如何检查列表中包含的类型？

Question

我要做到以下几点没有定义的函数：

if isinstance(x,(list,tuple)) and every_element_isinstance(x,basestring): 
    foobar 

即：implementing type checking

是否有速记/ builtin这个？

Answer 1

我认为这是最好的解决方案（如果我理解这个问题）

if isinstance(x, (list, tuple)) and all(isinstance(i, basestring) for i in x): 
    #do whatever 

Answer 2

if isinstance(x, (list, tuple)) and all([isinstance(i, basestring) for i in x]): 
    foobar 

出人意料的是，与[ ... ]列表理解更快这里比一个没有，都是具有短期和长期的名单：

短名单：

>>> timeit('isinstance(x, (list, tuple)) and all(isinstance(i, basestring) for i in x)', "x=['a','b','c']") 
2.7594685942680144 
>>> timeit('isinstance(x, (list, tuple)) and all(isinstance(i, basestring) for i in x)', "x=['a','b','c']") 
2.8013695153947538 
>>> timeit('isinstance(x, (list, tuple)) and all([isinstance(i, basestring) for i in x])', "x=['a','b','c']") 
2.4351678506033068 
>>> timeit('isinstance(x, (list, tuple)) and all([isinstance(i, basestring) for i in x])', "x=['a','b','c']") 
2.4491469896721583 

长的列表：

>>> timeit('isinstance(x, (list, tuple)) and all(isinstance(i, basestring) for i in x)', "x=['a','b','c'] * 1000", number=1000) 
1.3357901657891489 
>>> timeit('isinstance(x, (list, tuple)) and all(isinstance(i, basestring) for i in x)', "x=['a','b','c'] * 1000", number=1000) 
1.3305278872818462 
>>> timeit('isinstance(x, (list, tuple)) and all([isinstance(i, basestring) for i in x])', "x=['a','b','c'] * 1000", number=1000) 
1.2626525921055531 
>>> timeit('isinstance(x, (list, tuple)) and all([isinstance(i, basestring) for i in x])', "x=['a','b','c'] * 1000", number=1000) 
1.2881240045551863 

Answer 3

有没有内置定义generic types。但有a lot of validation libraries，它可以模仿此功能。

示例使用https://github.com/alecthomas/voluptuous：

>>> from voluptuous import Schema 
>>> s_list = Schema([basestring]) # only strings in a list are allowed 
>>> s_list("hello") 
... 
voluptuous.InvalidList: expected a list 
>>> s_list([123]) 
... 
voluptuous.InvalidList: invalid list value @ data[0] 
>>> s_list(["correct"]) 
["correct"] # returns the object, if validation was successful 

前几天，support for tuple was added到该库：

>>> s_tuple = voluptuous.Schema((basestring,)) 

现在将二者结合起来，让您的结果：

>>> from voluptuous import any 

# - this is now equivalent to your code 
# - raises Exceptions on invalid input 
>>> schema = Schema(any(s_list, s_tuple)) 

它甚至稍微快一点，即双倍isinstance：

>>> from timeit import timeit 

>>> timeit('(schema(i) for i in x)', "x=['a','b','c']") 
0.679318904876709 

>>> timeit(""" 
     (isinstance(x, (list, tuple)) 
     and all(isinstance(i, basestring)) for i in x)""", "x=['a','b','c']") 
0.7801780700683594 

Answer 4

随着妖娆“0.8.7”您可以从miku更新的答案，跳过“元组的一部分”：

>>> from voluptuous import Schema 
>>> from timeit import timeit 
>>> s_list = Schema([basestring]) # only strings in a list are allowed 
>>> timeit('(s_list(i) for i in x)', "x=['a','b','c']") 
0.503572940826416 
>>> timeit("(isinstance(x, (list, tuple)) and all(isinstance(i, basestring)) for i in x)", "x=['a','b','c']") 
0.5400209426879883