检查两个列表是否与Python类型相同

Question

我想检查两个列表是否对每个索引都有相同类型的项目.例如,如果我有

y = [3, "a"]
x = [5, "b"] 
z = ["b", 5]

检查应True为x和y.检查应该是False因为y并且z因为相同位置的元素类型不相等.

Answer 1

只是map他们各自的元素,type并比较那些:

>>> x = [5, "b"] 
>>> y = [3, "a"]
>>> z = ["b", 5]
>>> map(type, x) == map(type, y)
True
>>> map(type, x) == map(type, z)
False

对于Python 3,您还必须map通过使用list函数或列表推导将生成器转换为正确的列表:

>>> list(map(type, x)) == list(map(type, y))
True
>>> [type(i) for i in x] == [type(i) for i in z]
False

我做了一些时序分析,将上述溶液进行比较来的@timgeb,使用all和izip与输入在不同的位置上的第一个非匹配类型.正如预期的那样,map对于每个输入,解决方案所花费的时间几乎完全相同,而all+ izip解决方案可能非常快或需要三倍的时间,具体取决于第一个差异的位置.

In [52]: x = [1] * 1000 + ["s"] * 1000
In [53]: y = [2] * 1000 + ["t"] * 1000 # same types as x
In [54]: z = ["u"] * 1000 + [3] * 1000 # difference at first element
In [55]: u = [4] * 2000                # difference after first half
In [56]: %timeit map(type, x) == map(type, y)
10000 loops, best of 3: 129 µs per loop
In [58]: %timeit all(type(i) == type(j) for i, j in izip(x, y))
1000 loops, best of 3: 342 µs per loop
In [59]: %timeit all(type(i) == type(j) for i, j in izip(x, z))
1000000 loops, best of 3: 748 ns per loop
In [60]: %timeit all(type(i) == type(j) for i, j in izip(x, u))
10000 loops, best of 3: 174 µs per loop

Answer 2

懒惰的评价all:

>>> from itertools import izip
>>> all(type(a) == type(b) for a,b in izip(x,y))
True

zip在Python 3中使用常规,它已经返回一个生成器.

如果列表可以有不同的长度,只需预先检查长度.检查长度是一个非常快速的O(1)操作:

>>> len(x) == len(y) and all(type(a) == type(b) for a,b in izip(x,y))
True
>>> x = [5,"b",'foo']
>>> len(x) == len(y) and all(type(a) == type(b) for a,b in izip(x,y))
False

在and将短路评价,这意味着all甚至不会被如果长度不同调用.