python：用元素散布列表的最优雅方式

Question

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 ?

Answer 1

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

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

Answer 2

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

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"]

Answer 3

Another option that works for sequences:

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

Answer 4

Solution is trivial using more_itertools.intersperse :

>>> 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.

Answer 5

I would go with a simple generator.

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:

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

alternatively you could do:

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

Answer 6

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:

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']

Answer 7

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

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

Answer 8

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

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]

Answer 9

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

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

Answer 10

How about:

from itertools import chain,izip_longest

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

Answer 11

The basic and easy you could do is:

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

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

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

output:

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

Answer 12

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] .

Answer 13

Seems general and efficient:

def intersperse(lst, fill=...):
    """
    >>> list(intersperse([1,2,3,4]))
    [1, Ellipsis, 2, Ellipsis, 3, Ellipsis, 4]
    """
    return chain(*zip(lst[:-1], repeat(fill)), [lst[-1]])

Answer 14

You can use Python's list comprehension:

def intersperse(iterable, element):
    return [iterable[i // 2] if i % 2 == 0 else element for i in range(2 * len(iterable) - 1)]

Answer 15

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

Should work for lists or generators.