问题:如何检查对象是列表还是元组(而不是字符串)?
这就是我通常做,以确定输入是一个list
/ tuple
-但不是str
。因为很多时候我偶然发现了一个错误,即一个函数str
错误地传递了一个对象,而目标函数确实for x in lst
假定这lst
实际上是一个list
or tuple
。
assert isinstance(lst, (list, tuple))
我的问题是:是否有更好的方法来实现这一目标?
This is what I normally do in order to ascertain that the input is a list
/tuple
– but not a str
. Because many times I stumbled upon bugs where a function passes a str
object by mistake, and the target function does for x in lst
assuming that lst
is actually a list
or tuple
.
assert isinstance(lst, (list, tuple))
My question is: is there a better way of achieving this?
回答 0
仅在python 2中(不是python 3):
assert not isinstance(lst, basestring)
实际上就是您想要的,否则您会错过很多像列表一样的东西,但它们不是list
or的子类tuple
。
In python 2 only (not python 3):
assert not isinstance(lst, basestring)
Is actually what you want, otherwise you’ll miss out on a lot of things which act like lists, but aren’t subclasses of list
or tuple
.
回答 1
请记住,在Python中,我们要使用“鸭子类型”。因此,任何类似列表的行为都可以视为列表。因此,不要检查列表的类型,只看它是否像列表一样。
但是字符串也像列表一样,通常这不是我们想要的。有时甚至是一个问题!因此,显式检查字符串,然后使用鸭子类型。
这是我写的一个有趣的函数。这是它的特殊版本,repr()
可以在尖括号('<‘,’>’)中打印任何序列。
def srepr(arg):
if isinstance(arg, basestring): # Python 3: isinstance(arg, str)
return repr(arg)
try:
return '<' + ", ".join(srepr(x) for x in arg) + '>'
except TypeError: # catch when for loop fails
return repr(arg) # not a sequence so just return repr
总体而言,这是干净优雅的。但是那张isinstance()
支票在那里做什么?这是一种hack。但这是必不可少的。
该函数以递归方式调用类似于列表的任何对象。如果我们不专门处理字符串,则将其视为列表,并一次拆分一个字符。但是,然后递归调用将尝试将每个字符视为一个列表-它将起作用!即使是一个字符的字符串也可以作为列表!该函数将继续递归调用自身,直到堆栈溢出为止。
像这样的函数,依赖于每个递归调用来分解要完成的工作,必须使用特殊情况的字符串-因为您不能将字符串分解为一个字符以下的字符串,甚至不能分解为一个以下的字符串-字符字符串的作用类似于列表。
注意:try
/ except
是表达我们意图的最干净的方法。但是,如果这段代码在某种程度上对时间很紧迫,我们可能要用某种测试来替换它,看看是否arg
是一个序列。除了测试类型,我们可能应该测试行为。如果它有一个.strip()
方法,它是一个字符串,所以不要认为它是一个序列。否则,如果它是可索引的或可迭代的,则它是一个序列:
def is_sequence(arg):
return (not hasattr(arg, "strip") and
hasattr(arg, "__getitem__") or
hasattr(arg, "__iter__"))
def srepr(arg):
if is_sequence(arg):
return '<' + ", ".join(srepr(x) for x in arg) + '>'
return repr(arg)
编辑:我最初写上面检查,__getslice__()
但我注意到在collections
模块文档中,有趣的方法是__getitem__()
; 这很有意义,这就是您索引对象的方式。这似乎比根本,__getslice__()
因此我更改了上面的内容。
Remember that in Python we want to use “duck typing”. So, anything that acts like a list can be treated as a list. So, don’t check for the type of a list, just see if it acts like a list.
But strings act like a list too, and often that is not what we want. There are times when it is even a problem! So, check explicitly for a string, but then use duck typing.
Here is a function I wrote for fun. It is a special version of repr()
that prints any sequence in angle brackets (‘<‘, ‘>’).
def srepr(arg):
if isinstance(arg, basestring): # Python 3: isinstance(arg, str)
return repr(arg)
try:
return '<' + ", ".join(srepr(x) for x in arg) + '>'
except TypeError: # catch when for loop fails
return repr(arg) # not a sequence so just return repr
This is clean and elegant, overall. But what’s that isinstance()
check doing there? That’s kind of a hack. But it is essential.
This function calls itself recursively on anything that acts like a list. If we didn’t handle the string specially, then it would be treated like a list, and split up one character at a time. But then the recursive call would try to treat each character as a list — and it would work! Even a one-character string works as a list! The function would keep on calling itself recursively until stack overflow.
Functions like this one, that depend on each recursive call breaking down the work to be done, have to special-case strings–because you can’t break down a string below the level of a one-character string, and even a one-character string acts like a list.
