Failed to wrap function lambda

Question

I want to wrap a function with the specified arguments, for example functools.partial , but it does not work properly:

source_codes = (0, 1, 2) def callback(source, *args): print 'callback from source: ', source funcs = [] for source in source_codes: funcs.append(lambda *args: callback(source, *args)) for i, func in enumerate(funcs): print 'source expected: ', i func() print 

output:

 source expected: 0 callback from source: 2 source expected: 1 callback from source: 2 source expected: 2 callback from source: 2 

But I want:

 source expected: 0 callback from source: 0 source expected: 1 callback from source: 1 source expected: 2 callback from source: 2 

I know this works if I use functools.partial , but I want to know the real problem in my code ... Does the lambda shell use the global variable source ?

Answer 1

The problem in your code is that the lambda expression is not evaluated before it is called.

And then when it is called, it uses the last value of source . It does not bind the source value during lambda creation.

Sample code to show this behavior -

 >>> y = lambda: z >>> z = 0 >>> y() 0 >>> z = 1 >>> y() 1 

Answer 2

partial already "freezes" the function arguments and / or keywords that you passed to it, so you can simply exclude the lambda expression:

 source_codes = (0, 1, 2) def callback(source, *args): print 'callback from source: ', source funcs = [] for source in source_codes: funcs.append(partial(callback, source)) source = 30 # don't have any effects over partial function. for i, func in enumerate(funcs): print 'source expected: ', i func() print