python - 使用列表推导构建元组

Question

如何使用列表推导从列表中构建 2 元组的元组。它相当于

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

Answer 1

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

从技术上讲，它是 generator expression .这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。

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:attrgetter 并不快(alist 这里有一百万条):

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