Python如何优雅的交错合并两个列表

Question

比如a = [1, 2, 3], b = [4, 5, 6]
合并为[1, 4, 2, 5, 3, 6]
你觉得怎么写比较优雅？

Answer 1

pythonfrom itertools import chain
list(chain.from_iterable(zip(a, b)))

# py2
list(chain(*zip(a, b)))

Answer 2

Correction: There is a problem with the previous code, update it once.

I don’t know if it’s elegant, but it should save memory:

def xmerge(a, b):
    alen, blen = len(a), len(b)
    mlen = min(alen, blen)
    for i in xrange(mlen):
        yield a[i]
        yield b[i]

    if alen > blen:
        for i in xrange(mlen, alen):
            yield a[i]
    else:
        for i in xrange(mlen, blen):
            yield b[i]

a = [1, 2, 3]
b = [5, 6, 7, 8, 9, 10]

c = [i for i in xmerge(a, b)]
print c

c = [i for i in xmerge(b, a)]
print c

Answer 3

It turns out that stackoverflow has already discussed it, and the writing method is very awkward. I personally love this:
The cycle/islice functions called are all from itertools

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    pending = len(iterables)
    nexts = cycle(iter(it).next for it in iterables)
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

Answer 4

pythondef xmerge(a, b):
  tmp = (list(a), list(b));
  return [tmp[i%2].pop(0) if tmp[i%2] else tmp[1-i%2].pop(0) for i in xrange(0, len(a) + len(b))]


print xmerge([1,2,3], [5,6,7,8,9])
print xmerge([1,2,3,4,5], [7,8,9])

Is this so?

//Halfway through writing, I searched for ternary expressions in python and remembered that I used python to feel like this kind of mythical beast flying by...
//But there is also a language with three elements written as if a then b else c, which I can still understand now...

Answer 5

python3.3
result = [list(zip(a, b))[i][j] for i in range(len(a)) for j in range(len(list(zip(a, b))[0]))]

虽然是一行，但是有点牵强，小括号太多

Answer 6

python# Python 3
def interpolate(*seqs):
    for items in zip(*seqs):
        yield from items

>>> list(interpoltae('string', 'asdfgh'))
['s', 'a', 't', 's', 'r', 'd', 'i', 'f', 'n', 'g', 'g', 'h']

There is a question, what should I do if these sequences are not all of equal length. The above solution is based on the smallest length:

python>>> list(interpolate('string', 'asdf'))
['s', 'a', 't', 's', 'r', 'd', 'i', 'f']

But what if you don’t choose the smallest one? Complementary characters? What characters should be added?

Answer 7

For elegant data processing, scipy series libraries are still needed.
There is a ready-made flatten function in matplotlib that can be used.

from matplotlib.cbook import flatten
a = [1, 2, 3]
b = [4, 5, 6]
list(flatten(zip(a,b)))

Answer 8

a.extend(b)

Answer 9

a = [1, 2, 3]
b = [4, 5, 6]

一般的方法

def slove(a, b):
c = []
i = 0
j = 0
while i<len(a) and j<len(b):
c.append(a[i])
c.append(b[j])
i += 1
j += 1
while i < len(a):
c.append(a[i])
i += 1
while j < len(b):
c.append(b[j])
j += 1

print(c)

if name == 'main':
slove(a, b)

Answer 10

a=[1,2,3]
b=[4,5,6]
a=set(a)
b=set(b)
c=list(a|b)

If it is just a merge of lists, can it be converted into set() and then the intersection operation
Set is faster than list when doing list merge.
But @lohocla4dam helped point out the shortcomings