Define a list of lambdas in Python that is eagerly evaluated

Question

I'd like to programmatically create a list of functions, each of which returns a dictionary with a specific key, and always the same value. That is:

l1 = [lambda _: {0: 42}, lambda _: {2: 42}, lambda _: {3: 42}]

I'm trying to achieve the same result programmatically. To try and avoid the usual problem with lazy evaluation, I thought I'd only need to create a generator for generating the functions on the fly:

gen = (lambda _: {f: 42} for f in range(3))

And then use a simple list comprehension to get the list:

l2 = [f for f in gen]

That's not sufficient. In fact, if you compare func_clousures with one another, they're all the same:

x[0].func_closure == x[1].func_closure == x[2].func_closure  # this is True

What's the best way to get the result I'm looking for?

Maybe I'm misunderstanding but how is `create a list of functions, each of which returns a dictionary with a specific key, and always the same value` any different to a list of dicts? — SuperBiasedMan, May 18 '15 at 11:54
This may help http://stackoverflow.com/questions/938429/scope-of-python-lambda-functions-and-their-parameters — , May 18 '15 at 11:56
@SuperBiasedMan The main difference is I'm looking for a list of functions (callable objects, really) whereas a list of dicts is not callable. Put it another way, I need to be able to call `l2[0]('something')`. — Jir, May 18 '15 at 11:57
Well known Python scoping gotcha. Short answer : `gen = (lambda _, f=f: {f: 42} for f in range(3))`, long answer here http://stackoverflow.com/questions/12423614/local-variables-in-python-nested-functions — bruno desthuilliers, May 18 '15 at 11:58
@brunodesthuilliers you're the one who nailed it (2 args needed). If you turn your comment as an answer, I'll mark it as the one. — Jir, May 18 '15 at 12:04
Side note: `[f for f in gen]` is written more directly as `list(gen)`. — Eric O. Lebigot, May 18 '15 at 12:10
@Jir: I voted to close this question as duplicate (cf the link) so I don't see the point in posting it as a full answer - the question I linked to already explain the whole thing ;) — bruno desthuilliers, May 18 '15 at 12:13

score 3 · Accepted Answer · edited May 29 '15 at 13:18

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You are assigning the last value of f to all the lambdas:

gen = (lambda _,f=f: {f: 42} for f in range(3))

As @Jon Clements so kindly pointed out the term is lazy binding, which is described in Common_Gotchas_In_Python

edited May 29 '15 at 13:18

Jir

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answered May 18 '15 at 11:54

Padraic Cunningham

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You can nick & elaborate on the sentence from "Lazy Binding" on [sopython](http://sopython.com/wiki/Common_Gotchas_In_Python/) if you want... :p – Jon Clements May 18 '15 at 11:59
@JonClements, nicked ;) – Padraic Cunningham May 18 '15 at 12:01

score 0 · Answer 2 · answered May 18 '15 at 11:56

First it depends on which python version you are using. Pyhton3 creates lazy generators for most of iterators. So you could do something like:

l2 = list([lambda: {f: 42} for f in range(3)])

or if you your lambda needs an input parameter

l2 = list([lambda f: {f: 42} for f in range(3)])

Define a list of lambdas in Python that is eagerly evaluated

2 Answers2