Dictionary comprehension with lambda expression fails to produce desired result

Question

I'm creating a one liner to map the string of int to a function testing if the values are matched. Ideally, the result dictionary d behaves like d['0'](0) is True and d['0'](1) is False. But instead, I get the following output:

>>> d = { str(i): lambda v: v == i for i in range(3) }
>>> d['0'](0)
False
>>> d['0'](2)
True

I'm guessing the reason being lazy evaluation. I guess I could build the dictionary with a for loop correctly but I want a one line expression instead.

Can anyone explain why this approach fails and how I do it right?

Also have a look at: http://docs.python-guide.org/en/latest/writing/gotchas/#late-binding-closures — moooeeeep, Mar 12 '15 at 07:50
Not really anything to do with lazy evaluation, just how function namespaces work in Python. — zehnpaard, Mar 12 '15 at 08:10
The correct term is _late binding_. Certainly, a good answer would address that. — moooeeeep, Mar 12 '15 at 08:30
"I guess I could build the dictionary with a for loop correctly but I want a one line expression instead." It would work the same way. — Karl Knechtel, Aug 16 '22 at 00:59

score 4 · Accepted Answer · answered Mar 12 '15 at 07:43

4

You need to capture the current value of i for each lambda which can be done via the default argument i=i. See:

>>> d = { str(i): lambda v, i=i: v == i for i in range(3) }
>>> d['0'](0)
True
>>> d['0'](2)
False

answered Mar 12 '15 at 07:43

Dan D.

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score 0 · Answer 2 · answered Mar 12 '15 at 11:06

0

Just an alternative method to see things more logically.
d=dict((str(i),lambda v, i=i: v==i ) for i in range(3))

this will give the same results as above

answered Mar 12 '15 at 11:06

fazkan

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Dictionary comprehension with lambda expression fails to produce desired result

2 Answers2