如果你需要转换为一个嵌套的列表，答案将是:

a = [(1, u'abc'), (2, u'def')]
print([list(i[0]) for i in a])

输出:

[[1], [2]]

使用zip函数解耦元素:

>>> inpt = [(1, u'abc'), (2, u'def')]
>>> unzipped = zip(*inpt)
>>> print unzipped
[(1, 2), (u'abc', u'def')]
>>> print list(unzipped[0])
[1, 2]

编辑(@BradSolomon): 以上内容适用于Python 2。X，其中zip返回一个列表。

在Python 3中。X, zip返回一个迭代器，下面的等价于上面的:

>>> print(list(list(zip(*inpt))[0]))
[1, 2]

>>> a = [(1, u'abc'), (2, u'def')]
>>> [i[0] for i in a]
[1, 2]

如果元组是唯一的，那么这可以工作

>>> a = [(1, u'abc'), (2, u'def')]
>>> a
[(1, u'abc'), (2, u'def')]
>>> dict(a).keys()
[1, 2]
>>> dict(a).values()
[u'abc', u'def']
>>> 

从性能的角度来看，在python3中。X

[i[0] for i in a]和list(zip(*a))[0]等价 它们比list(map(operator.itemgetter(0)， a))更快

Code

import timeit


iterations = 100000
init_time = timeit.timeit('''a = [(i, u'abc') for i in range(1000)]''', number=iterations)/iterations
print(timeit.timeit('''a = [(i, u'abc') for i in range(1000)]\nb = [i[0] for i in a]''', number=iterations)/iterations - init_time)
print(timeit.timeit('''a = [(i, u'abc') for i in range(1000)]\nb = list(zip(*a))[0]''', number=iterations)/iterations - init_time)

输出

3.491014136001468 e-05

3.422205176000717 e-05

要获取列表或元组的元素，可以遍历列表或元组

a = [(1, u'abc'), (2, u'def')]

list1 = [a[i][0] for i in range(len(a))]

print(list1)