使用列表内涵构建 tuple

我如何使用列表内涵从一个列表中构建一个由2个元组组成的元组

tup = ()
for element in alist:
tup = tup + ((element.foo, element.bar),)
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最佳答案 
tup = tuple((element.foo, element.bar) for element in alist)

Technically, it's a generator expression. It's like a list comprehension, but it's evaluated lazily and won't need to allocate memory for an intermediate list.

For completeness, the list comprehension would look like this:

tup = tuple([(element.foo, element.bar) for element in alist])

 

PS: attrgetter is not faster (alist has a million items here):

In [37]: %timeit tuple([(element.foo, element.bar) for element in alist])
1 loops, best of 3: 165 ms per loop


In [38]: %timeit tuple((element.foo, element.bar) for element in alist)
10 loops, best of 3: 155 ms per loop


In [39]: %timeit tuple(map(operator.attrgetter('foo','bar'), alist))
1 loops, best of 3: 283 ms per loop


In [40]: getter = operator.attrgetter('foo','bar')


In [41]: %timeit tuple(map(getter, alist))
1 loops, best of 3: 284 ms per loop


In [46]: %timeit tuple(imap(getter, alist))
1 loops, best of 3: 264 ms per loop
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You may use the following expression

tup = *[(element.foo, element.bar) for element in alist]

This will first of all generate a list of tuples and then it will convert that list of tuples into a tuple of tuples.

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Despite the already working answer:

tup = tuple((element.foo, element.bar) for element in alist)

The shortest way is (comma at end is mandatory!):

tup = *((elements.foo, elements.bar) for elements in alist),

It' s arguable which solution is the more readable or more pythonic.

Explanation of: *(...),


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