python：最优雅的方式来散布带有元素的列表

Input:

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

Output:

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

What's the most elegant (read: Pythonic) way to write intersperse?

什么是最优雅的(阅读:Pythonic)写散布的方式?

12 个解决方案

#1

I would have written a generator myself, but like this:

我自己会写一个发电机,但是这样:

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

#2

itertools to the rescue
- or -
How many itertools functions can you use in one line?

itertools拯救 - 或者 - 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)

Usage:

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

#3

Another option that works for sequences:

另一个适用于序列的选项:

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

#4

I would go with a simple generator.

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

and then you can get your list like so:

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']

alternatively you could do:

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

I'm not sure which is more pythonic

我不确定哪个更pythonic

#5

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]

[Edit] Corrected code below:

[编辑]修改后的代码如下:

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']

#6

How about:

from itertools import chain,izip_longest

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

#7

Dunno if it's pythonic, but it's pretty simple:

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]

#8

Solution is trivial using more_itertools.intersperse:

使用more_itertools.intersperse解决方案是微不足道的:

>>> 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']

Technically, this answer isn't "writing" intersperse, it's just using it from another library. But it might save others from having to reinvent the wheel.

从技术上讲,这个答案不是“写”散布,它只是从另一个库中使用它。但它可能会让其他人不必重新发明*。

#9

This works:

>>> 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)

If you don't want a trailing 666, then return reduce(...)[:-1].

如果你不想要尾随666,那么返回reduce(...)[: - 1]。

#10

I just came up with this now, googled to see if there was something better... and IMHO there wasn't :-)

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]

#11

I believe this one looks pretty nice and easy to grasp compared to the yield next(iterator) or itertools.iterator_magic() one :)

我认为与yield next(iterator)或itertools.iterator_magic()之一相比,这个看起来相当不错且容易掌握:)

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]

#12

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

Should work for lists or generators.

应该适用于列表或生成器。