Question

Background

The algorithm manipulates financial analytics. There are multiple lists of the same size and they are filtered into other lists for analysis. I am doing the same filtering on different by parallel lists. I could set it up so that a1,b1,c2 occur as a tuple in a list but then the analytics have to stripe the tuples the other way to do analysis (regression of one list against the other, beta, etc.).

What I want to do

I want to generate two different lists based on a third list:

>>> a = list(range(10))
>>> b = list(range(10,20))
>>> c = list(i & 1 for i in range(10))
>>>
>>> aprime = [a1 for a1, c1 in zip(a,c) if c1 == 0]
>>> bprime = [b1 for b1, c1 in zip(b,c) if c1 == 0]
>>> aprime
[0, 2, 4, 6, 8]
>>> bprime
[10, 12, 14, 16, 18]

It seems there should be a pythonic/functional programming/itertools way to create the two lists and iterate over the three lists only once. Something like:

aprime, bprime = [a1, b1 for a1, b1, c1 in zip(a,b,c) if c1 == 0]

But of course this generates a syntax error.

The question

Is there a pythonic way?

Micro-optimization shootout

The ugly but pythonic-to-the-max one-liner edges out the “just use a for-loop” solution and my original code in the ever popular timeit cage match:

>>> import timeit
>>> timeit.timeit("z2(a,b,c)", "n=100;a = list(range(n)); b = list(range(10,10+n)); c = list(i & 1 for i in range(n));\ndef z2(a,b,c):\n\treturn zip(*[(a1,b1) for a1,b1,c1 in zip(a,b,c) if c1==0])\n")
26.977873025761482
>>> timeit.timeit("z2(a,b,c)", "n=100;a = list(range(n)); b = list(range(10,10+n)); c = list(i & 1 for i in range(n));\ndef z2(a,b,c):\n\taprime, bprime = [], [];\n\tfor a1, b1, c1 in zip(a, b, c):\n\t\tif c1 == 0:\n\t\t\taprime.append(a1);  bprime.append(b1);\n\treturn aprime, bprime\n")
32.232914169258947
>>> timeit.timeit("z2(a,b,c)", "n=100;a = list(range(n)); b = list(range(10,10+n)); c = list(i & 1 for i in range(n));\ndef z2(a,b,c):\n\treturn [a1 for a1, c1 in zip(a,c) if c1 == 0], [b1 for b1, c1 in zip(b,c) if c1 == 0]\n")
32.37302275847901

Answer 1

This might win the ugliest code award, but it works in one line:

aprime, bprime = zip(*[(a1,b1) for a1,b1,c1 in zip(a,b,c) if c1==0])