Reputation: 549
I have a list look like this:
Lux = ([ 0 0 0 0 0 120 120 120 120 120 240 240 240 240 240 0 0 0 0 0 120 120 120 120 120])
I want to count how many zeros I have, but only from the 14 place, and let say until 16 place The answer will be in this case 2. I know that count function count all the appearance. How can I do it, but with out loop? I want this when I'm already in two loops, and don't want to add another one.
Thank you.
Upvotes: 3
Views: 784
Reputation: 251146
Use list.count
and slicing
:
>>> lis = [0, 0, 0, 0, 0, 120, 120, 120, 120, 120, 240, 240, 240, 240, 240, 0, 0, 0, 0, 0, 120, 120, 120, 120, 120]
>>> lis[14:].count(0)
5
>>> lis[14:17].count(0)
2
Another option is to use sum
and a generator expression, but this is going to be very slow compared to list.count
:
>>> sum(1 for i in xrange(14, len(lis)) if lis[i]==0)
5
>>> sum(1 for i in xrange(14, 17) if lis[i]==0)
2
Timing comparisons:
In [4]: lis = [0, 0, 0, 0, 0, 120, 120, 120, 120, 120, 240, 240, 240, 240, 240, 0, 0, 0, 0, 0, 12
0, 120, 120, 120, 120]*10**5
In [5]: %timeit lis[14:].count(0)
10 loops, best of 3: 64.7 ms per loop
In [6]: %timeit sum(1 for i in xrange(14, len(lis)) if lis[i]==0)
1 loops, best of 3: 307 ms per loop
Upvotes: 10