Note: the try
/except
is the cleanest way to express our intentions. But if this code were somehow time-critical, we might want to replace it with some sort of test to see if arg
is a sequence. Rather than testing the type, we should probably test behaviors. If it has a .strip()
method, it’s a string, so don’t consider it a sequence; otherwise, if it is indexable or iterable, it’s a sequence:
def is_sequence(arg):
return (not hasattr(arg, "strip") and
hasattr(arg, "__getitem__") or
hasattr(arg, "__iter__"))
def srepr(arg):
if is_sequence(arg):
return '<' + ", ".join(srepr(x) for x in arg) + '>'
return repr(arg)
EDIT: I originally wrote the above with a check for __getslice__()
but I noticed that in the collections
module documentation, the interesting method is __getitem__()
; this makes sense, that’s how you index an object. That seems more fundamental than __getslice__()
so I changed the above.
回答 2
H = "Hello"
if type(H) is list or type(H) is tuple:
## Do Something.
else
## Do Something.
H = "Hello"
if type(H) is list or type(H) is tuple:
## Do Something.
else
## Do Something.
回答 3
对于Python 3:
import collections.abc
if isinstance(obj, collections.abc.Sequence) and not isinstance(obj, str):
print("obj is a sequence (list, tuple, etc) but not a string")
在版本3.3中进行了更改:将集合抽象基类移至collections.abc模块。为了向后兼容,它们在此模块中也将继续可见,直到3.8版将停止工作为止。
对于Python 2:
import collections
if isinstance(obj, collections.Sequence) and not isinstance(obj, basestring):
print "obj is a sequence (list, tuple, etc) but not a string or unicode"
For Python 3:
import collections.abc
if isinstance(obj, collections.abc.Sequence) and not isinstance(obj, str):
print("obj is a sequence (list, tuple, etc) but not a string")
Changed in version 3.3: Moved Collections Abstract Base Classes to the collections.abc module. For backwards compatibility, they will continue to be visible in this module as well until version 3.8 where it will stop working.
For Python 2:
import collections
if isinstance(obj, collections.Sequence) and not isinstance(obj, basestring):
print "obj is a sequence (list, tuple, etc) but not a string or unicode"
回答 4
具有PHP风格的Python:
def is_array(var):
return isinstance(var, (list, tuple))
Python with PHP flavor:
def is_array(var):
return isinstance(var, (list, tuple))
回答 5
一般来说,在对象上进行迭代的函数不仅可以处理错误,还可以处理字符串,元组和列表。您当然可以使用isinstance
或鸭式输入来检查参数,但是为什么要这么做呢?
这听起来像是个反问,但事实并非如此。答案为“为什么我应该检查参数的类型?” 可能会建议解决实际问题,而不是感知到的问题。将字符串传递给函数时,为什么会出错?另外:如果将字符串传递给此函数是一个错误,是否将其他非列表/元组可迭代传递给它也是一个错误吗?为什么或者为什么不?
我认为这个问题的最常见答案可能是 f("abc")
期望该函数的行为就像编写的一样f(["abc"])
。在某些情况下,保护开发人员免受自身侵害比支持对字符串中的字符进行迭代的用例更有意义。但是我首先会考虑很长时间。
Generally speaking, the fact that a function which iterates over an object works on strings as well as tuples and lists is more feature than bug. You certainly can use isinstance
or duck typing to check an argument, but why should you?
That sounds like a rhetorical question, but it isn’t. The answer to “why should I check the argument’s type?” is probably going to suggest a solution to the real problem, not the perceived problem. Why is it a bug when a string is passed to the function? Also: if it’s a bug when a string is passed to this function, is it also a bug if some other non-list/tuple iterable is passed to it? Why, or why not?
I think that the most common answer to the question is likely to be that developers who write f("abc")
are expecting the function to behave as though they’d written f(["abc"])
. There are probably circumstances where it makes more sense to protect developers from themselves than it does to support the use case of iterating across the characters in a string. But I’d think long and hard about it first.
回答 6
尝试此操作以提高可读性和最佳做法:
Python2
import types
if isinstance(lst, types.ListType) or isinstance(lst, types.TupleType):
# Do something
Python3
import typing
if isinstance(lst, typing.List) or isinstance(lst, typing.Tuple):
# Do something
希望能帮助到你。
Try this for readability and best practices:
Python2
import types
if isinstance(lst, types.ListType) or isinstance(lst, types.TupleType):
# Do something
Python3
import typing
if isinstance(lst, typing.List) or isinstance(lst, typing.Tuple):
# Do something
Hope it helps.
回答 7
该str
对象没有__iter__
属性
>>> hasattr('', '__iter__')
False
所以你可以检查一下
assert hasattr(x, '__iter__')
这也AssertionError
将为其他任何不可迭代的对象带来好处。
编辑: 正如蒂姆在评论中提到的那样,这仅适用于python 2.x,而不是3.x
The str
object doesn’t have an __iter__
attribute
>>> hasattr('', '__iter__')
False
so you can do a check
assert hasattr(x, '__iter__')
and this will also raise a nice AssertionError
for any other non-iterable object too.
