python:最优雅的方式来散布带有元素的列表

时间:2011-04-13 21:13:05

标签: python list

输入:

intersperse(666, ["once", "upon", "a", 90, None, "time"])

输出:

["once", 666, "upon", 666, "a", 666, 90, 666, None, 666, "time"]

intersperse的最优雅(读取:Pythonic)方法是什么?

13 个答案:

答案 0 :(得分:30)

我自己会写一个生成器,但是像这样:

def joinit(iterable, delimiter):
    it = iter(iterable)
    yield next(it)
    for x in it:
        yield delimiter
        yield x

答案 1 :(得分:17)

itertools救援 - 或 -
你可以在一行中使用多少个itertools函数?

from itertools import chain, izip, repeat, islice

def intersperse(delimiter, seq):
    return islice(chain.from_iterable(izip(repeat(delimiter), seq)), 1, None)

用法:

>>> list(intersperse(666, ["once", "upon", "a", 90, None, "time"])
["once", 666, "upon", 666, "a", 666, 90, 666, None, 666, "time"]

答案 2 :(得分:10)

适用于序列的另一个选项:

def intersperse(seq, value):
    res = [value] * (2 * len(seq) - 1)
    res[::2] = seq
    return res

答案 3 :(得分:3)

我会选择一个简单的发电机。

def intersperse(val, sequence):
    first = True
    for item in sequence:
        if not first:
            yield val
        yield item
        first = False

然后你就可以得到你的清单:

>>> list(intersperse(666, ["once", "upon", "a", 90, None, "time"]))
['once', 666, 'upon', 666, 'a', 666, 90, 666, None, 666, 'time']

或者你可以这样做:

def intersperse(val, sequence):
    for i, item in enumerate(sequence):
        if i != 0:
            yield val
        yield item

我不确定哪个更像pythonic

答案 4 :(得分:2)

def intersperse(word,your_list):
    x = [j for i in your_list for j in [i,word]]

>>> intersperse(666, ["once", "upon", "a", 90, None, "time"])
['once', 666, 'upon', 666, 'a', 666, 90, 666, None, 666, 'time', 666]

[修改] 以下修正后的代码:

def intersperse(word,your_list):
    x = [j for i in your_list for j in [i,word]]
    x.pop()
    return x

>>> intersperse(666, ["once", "upon", "a", 90, None, "time"])
['once', 666, 'upon', 666, 'a', 666, 90, 666, None, 666, 'time']

答案 5 :(得分:2)

使用more_itertools.intersperse解决方案很简单:

>>> from more_itertools import intersperse
>>> list(intersperse(666, ["once", "upon", "a", 90, None, "time"]))
['once', 666, 'upon', 666, 'a', 666, 90, 666, None, 666, 'time']

从技术上讲,这个答案不是"写" intersperse,它只是从其他图书馆使用它。但它可能会让其他人不必重新发明轮子。

答案 6 :(得分:1)

Dunno,如果它是pythonic,但它很简单:

def intersperse(elem, list):
    result = []
    for e in list:
      result.extend([e, elem])
    return result[:-1]

答案 7 :(得分:1)

怎么样:

from itertools import chain,izip_longest

def intersperse(x,y):
     return list(chain(*izip_longest(x,[],fillvalue=y)))

答案 8 :(得分:1)

我现在想出了这个,用谷歌搜索看看是否有更好的东西......而恕我直言那里没有: - )

def intersperse(e, l):    
    return list(itertools.chain(*[(i, e) for i in l]))[0:-1]

答案 9 :(得分:1)

您可以做的基本且简单的事情是:

a = ['abc','def','ghi','jkl']

# my separator is : || separator ||
# hack is extra thing : --

'--|| separator ||--'.join(a).split('--')

输出:

['abc','|| separator ||','def','|| separator ||','ghi','|| separator ||','jkl']

答案 10 :(得分:0)

这有效:

>>> def intersperse(e, l):
...    return reduce(lambda x,y: x+y, zip(l, [e]*len(l)))
>>> intersperse(666, ["once", "upon", "a", 90, None, "time"])
('once', 666, 'upon', 666, 'a', 666, 90, 666, None, 666, 'time', 666)

如果您不想跟踪666,请return reduce(...)[:-1]

答案 11 :(得分:0)

yield next(iterator)itertools.iterator_magic()相比,我认为它看起来不错并且易于掌握:)

def list_join_seq(seq, sep):
  for i, elem in enumerate(seq):
    if i > 0: yield sep
    yield elem

print(list(list_join_seq([1, 2, 3], 0)))  # [1, 0, 2, 0, 3]

答案 12 :(得分:-1)

def intersperse(items, delim):
    i = iter(items)
    return reduce(lambda x, y: x + [delim, y], i, [i.next()])

应该适用于列表或生成器。