Python 使用列表理解构建元组
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Use list comprehension to build a tuple
提问by Lim H.
How can I use list comprehension to build a tuple of 2-tuple from a list. It would be equivalent to
如何使用列表理解从列表中构建 2 元组的元组。这将相当于
tup = ()
for element in alist:
tup = tup + ((element.foo, element.bar),)
采纳答案by Pavel Anossov
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.
从技术上讲，它是一个生成器表达式。它就像一个列表推导式，但它是惰性求值的，不需要为中间列表分配内存。
For completeness, the list comprehension would look like this:
为了完整起见，列表推导式如下所示：
tup = tuple([(element.foo, element.bar) for element in alist])
PS: attrgetteris not faster (alisthas a million items here):
PS：attrgetter不是更快（alist这里有一百万个项目）：
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
