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如何检查变量是否为具有python 2和3兼容性的字符串

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问题：如何检查变量是否为具有python 2和3兼容性的字符串
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问题：如何检查变量是否为具有python 2和3兼容性的字符串

我知道我可以isinstance(x, str)在python-3.x中使用：但我还需要检查python-2.x中是否还有字符串。会isinstance(x, str)按预期在python-2.x的？还是我需要检查版本并使用isinstance(x, basestr)？

具体来说，在python-2.x中：

>>>isinstance(u"test", str)
False

和python-3.x没有 u"foo"

点击查看英文原文

I’m aware that I can use: isinstance(x, str) in python-3.x but I need to check if something is a string in python-2.x as well. Will isinstance(x, str) work as expected in python-2.x? Or will I need to check the version and use isinstance(x, basestr)?

Specifically, in python-2.x:

>>>isinstance(u"test", str)
False

and python-3.x does not have u"foo"

回答 0

如果要编写兼容2.x和3.x的代码，则可能要使用六个：

from six import string_types
isinstance(s, string_types)
点击查看英文原文

If you’re writing 2.x-and-3.x-compatible code, you’ll probably want to use six:

from six import string_types
isinstance(s, string_types)

回答 1

我发现不依赖六个软件包的最简洁的方法是：

try:
  basestring
except NameError:
  basestring = str

然后，假设您以最通用的方式检查了Python 2中的字符串，

isinstance(s, basestring)

现在也将适用于Python 3+。

点击查看英文原文

The most terse approach I’ve found without relying on packages like six, is:

try:
  basestring
except NameError:
  basestring = str

then, assuming you’ve been checking for strings in Python 2 in the most generic manner,

isinstance(s, basestring)

will now also work for Python 3+.

回答 2

那在所有情况下都可行吗？

isinstance(x, ("".__class__, u"".__class__))
点击查看英文原文

What about this, works in all cases?

isinstance(x, ("".__class__, u"".__class__))

回答 3

这是@Lev Levitsky的答案，重写了一下。

try:
    isinstance("", basestring)
    def isstr(s):
        return isinstance(s, basestring)
except NameError:
    def isstr(s):
        return isinstance(s, str)

该try/ except测试只进行一次，然后定义总是工作，并尽可能快的功能。

编辑：实际上，我们甚至不需要调用isinstance()；我们只需要评估一下basestring，看看是否得到NameError：

try:
    basestring  # attempt to evaluate basestring
    def isstr(s):
        return isinstance(s, basestring)
except NameError:
    def isstr(s):
        return isinstance(s, str)

我认为，调用会更容易isinstance()。

点击查看英文原文

This is @Lev Levitsky’s answer, re-written a bit.

try:
    isinstance("", basestring)
    def isstr(s):
        return isinstance(s, basestring)
except NameError:
    def isstr(s):
        return isinstance(s, str)

The try/except test is done once, and then defines a function that always works and is as fast as possible.

EDIT: Actually, we don’t even need to call isinstance(); we just need to evaluate basestring and see if we get a NameError:

try:
    basestring  # attempt to evaluate basestring
    def isstr(s):
        return isinstance(s, basestring)
except NameError:
    def isstr(s):
        return isinstance(s, str)

I think it is easier to follow with the call to isinstance(), though.

回答 4

该future库添加了（兼容 Python 2）兼容名称，因此您可以继续编写Python 3。您可以简单地执行以下操作：

from builtins import str
isinstance(x, str)

要安装它，只需执行pip install future。

作为一个警告，它仅支持python>=2.6，>=3.3但它是更现代的比six，这是唯一的建议，如果使用python 2.5

点击查看英文原文

The future library adds (to Python 2) compatible names, so you can continue writing Python 3. You can simple do the following:

from builtins import str
isinstance(x, str)

To install it, just execute pip install future.

As a caveat, it only support python>=2.6,>=3.3, but it is more modern than six, which is only recommended if using python 2.5

回答 5

也许使用类似的解决方法

def isstr(s):
    try:
        return isinstance(s, basestring)
    except NameError:
        return isinstance(s, str)
点击查看英文原文

Maybe use a workaround like

def isstr(s):
    try:
        return isinstance(s, basestring)
    except NameError:
        return isinstance(s, str)

回答 6

您可以通过调用来获取对象的类object.__class__，以便检查object是否为默认字符串类型：

    isinstance(object,"".__class__)

而且您可以将以下内容放在代码的顶部，以便用引号括起来的字符串在python 2中的unicode中：

    from __future__ import unicode_literals
点击查看英文原文

You can get the class of an object by calling object.__class__, so in order to check if object is the default string type:

    isinstance(object,"".__class__)

And You can place the following in the top of Your code so that strings enclosed by quotes are in unicode in python 2:

    from __future__ import unicode_literals

回答 7

您可以在代码的开头尝试此操作：

from __future__ import print_function
import sys
if sys.version[0] == "2":
    py3 = False
else:
    py3 = True
if py3: 
    basstring = str
else:
    basstring = basestring

然后在代码中：

anystring = "test"
# anystring = 1
if isinstance(anystring, basstring):
    print("This is a string")
else:
    print("No string")
点击查看英文原文

You can try this at the beginning of your code:

from __future__ import print_function
import sys
if sys.version[0] == "2":
    py3 = False
else:
    py3 = True
if py3: 
    basstring = str
else:
    basstring = basestring

and later in the code:

anystring = "test"
# anystring = 1
if isinstance(anystring, basstring):
    print("This is a string")
else:
    print("No string")

回答 8

小心！在Python 2，str并且bytes基本上是相同的。如果您试图区分两者，则可能会导致错误。

>>> size = 5    
>>> byte_arr = bytes(size)
>>> isinstance(byte_arr, bytes)
True
>>> isinstance(byte_arr, str)
True
点击查看英文原文

Be careful! In python 2, str and bytes are essentially the same. This can cause a bug if you are trying to distinguish between the two. 

>>> size = 5    
>>> byte_arr = bytes(size)
>>> isinstance(byte_arr, bytes)
True
>>> isinstance(byte_arr, str)
True

回答 9

类型（字符串）== str

如果它是一个字符串，则返回true；否则，则返回false

点击查看英文原文

type(string) == str

returns true if its a string, and false if not