Consider the following:
>>> a=2
>>> f=lambda x: x**a
>>> f(3)
9
>>> a=4
>>> f(3)
81
I would like for f not to change when a is changed. What is the nicest way to do this?
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2 Answers
You need to bind a to a keyword argument when defining the lambda:
f = lambda x, a=a: x**a
Now a is a local (bound as an argument) instead of a global name.
Demo:
>>> a = 2
>>> f = lambda x, a=a: x**a
>>> f(3)
9
>>> a = 4
>>> f(3)
9
answered Nov 27, 2013 at 13:47
Martijn Pieters
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Another option is to create a closure:
>>> a=2
>>> f = (lambda a: lambda x: x**a)(a)
>>> f(3)
9
>>> a=4
>>> f(3)
9
This is especially useful when you have more than one argument:
f = (lambda a, b, c: lambda x: a + b * c - x)(a, b, c)
or even
f = (lambda a, b, c, **rest: lambda x: a + b * c - x)(**locals())
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