Edit: As Tim mentions in the comments, this will only work in python 2.x, not 3.x
回答 8
这并不是要直接回答OP,而是要分享一些相关想法。
我对上面的@steveha回答非常感兴趣,这似乎举了一个鸭子输入似乎中断的示例。换个角度说,他的例子表明鸭子的分类很难遵循,但是并不能说明str
值得任何特殊处理。
毕竟,非str
类型(例如,维护一些复杂的递归结构的用户定义类型)可能导致@steveha srepr
函数引起无限递归。尽管这确实不太可能,但我们不能忽略这种可能性。因此,与其特殊外壳str
中srepr
,我们应该明确,我们想要什么srepr
在无限递归产生时的事情情况。
似乎一种合理的方法是srepr
暂时中断当前递归list(arg) == [arg]
。这,其实,彻底解决这个问题str
,没有任何isinstance
。
但是,真正复杂的递归结构可能会导致无限循环,list(arg) == [arg]
永远不会发生。因此,尽管上面的检查很有用,但还不够。我们需要对递归深度进行严格限制。
我的观点是,如果您打算处理任意参数类型,则str
通过鸭子类型进行处理要比处理(理论上)遇到的更通用类型容易得多。因此,如果您需要排除str
实例,则应该要求该参数是您明确指定的几种类型之一的实例。
This is not intended to directly answer the OP, but I wanted to share some related ideas.
I was very interested in @steveha answer above, which seemed to give an example where duck typing seems to break. On second thought, however, his example suggests that duck typing is hard to conform to, but it does not suggest that str
deserves any special handling.
After all, a non-str
type (e.g., a user-defined type that maintains some complicated recursive structures) may cause @steveha srepr
function to cause an infinite recursion. While this is admittedly rather unlikely, we can’t ignore this possibility. Therefore, rather than special-casing str
in srepr
, we should clarify what we want srepr
to do when an infinite recursion results.
It may seem that one reasonable approach is to simply break the recursion in srepr
the moment list(arg) == [arg]
. This would, in fact, completely solve the problem with str
, without any isinstance
.
However, a really complicated recursive structure may cause an infinite loop where list(arg) == [arg]
never happens. Therefore, while the above check is useful, it’s not sufficient. We need something like a hard limit on the recursion depth.
My point is that if you plan to handle arbitrary argument types, handling str
via duck typing is far, far easier than handling the more general types you may (theoretically) encounter. So if you feel the need to exclude str
instances, you should instead demand that the argument is an instance of one of the few types that you explicitly specify.
回答 9
我在tensorflow中找到了一个名为is_sequence的函数。
def is_sequence(seq):
"""Returns a true if its input is a collections.Sequence (except strings).
Args:
seq: an input sequence.
Returns:
True if the sequence is a not a string and is a collections.Sequence.
"""
return (isinstance(seq, collections.Sequence)
and not isinstance(seq, six.string_types))
而且我已经证实它可以满足您的需求。
I find such a function named is_sequence in tensorflow.
def is_sequence(seq):
"""Returns a true if its input is a collections.Sequence (except strings).
Args:
seq: an input sequence.
Returns:
True if the sequence is a not a string and is a collections.Sequence.
"""
return (isinstance(seq, collections.Sequence)
and not isinstance(seq, six.string_types))
And I have verified that it meets your needs.
回答 10
我在测试用例中执行此操作。
def assertIsIterable(self, item):
#add types here you don't want to mistake as iterables
if isinstance(item, basestring):
raise AssertionError("type %s is not iterable" % type(item))
#Fake an iteration.
try:
for x in item:
break;
except TypeError:
raise AssertionError("type %s is not iterable" % type(item))
未经生成器测试,我认为如果通过生成器,您将处于下一个“收益”状态,这可能会使下游情况恶化。但是再说一次,这是一个“单元测试”
I do this in my testcases.
def assertIsIterable(self, item):
#add types here you don't want to mistake as iterables
if isinstance(item, basestring):
raise AssertionError("type %s is not iterable" % type(item))
#Fake an iteration.
try:
for x in item:
break;
except TypeError:
raise AssertionError("type %s is not iterable" % type(item))
Untested on generators, I think you are left at the next ‘yield’ if passed in a generator, which may screw things up downstream. But then again, this is a ‘unittest’
回答 11
以“鸭子打字”的方式
try:
lst = lst + []
except TypeError:
#it's not a list
要么
try:
lst = lst + ()
except TypeError:
#it's not a tuple
分别。这避免了isinstance
/ hasattr
内省的东西。
您也可以反之亦然:
try:
lst = lst + ''
except TypeError:
#it's not (base)string
所有变体实际上都不会更改变量的内容,而是暗示了重新分配。我不确定这在某些情况下是否不受欢迎。
有趣的是,在任何情况下,如果是列表(不是元组),则在“就地”赋值时都不会引发+=
no 。这就是为什么以这种方式完成分配的原因。也许有人可以阐明原因。TypeError
lst
In “duck typing” manner, how about
try:
lst = lst + []
except TypeError:
#it's not a list
or
try:
lst = lst + ()
except TypeError:
#it's not a tuple
respectively. This avoids the isinstance
/ hasattr
introspection stuff.
