Question

这一点很清楚：

a = ['a', 'b', 'c']
x = [v for v in a]

但我不知道该怎么做：

sep = 5
# ??? y = [v + sep for v in a]
print y # expected ['a', 5, 'b', 5, 'c', 5]

如何编写每个源元素包含多个元素的列表推导？

我对此代码的优化不感兴趣：请不要在这些代码中引用[:]运算符或join方法或其他内容。我的代码需要列表理解。我目前唯一的选择是4行for循环，这很不方便：

y = []
for v in a:
    y.append(v)
    y.append(sep)

Answer 1

使用嵌套列表理解：

>>> a = ['a','b','c']
>>> [item  for x in a for item in (x, 5)]
['a', 5, 'b', 5, 'c', 5]

Answer 2

您可以先构建list tuples，然后使用itertools.chain.from_iterable展平结果列表：

>>> sep = 5
>>> a = ['a', 'b', 'c']
>>> 
>>> import itertools
>>> list(itertools.chain.from_iterable((elem, sep) for elem in a))
['a', 5, 'b', 5, 'c', 5]

Answer 3

只需使用功能sum()来展平列表，[[v,sep] for v in a]将生成[['a', 5], ['b', 5], ['c', 5]]

a = ['a', 'b', 'c']
x = [v for v in a]
sep = 5
y = sum([[v,sep] for v in a],[])
print y
#['a', 5, 'b', 5, 'c', 5]

Answer 4

这是展示Python的itertool包的强大功能的完美问题

repeat:创建一个对象的无限（或极限）迭代

izip:转换迭代。

chain展开可迭代的

>>> from itertools import izip, chain, repeat
>>> a = ['a', 'b', 'c']
>>> list(chain.from_iterable(izip(a, repeat(5))))
['a', 5, 'b', 5, 'c', 5]

如果你有微观优化的诀窍，你可能有兴趣知道，这比列表理解更快

>>> stmt1 = """
list(chain.from_iterable((elem, 5) for elem in a))
"""
>>> stmt2 = """
list(chain.from_iterable(izip(a, repeat(5))))
"""
>>> timeit.timeit(stmt=stmt1, setup="from __main__ import chain, a", number=100000)
1.3136643729533688
>>> timeit.timeit(stmt=stmt2, setup="from __main__ import chain, izip, repeat, a", number=100000)
0.8959859753707633
>>>

Answer 5

result = [5 if i % 2 else a[i/2] for i in range(2*len(a))]

（并不是说你应该把这样的代码投入生产）

Answer 6

>>> a = ['a', 'b', 'c']
>>> [item for sublist in zip(a, [5]*len(a)) for item in sublist]
['a', 5, 'b', 5, 'c', 5]

要打破它：

>>> [5]*len(a)
[5, 5, 5]

>>> zip(a, [5]*len(a))
[('a', 5), ('b', 5), ('c', 5)]

然后使用以下习语展平元组列表：

[item for sublist in l for item in sublist]

使用timeit，这种方法比最快的itertools方法快15％

>>> stmt3 = """
[item for sublist in zip(a, [5]*len(a)) for item in sublist]
"""
>>> timeit.timeit(stmt=stmt3, setup="a = ['a', 'b', 'c']", number=100000)

列表理解每个条目有几个元素

6 个答案: