Lambda evaluation

[Joshua Ginsberg]

So this part makes total sense to me:

d = {}
for x in [1,2,3]:

.... d[x] = lambda y: y*x
....

d[1](3)

9

Because x in the lambda definition isn't evaluated until the lambda is
executed, at which point x is 3.

Is there a way to specifically hard code into that lambda definition the
contemporary value of an external variable? In other words, is there a
way to rewrite the line "d[x] = lambda y: y*x" so that it is always the
case that d[1](3) = 3?

Thanks!

-jag

 

[Steve M]

Here's another one:

d = {}
for x in [1,2,3]:

.... d[x] = (lambda z: lambda y: y * z) (x)
....

d[1](3) 3
d[2](3)

6