# count - counting number of gaps in python

how can I calculate the number of gaps in sequences:

for example:

``s1='G _ A A T T C A G T T A's2='G G _ A _ T C _ G _ _ A's3='G A A T T C A G T _ T _'``

her the number of `'_'` is 8

I try the following:

``def count():gap=0for i in range(0, len(s1), 3):for x,y,z in zip(s1,s2,s3):if (x=='_') or (y=='_')or (z=='_') :gap=gap+1return gap``

it gives 6 not 8

Your code returns 7 which is the total count of all the underscores minus the extra underscore in the third to last position. You can fix that by removing the or-test (which short-circuits the tests when a match is found).

Also note there is no need to triple-zip the code or to loop with a stride-of-three.

Here is a cleaned-up version of your original code:

``````def count():
gap=0
for x,y,z in zip(s1,s2,s3):
if (x == '_'):               # these if-stmts don't short-circuit
gap += 1
if (y == '_'):
gap += 1
if (z == '_'):
gap += 1
return gap
``````

There are other ways to do this faster (i.e. the str.count method) but I wanted to show you how to repair and clean-up your original logic. That ought to put you on the right track when you do other analytics.

Strings have a count() method:

``````s1.count('_') + s2.count('_') + s3.count('_')
``````

The two `_`'s in the 10th position only get counted twice. You should get 7, rather than 6.

The simple solution is `sum([item.count('_') for item in [s1,s2,s3]])`