List argument for a lambda function in python

Question

I work with scipy optimize minimize package. In order to provide constraints for my problem I need to create a massive tuple. I do it in the following way:

c0 = [];
count = 0;
for i in range(0, Nx):
    for j in range(0, Ny):
        c0.append({'type': 'eq', 'fun': lambda x:  x[count]*x[count] + x[T + count]*x[T + count] + x[2*T + count]*x[2*T + count] - 1.});
        count+=1;
cons = tuple(c0);

But when the minimizer takes them to use, it always takes the terminal value of count, which obviously results into an index out of bounds error. Trying del(count) resulted into another error, so I guess there is something wrong with my understanding of the python way of lambda functions usage. Maybe there is a better, python-style way using slices and stuff? Would appreciate any help.

Names in lambdas will evaluate to the value the name has when the lambda runs, not when it was created. This explains your terminal counts problem. See [Local variables in Python nested functions](http://stackoverflow.com/q/12423614/953482) for more information. — Kevin, Jul 10 '15 at 12:32

score 4 · Accepted Answer · answered Jul 10 '15 at 12:38

The count variable in your variable is evaluated when you call the lambda function, not when you create it (unlike in some other languages). What you need to do is force a copy of this variable, one way could be to use a closure:

c0.append({
        'type': 'eq', 
        'fun': (lambda count: lambda x:  x[count]*x[count] + x[T + count]*x[T + count] + x[2*T + count]*x[2*T + count] - 1.) (count)
});

Another way (tricky way IMO) is to use as the default value for an argument:

c0.append({
    'type': 'eq', 
    'fun': lambda x, count = count:  x[count]*x[count] + x[T + count]*x[T + count] + x[2*T + count]*x[2*T + count] - 1.
});

Thank you very much for the quick response! That is exactly what I wanted! — Eugene B, Jul 10 '15 at 13:38

score 0 · Answer 2 · answered Jul 10 '15 at 13:11

What you've done there is create a function object which references a variable, it does not create its own copy, so that when you go to execute that function, it looks up the variable, which now has the last value from the loop.

@Holt closure is probably the safest way to a accomplish the task, albeit a bit more convoluted. I would change the name of my variable within the closure to make it clearer what is happening.

What this does is create a separate scope, so that once execution leaves the outer lambda function (having called it and returned a value which is the function object you actually want) the count variable in your function object and the count variable in your loop no longer reference the same place in memory, and you can increment the counter without affecting the variable referenced by the function.

List argument for a lambda function in python

2 Answers2