type-hinting

问题 I have very simple Python function that subtracts scalar value from pandas Series: def sub(x: pd.Series, a: int) -> pd.Series: return x - a It works as expected. I'm using type hints to enable type checker directly in my PyCharm IDE. The issue here is I get this warning message: Expected type 'Series', got 'int' instead As you can see in the image below: I understand that Python is dynamically typed language so in some cases type checking with type hints has its own limitations. But this

问题 foo.py : kwargs = {"a": 1, "b": "c"} def consume(*, a: int, b: str) -> None: pass consume(**kwargs) mypy foo.py : error: Argument 1 to "consume" has incompatible type "**Dict[str, object]"; expected "int" error: Argument 1 to "consume" has incompatible type "**Dict[str, object]"; expected "str" This is because object is a supertype of int and str , and is therefore inferred. If I declare: from typing import TypedDict class KWArgs(TypedDict): a: int b: str and then annotate kwargs as KWArgs ,

问题 Can someone explain why inheriting from unparameterized and parameterized Callable : from typing import Callable from typing import NoReturn from typing import TypeVar T = TypeVar('T', str, int) C = Callable[[T], NoReturn] class Foo(Callable): def __call__(self, t: T): pass class Bar(C): def __call__(self, t: T): pass when passed to mypy raises errors for both Foo and Bar : tmp.py:13: error: Invalid base class tmp.py:19: error: Invalid base class 回答1: This is in part because classes at

问题 Why am I receiving this error when I run this code? Traceback (most recent call last): File "main.py", line 13, in <module> def twoSum(self, nums: list[int], target: int) -> list[int]: TypeError: 'type' object is not subscriptable nums = [4,5,6,7,8,9] target = 13 def twoSum(self, nums: list[int], target: int) -> list[int]: dictionary = {} answer = [] for i in range(len(nums)): secondNumber = target-nums[i] if(secondNumber in dictionary.keys()): secondIndex = nums.index(secondNumber) if(i !=

问题 This question already has answers here : What are type hints in Python 3.5? (5 answers) Closed 4 years ago . In other languages anything like the example throws a type error. Why not in Python? >>> def foo(a:int) -> str: return a+1 >>> foo(5) 6 回答1: Type hinting in Python is an optional addition to aid static code analysis and editors. From the PEP 484 -- Type Hints specification: Note that this PEP still explicitly does NOT prevent other uses of annotations, nor does it require (or forbid)

问题 This question already has answers here : What are type hints in Python 3.5? (5 answers) Closed 4 years ago . In other languages anything like the example throws a type error. Why not in Python? >>> def foo(a:int) -> str: return a+1 >>> foo(5) 6 回答1: Type hinting in Python is an optional addition to aid static code analysis and editors. From the PEP 484 -- Type Hints specification: Note that this PEP still explicitly does NOT prevent other uses of annotations, nor does it require (or forbid)

问题 I have a decorator that takes a function and returns the same function with some added attributes: import functools from typing import * def decorator(func: Callable) -> Callable: func.attr1 = "spam" func.attr2 = "eggs" return func How do I type hint the return value of decorator ? I want the type hint to convey two pieces of information: the return value is a Callable the return value has attributes attr1 and attr2 If I write a protocol, class CallableWithAttrs(Protocol): attr1: str attr2:

问题 I have a decorator that takes a function and returns the same function with some added attributes: import functools from typing import * def decorator(func: Callable) -> Callable: func.attr1 = "spam" func.attr2 = "eggs" return func How do I type hint the return value of decorator ? I want the type hint to convey two pieces of information: the return value is a Callable the return value has attributes attr1 and attr2 If I write a protocol, class CallableWithAttrs(Protocol): attr1: str attr2:

问题 How significant is the performance overhead of type-hinting in PHP - is it significant enough to be a consideration in the decision to use it or not? 回答1: No, it's not significant. If you need to do something algorithmically intensive, like sound-processing or 3D-programming, you should use another programming language. if you want hard data, make a benchmark... <?php function with_typehint(array $bla) { if(count($bla) > 55) { die("Array to big"); } } function dont_typehint($bla) { if(count(

问题 I have a function that can accept as input any variable that can be indexed, such as a list of a tuple. How do I indicate this in the type-hint of the function? 回答1: Your method is accepting a sequence, so use typing.Sequence. That's a generic, so you can specify what type of object(s) the sequence must contain: from typing import Sequence def foo(bar: Sequence[int]): # bar is a sequence of integers Quoting the Python glossary: An iterable which supports efficient element access using integer