  1 3 5 7 9
    t1, t2 = tee(iterable)
    return ifilterfalse(pred, t1), ifilter(pred, t2)

def is_odd(n):
    return bool(n%2)

evens, odds = partition(is_odd, range(51))

print list(evens)
print list(odds)

实际上，docstring中的示例正好解释了这种情况的用法。它返回迭代器，因此在打印时需要使用list。

Answer 6

如果这是一个简化的问题，方便使用数字，那么我的答案是不相关的，但最简单的方法是做你所要求的是：

>>> odd, even = range(1, 51, 2), range(0, 51, 2)

这给了你：

>>> list(odd)
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49]
>>> list(even)
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50]

Answer 7

解决一般问题：

给定谓词函数predicate在给定对象时返回True或False，根据谓词将列表分成两个子列表，同时保留顺序

然后我们可以使用sorted功能稳定的事实，因此如果我们将predicate作为关键字进行排序，则列表有效地分为两部分保留每个部分内的顺序时的未知点，其左侧的谓词为False，右侧为True的值。

现在我们可以使用itertools.groupby使用相同的键在此未知点进行拆分。

from itertools import groupby

predicate = lambda x: x % 2 == 0

def group_by_predicate(L, pred):
    return [list(i) for k, i in groupby(sorted(L, key=pred), key=pred)]

odd, even = group_by_predicate(range(51), predicate)

即使predicate为True的值不均匀，也会有效...例如，如果predicate为True，则为素数。< / p>

Answer 8

试试这个版本：对我来说优雅，紧凑，易于理解

even,odd = [],[]
for x in range(51): x%2 and even.append(x) or odd.append(x)

列表理解建议

8 个答案: