Python 条件合并列表

假设我有两张清单

[16, 0, 0, ';', 17, 0, 2, ';', 0, 2, 1, ';']
  [-1, 0, ';', 0, -2, ';', -2, -1, ';']

有没有更简单的方法在满足“；”条件时有条件地合并这些列表元素字符，而不是单独遍历它们并组合它们

输出应该是

[16, 0, 0, -1, 0, ';', 17, 0, 2, 0, -2, ';', 0, 2, 1, -2, -2,';']

现在，我们可以压缩您的列表：

merged = []
for l1,l2 in zip(get_part(list1),get_part(list2)):
    merged.extend(l1)
    merged.extend(l2)
    merged.append(';')

当然，在它的核心，这本质上是对它们进行迭代并结合。。。所以在回答你的问题时，我认为没有更好的办法了

---

存储列表的更好方法可能是将它们存储为列表列表--例如：

这样存储，您只需执行以下操作：

merged = [ l1 + l2 for l1,l2 in zip(list1,list2) ]

---

更多itertools疯狂：

from itertools import groupby, chain, izip, repeat

a = [16, 0, 0, ';', 17, 0, 2, ';', 0, 2, 1, ';']
b = [-1, 0, ';', 0, -2, ';', -2, -1, ';']

wanted = [16, 0, 0, -1, 0, ";",
          17, 0, 2, 0, -2, ';',
          0, 2, 1, -2, -1,";"]

def split(items, sep=";"):
    return (group for key, group in
            groupby(items, lambda item: item != sep) if key)

got = list(
    chain.from_iterable(
        chain.from_iterable(
            izip(split(a), split(b), repeat([";"])))))
assert got == wanted

---

当然，谢谢你指出这一点

merged = [ l1 + l2 for l1,l2 in zip(list1,list2) ]

>>> from itertools import groupby, izip, chain
>>> L1=[16, 0, 0, ';', 17, 0, 2, ';', 0, 2, 1, ';']
>>> L2=[-1, 0, ';', 0, -2, ';', -2, -1, ';']    
>>> g1 = groupby(L1, key=';'.__eq__)
>>> g2 = groupby(L2, key=';'.__eq__)
>>> [i for i1,i2 in izip(g1, g2) for i in ([';'] if i1[0] is True else chain(i1[1], i2[1]))]
[16, 0, 0, -1, 0, ';', 17, 0, 2, 0, -2, ';', 0, 2, 1, -2, -1, ';']

from itertools import groupby, chain, izip, repeat

a = [16, 0, 0, ';', 17, 0, 2, ';', 0, 2, 1, ';']
b = [-1, 0, ';', 0, -2, ';', -2, -1, ';']

wanted = [16, 0, 0, -1, 0, ";",
          17, 0, 2, 0, -2, ';',
          0, 2, 1, -2, -1,";"]

def split(items, sep=";"):
    return (group for key, group in
            groupby(items, lambda item: item != sep) if key)

got = list(
    chain.from_iterable(
        chain.from_iterable(
            izip(split(a), split(b), repeat([";"])))))
assert got == wanted