Python 3.11+operatorfunction

operator.call

calls and access

Call an object: call(f, *args, **kwargs) == f(*args, **kwargs) (3.11+).

Syntax

operator.call(f, *args, **kwargs)

It lets you invoke a set of functions in one line: list(map(operator.call, callbacks)) runs each one with no arguments. With arguments it pairs well with itertools.starmap. It was added in Python 3.11; on older versions lambda f, *a: f(*a) does the same. Order matters: the callable comes first.

Examples

import operator
print(operator.call(abs, -5))   # → 5
print(operator.call(abs, -5) == abs(-5))   # → True
print(operator.call(round, 3.14159, ndigits=2))   # → 3.14
print([operator.call(f) for f in (list, dict, tuple)])   # → [[], {}, ()]
print(operator.call(5))   # → TypeError

See also

Common errors

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