Deferred evaluation with lambda in Python [duplicate]

Question

In a loop, I am trying to defer the comparison the two value()s of two Nodes to a later time.

class Node():
    def __init__(self, v):
        self.v = v
    def value(self):
        return self.v

nodes = [Node(0), Node(1), Node(2), Node(3), Node(4), Node(2)]
results = []
for i in [0, 1, 2]:
    j = i + 3
    results.append(lambda: nodes[i].value() == nodes[j].value())

for result in results:
    print result

The results are all True (because i,j==2,5 for all the lambdas). How can I defer the execution of the lambda until it is actually called, but with the correct variable bindings? And the expressions in the lambda are not all necessarily equality... there are a bunch of other more involved expressions.

Thanks for any help!

Fragrance

I'm not really sure what you're trying to do. The lambda expression seems unnecessary to me here. Why can't you just do results.append(nodes[i].value() == nodes[j].value())? – Singband

Answer 1

To bind the current values of i and j to the function instead of having it look in the outer scope, you can use either a closure or default argument values. The easiest way to do this is to use default argument values in your lambda:

for i in [0, 1, 2]:
    j = i + 3
    results.append(lambda i=i, j=j: nodes[i].value() == nodes[j].value())

Here is how it would look as a closure:

def make_comp_func(i, j):
    return lambda: nodes[i].value() == nodes[j].value()

for i in [0, 1, 2]:
    j = i + 3
    results.append(make_comp_func(i, j))

Fanni

Answer 2

Wrap it in another lambda:

results.append((lambda x, y: lambda: nodes[x].value() == nodes[y].value()) (i, j))

or in a nicer way, with partial:

from functools import partial

results.append(partial(lambda x, y: nodes[x].value() == nodes[y].value(), i, j))

Default arguments trick is, well... a trick, and I'd suggest to avoid it.

Without

Answer 3

The idiomatic way is to use a default argument:

[f() for f in [lambda: i for i in range(3)]]
[2, 2, 2]

Change this to:

[f() for f in [lambda i=i: i for i in range(3)]]
[0, 1, 2]

Bejewel