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
>
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.
The expected output of this code is
11
12
11
12
However, what we get instead is
12
12
11
12
The problem is that the two functions return
