Python – How to tell PyCharm what type a parameter is expected to be

code-completionpycharmpythonpython-typingtype-hinting

When it comes to constructors, and assignments, and method calls, the PyCharm IDE is pretty good at analyzing my source code and figuring out what type each variable should be. I like it when it's right, because it gives me good code-completion and parameter info, and it gives me warnings if I try to access an attribute that doesn't exist.

But when it comes to parameters, it knows nothing. The code-completion dropdowns can't show anything, because they don't know what type the parameter will be. The code analysis can't look for warnings.

class Person:
    def __init__(self, name, age):
        self.name = name
        self.age = age

peasant = Person("Dennis", 37)
# PyCharm knows that the "peasant" variable is of type Person
peasant.dig_filth()   # shows warning -- Person doesn't have a dig_filth method

class King:
    def repress(self, peasant):
        # PyCharm has no idea what type the "peasant" parameter should be
        peasant.knock_over()   # no warning even though knock_over doesn't exist

King().repress(peasant)
# Even if I call the method once with a Person instance, PyCharm doesn't
# consider that to mean that the "peasant" parameter should always be a Person

This makes a certain amount of sense. Other call sites could pass anything for that parameter. But if my method expects a parameter to be of type, say, pygame.Surface, I'd like to be able to indicate that to PyCharm somehow, so it can show me all of Surface's attributes in its code-completion dropdown, and highlight warnings if I call the wrong method, and so on.

Is there a way I can give PyCharm a hint, and say "psst, this parameter is supposed to be of type X"? (Or perhaps, in the spirit of dynamic languages, "this parameter is supposed to quack like an X"? I'd be fine with that.)

EDIT: CrazyCoder's answer, below, does the trick. For any newcomers like me who want the quick summary, here it is:

class King:
    def repress(self, peasant):
        """
        Exploit the workers by hanging on to outdated imperialist dogma which
        perpetuates the economic and social differences in our society.

        @type peasant: Person
        @param peasant: Person to repress.
        """
        peasant.knock_over()   # Shows a warning. And there was much rejoicing.

The relevant part is the @type peasant: Person line of the docstring.

If you also go to File > Settings > Python Integrated Tools and set "Docstring format" to "Epytext", then PyCharm's View > Quick Documentation Lookup will pretty-print the parameter information instead of just printing all the @-lines as-is.

Best Answer

Yes, you can use special documentation format for methods and their parameters so that PyCharm can know the type. Recent PyCharm version supports most common doc formats.

For example, PyCharm extracts types from @param style comments.

See also reStructuredText and docstring conventions (PEP 257).

Another option is Python 3 annotations.

Please refer to the PyCharm documentation section for more details and samples.

Related Solutions

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.

Python – How to remove a trailing newline

Try the method rstrip() (see doc Python 2 and Python 3)

>>> 'test string\n'.rstrip()
'test string'

Python's rstrip() method strips all kinds of trailing whitespace by default, not just one newline as Perl does with chomp.

>>> 'test string \n \r\n\n\r \n\n'.rstrip()
'test string'

To strip only newlines:

>>> 'test string \n \r\n\n\r \n\n'.rstrip('\n')
'test string \n \r\n\n\r '

In addition to rstrip(), there are also the methods strip() and lstrip(). Here is an example with the three of them:

>>> s = "   \n\r\n  \n  abc   def \n\r\n  \n  "
>>> s.strip()
'abc   def'
>>> s.lstrip()
'abc   def \n\r\n  \n  '
>>> s.rstrip()
'   \n\r\n  \n  abc   def'

Best Answer

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Python – How to safely create a nested directory in Python

Python – How to remove a trailing newline

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