Python: can function be defined inside of another function's argument list?

Question

Is there a way to achieve something like this in python?

another_function( function(x) {return 2*x} )

What does this code do? If called as another_function(9) is 18 what is actually passed into another_function? — Ethan Furman
– Ethan Furman, Commented Oct 10, 2011 at 16:59

Ethan Furman · Accepted Answer · 2011-10-11 06:30:45Z

9

Yes:

another_function( lambda x: 2*x )

To be clear: this is taking place when another_function is called, not when it is defined.

edited Oct 11, 2011 at 6:30

Ethan Furman

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answered Oct 10, 2011 at 13:13

hamstergene

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4 Comments

Ethan Furman Over a year ago

This gives SyntaxError: invalid syntax

Ethan Furman Over a year ago

No typo -- thought question was about function definition time, whereas your answer is at function call time.

hamstergene Over a year ago

@EthanFurman the question was exactly about call time (doesn't the fact that the answer is accepted ring a bell?). The author has used anonymous function syntax from Javascript and other ecmascript-based languages.

Ethan Furman Over a year ago

Amazingly enough, not everybody speaks Javascript or other ecmascript-based languages (and unfortunately I've seen bad answers accepted); I've changed my vote, though.

Ethan Furman · Accepted Answer · 2011-10-11 06:15:14Z

0

def another_function( function=lambda x: 2*x ):
    print(function(10))  #an example

I'm not sure what happens with the example code you posted, but with the solution shown if you call another_function it will call function(10) and print 20.

More to the point, you cannot call another_function(7) and get 14 because 7 will get assigned to function and 7(10)' will get youTypeError: 'int' object is not callable`.

edited Oct 11, 2011 at 6:15

answered Oct 10, 2011 at 16:56

Ethan Furman

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