zip lists with condition

[3Pinter]

Guys,

Can I zip lists but ONLY if a list isn't empty?

a = [1,2,3,4,5]
b = [ [], [], ["a","b"], ["e","h"], []] 
c = zip(a, b)
d = []
for i in c:
    for j in i:
    	if anyj: 
            d.append(i)
    	else:
        	pass
print(d)

c returns: [(1, []), (2, []), (3, ['a', 'b']), (4, ['e', 'h']), (5, [])]
#BUT I don't care for c index 1, 2, 5 since they have a empty list.

Have do I alter "d[]"? Or does zip provide a conditonal zip?

[avorane]

Hello,

I suggest this:

a = [1,2,3,4,5]
b = [ [], [], ["a", "b"], ["e", "f"], [] ]
l = [elt for elt in zip(a,b) if elt[1]]

Best Regards,

Nicolas TATARENKO

[3Pinter]

@Nicolas TATARENKO,

Awesome. I knew it could be handled much easier!

[buran]

if the first list is just index/list of consecutive numbers, you can use enumerate()

[DeaD_EyE]

burans sugesition:

b = [ [], [], ["a", "b"], ["e", "f"], [] ]
l = [elt for elt in enumerate(b) if elt[1]]

Each elt object is a tuple of index and element(s) from b.

[michalmonday]

keep in mind that you can unpack the tuple. (so it is more readable and indices are not required)

l = ['a', 'b']
for i, val in enumerate(l): print(i,val)

Output:
0 a
1 b

[DeaD_EyE]

And if you want to start with 1:

enumerate(ITERABLE, 1)
Your indexing starts with 1