Python itertools product, but conditional?

Question

I have a function fun that takes several parameters p0,p1,.. For each parameter i give a list of possible values:

p0_list = ['a','b','c']
p1_list = [5,100]

I can now call my function for every combination of p0,p1

for i in itertools.product(*[p0,p1]):
    print fun(i)

Now comes the problem: What if i already know, that the parameter p1 only has an effect on the result of fun, if p0 is 'a' or 'c'? In this case i need my list of parameter combinations to look like:

[('a', 5), ('a',100), ('b', 5), ('c',5), ('c', 100)]

So ('b', 100) is just omitted, as it would be an unecessary evaluation of fun.

My final Solution:

param_lists = [['p0', ['a','b','c']],['p1', [5,100]]]
l = itertools.product(*[x[1] for x in param_lists])
l = [x for x in l if not x[0] == 'b' or x[1]==5]

I used this approach for 5 parameters and various conditions and it works fine. It's pretty easy to read as well. This code is inspired by Corley Brigmans' and nmcleans' answers.

Answer 1

You could generate it, and then suppress 'duplicates'. But probably just better to generate separately:

p0_list = ['a', 'b', 'c']
p0_noarg1 = ['b']
sp0_noarg1 = set(p0_noarg1)
p0_arg1 = [x for x in p0_list if x not in sp0_noarg1]
p1_list = [5, 100]

total_list = [x for x in itertools.product(p0_arg1, p1_list)] + [x for x in itertools.product(p0_noarg1, p1_list[:1])]

Answer 2

Here's a general filter function that could work for this:

def without_duplicate_item(groups, index, item):
    seen = False
    for group in groups:
        if group[index] == item:
            if seen:
                continue
            seen = True
        yield group

Usage:

param_groups = itertools.product(*[p0_list, p1_list])

param_groups = without_duplicate_item(param_groups, 0, "b")

You can of course keep adding filters for different parameters. This should be quite memory-efficient, compared to storing previous calls, because it essentially only stores one boolean value seen per filter.