Python – What’s the canonical way to check for type in Python

pythontypes

What is the best way to check whether a given object is of a given type? How about checking whether the object inherits from a given type?

Let's say I have an object o. How do I check whether it's a str?

Best Answer

To check if o is an instance of str or any subclass of str, use isinstance (this would be the "canonical" way):

if isinstance(o, str):

To check if the type of o is exactly str (exclude subclasses):

if type(o) is str:

The following also works, and can be useful in some cases:

if issubclass(type(o), str):

See Built-in Functions in the Python Library Reference for relevant information.

One more note: in this case, if you're using Python 2, you may actually want to use:

if isinstance(o, basestring):

because this will also catch Unicode strings (unicode is not a subclass of str; both str and unicode are subclasses of basestring). Note that basestring no longer exists in Python 3, where there's a strict separation of strings (str) and binary data (bytes).

Alternatively, isinstance accepts a tuple of classes. This will return True if o is an instance of any subclass of any of (str, unicode):

if isinstance(o, (str, unicode)):

Related Solutions

Python – Calling a function of a module by using its name (a string)

Assuming module foo with method bar:

import foo
method_to_call = getattr(foo, 'bar')
result = method_to_call()

You could shorten lines 2 and 3 to:

result = getattr(foo, 'bar')()

if that makes more sense for your use case.

You can use getattr in this fashion on class instance bound methods, module-level methods, class methods... the list goes on.

Python – How to safely create a nested directory in Python

On Python ≥ 3.5, use pathlib.Path.mkdir:

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:

Try os.path.exists, and consider os.makedirs for the creation.

import os
if not os.path.exists(directory):
    os.makedirs(directory)

As noted in comments and elsewhere, there's a race condition – if the directory is created between the os.path.exists and the os.makedirs calls, the os.makedirs will fail with an OSError. Unfortunately, blanket-catching OSError and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.

One option would be to trap the OSError and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

Alternatively, there could be a second os.path.exists, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.

Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.

Modern versions of Python improve this code quite a bit, both by exposing FileExistsError (in 3.3+)...

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

...and by allowing a keyword argument to os.makedirs called exist_ok (in 3.2+).

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

Best Answer

Related Solutions

Python – Calling a function of a module by using its name (a string)

Python – How to safely create a nested directory in Python

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