I have a string like this:
string = 'aaabbbcccddd'
and next I want to have a list that contains ALL the pieces that are 3 indices long, so:
aaa, aab, abb, bbb, bbc, bcc, ccc, ccd, cdd, ddd
How do I get there? Because re.finditer & re.findall won't take overlapping matches, which I do need.
Well, there's a simple way to do this:
>>> for a, b, c in zip(string[:], string[1:], string[2:]):
... print(a, b, c)
...
a a a
a a b
a b b
b b b
b b c
b c c
c c c
c c d
c d d
d d d
This using a list comprehension:
>>> ["".join(var) for var in zip(string, string[1:], string[2:])]
['aaa', 'aab', 'abb', 'bbb', 'bbc', 'bcc', 'ccc', 'ccd', 'cdd', 'ddd']
You want to create a sliding window over the string:
from itertools import islice
def window(seq, n=2):
"Returns a sliding window (of width n) over data from the iterable"
" s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ... "
it = iter(seq)
result = tuple(islice(it, n))
if len(result) == n:
yield result
for elem in it:
result = result[1:] + (elem,)
yield result
print [''.join(slice) for slice in window(string, 3)]
This produces:
>>> string = 'aaabbbcccddd'
>>> [''.join(slice) for slice in window(string, 3)]
['aaa', 'aab', 'abb', 'bbb', 'bbc', 'bcc', 'ccc', 'ccd', 'cdd', 'ddd']
An alternative that surely may be improved:
>>> s = 'aaabbbcccddd'
>>> [s[i:i+3] for i in range(len(s)-2)]
['aaa', 'aab', 'abb', 'bbb', 'bbc', 'bcc', 'ccc', 'ccd', 'cdd', 'ddd']
