Organize items of a list - python-2.7

Question

How to organize a list items? Suppose, if i have a list l = ['a','b','c','d','e','f','g','h','i']
Requirement is to get a,b,c to one list, d,e,f to other and g,h,i to another list.Current implementation is
list l = ['a','b','c','d','e','f','g','h','i']
m= list()
for i in l:
if (i.find("a")>=0) or (i.find("b")>=0) or (i.find("c")>=0):
m.append(i)
print m
and so as for next items.Is there any better logic to this? With current implementation Cyclomatic Complexity is high.

Answer 1

In your example, you must not use find in the list because:
you don't really need the index, so you just could use if "a" in l
find or even in in a list has a linear (O(n)) complexity, so this is not optimal. Not noticeable on a small list, but with a million elements, it is.
with has linear complexity, and loop on the searched items instead of the list itself.
in a set, elements are hashed (and must be unique, then) ensuring a much better search performance (and insert performance too but that's not the point).
l= set(['a','b','c','d','e','f'])
m=list()
for i in ['a','b','z','c']: # I have introduced an extra element
if i in l:
m.append(i)
print(m)
result:
['a', 'b', 'c']
What is funny is the above code is that it works with a set but with a list because in is shared by all collection objects. Only the performance varies.
You could replace the first line by l= ['a','b','c','d','e','f'] it would work but you'll get bad performance (well, not for 6 items, of course), just like your example in the question.
Proof for people still doubting about the power of the set object, here's a test to check if item is in the list. I have chosen the worst case for list, but can be done with another value.
import time
data=range(1000000)
start_time = time.time()
for i in range(1,1000):
999999 in data
print("list elapsed %f" % (time.time()-start_time))
data=set(data)
start_time = time.time()
for i in range(1,1000):
999999 in data
print("set elapsed %f" % (time.time()-start_time))
result:
list elapsed 17.284000
set elapsed 0.000000
not even close :) and you can reduce the searched value, the list value will decrease (but the set will always show 0)