You could also check vice versa:
try:
lst = lst + ''
except TypeError:
#it's not (base)string
All variants do not actually change the content of the variable, but imply a reassignment. I’m unsure whether this might be undesirable under some circumstances.
Interestingly, with the “in place” assignment +=
no TypeError
would be raised in any case if lst
is a list (not a tuple). That’s why the assignment is done this way. Maybe someone can shed light on why that is.
回答 12
最简单的方法…使用any
和isinstance
>>> console_routers = 'x'
>>> any([isinstance(console_routers, list), isinstance(console_routers, tuple)])
False
>>>
>>> console_routers = ('x',)
>>> any([isinstance(console_routers, list), isinstance(console_routers, tuple)])
True
>>> console_routers = list('x',)
>>> any([isinstance(console_routers, list), isinstance(console_routers, tuple)])
True
simplest way… using any
and isinstance
>>> console_routers = 'x'
>>> any([isinstance(console_routers, list), isinstance(console_routers, tuple)])
False
>>>
>>> console_routers = ('x',)
>>> any([isinstance(console_routers, list), isinstance(console_routers, tuple)])
True
>>> console_routers = list('x',)
>>> any([isinstance(console_routers, list), isinstance(console_routers, tuple)])
True
回答 13
鸭式打字的另一种形式,可以帮助区分类似字符串的对象和其他类似序列的对象。
类字符串对象的字符串表示形式是字符串本身,因此您可以检查是否从str
构造函数中返回了相等的对象:
# If a string was passed, convert it to a single-element sequence
if var == str(var):
my_list = [var]
# All other iterables
else:
my_list = list(var)
这应该适用于与str
所有可迭代对象兼容的所有对象。
Another version of duck-typing to help distinguish string-like objects from other sequence-like objects.
The string representation of string-like objects is the string itself, so you can check if you get an equal object back from the str
constructor:
# If a string was passed, convert it to a single-element sequence
if var == str(var):
my_list = [var]
# All other iterables
else:
my_list = list(var)
This should work for all objects compatible with str
and for all kinds of iterable objects.
回答 14
Python 3具有以下功能:
from typing import List
def isit(value):
return isinstance(value, List)
isit([1, 2, 3]) # True
isit("test") # False
isit({"Hello": "Mars"}) # False
isit((1, 2)) # False
因此,要同时检查列表和元组,将是:
from typing import List, Tuple
def isit(value):
return isinstance(value, List) or isinstance(value, Tuple)
Python 3 has this:
from typing import List
def isit(value):
return isinstance(value, List)
isit([1, 2, 3]) # True
isit("test") # False
isit({"Hello": "Mars"}) # False
isit((1, 2)) # False
So to check for both Lists and Tuples, it would be:
from typing import List, Tuple
def isit(value):
return isinstance(value, List) or isinstance(value, Tuple)
回答 15
assert (type(lst) == list) | (type(lst) == tuple), "Not a valid lst type, cannot be string"
assert (type(lst) == list) | (type(lst) == tuple), "Not a valid lst type, cannot be string"
回答 16
做这个
if type(lst) in (list, tuple):
# Do stuff
Just do this
if type(lst) in (list, tuple):
# Do stuff
回答 17
在python> 3.6中
import collections
isinstance(set(),collections.abc.Container)
True
isinstance([],collections.abc.Container)
True
isinstance({},collections.abc.Container)
True
isinstance((),collections.abc.Container)
True
isinstance(str,collections.abc.Container)
False
in python >3.6
import collections
isinstance(set(),collections.abc.Container)
True
isinstance([],collections.abc.Container)
True
isinstance({},collections.abc.Container)
True
isinstance((),collections.abc.Container)
True
isinstance(str,collections.abc.Container)
False
回答 18
我倾向于这样做(如果真的必须这样做的话):
for i in some_var:
if type(i) == type(list()):
#do something with a list
elif type(i) == type(tuple()):
#do something with a tuple
elif type(i) == type(str()):
#here's your string
I tend to do this (if I really, really had to):
for i in some_var:
if type(i) == type(list()):
#do something with a list
elif type(i) == type(tuple()):
#do something with a tuple
elif type(i) == type(str()):
#here's your string
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