Michal Ostrowski writes:
> def MakeLambdaBad():
> a = []
> for x in [1,2]:
> a.append(lambda q: x + q)
> return a
The problem here is that x is a free variable in the lambdas that you
put in a. When you actually evaluate those lambdas, they use whatever
the value of x happens to be a
Michal Ostrowski wrote:
...
[a,b] = MakeLambda()
print a(10)
print b(10)
Here is yet another way to solve the problem:
import functools
def AddPair(x, q):
return x + q
a, b = [functools.partial(AddPair, x) for x in [1, 2]]
print a(10)
print b(10)
Or even, since the
Michal Ostrowski wrote:
> def MakeLambdaBad():
> a = []
> for x in [1,2]:
> a.append(lambda q: x + q)
> return a
Two things to remember when using lambda:
1. You can always replace lambda with one line function returning the same
result. The only difference is that you have to give
In message , Michal
Ostrowski wrote:
> def MakeLambdaBad():
> a = []
> for x in [1,2]:
> a.append(lambda q: x + q)
> return a
Here's another form that should work:
def MakeLambdaGood2() :
a = []
for x in [1, 2] :
a.append((lambda x : lambda q : x + q)
On Sat, Oct 24, 2009 at 8:33 PM, Michal Ostrowski wrote:
> The snippet of code below uses two functions to dynamically create
> functions using lambda.
> Both of these uses should produce the same result, but they don't.
> def MakeLambdaGood():
> def DoLambda(x):
> return lambda q: x + q
>
