CODEWITHSUNDEEP
pythonpython-3.x
Menon A.
Menon A.

Reputation: 580

How to iterate over a string in groups of n characters instead of one character at a time?

Suppose - 

string = "abcdefgh"

If I do - 

for i in string:
    print (i)

I get - 

a
b
c
d
e
f
g

What I want is something like - 

ab
bc
cd
de
ef
fg

Or in any other grouping we specify. Is it possible to make a function for this keeping in mind the grouping we require? Thanks

Upvotes: 1

Views: 1219

Answers (4)

TerryA
TerryA

Reputation: 60014

You can use zip():

>>> for i, j in zip(string, string[1:]):
...     print(i+j)
... 
ab
bc
cd
de
ef
fg
gh

As a function:

def func(seq, n):
    return [''.join(item) for item in zip(*[seq[n:] for n in range(n)])]

Example:

>>> for item in func("abcdefgh", 3):
...     print(item)
... 
abc
bcd
cde
def
efg
fgh

Upvotes: 5

pyInTheSky
pyInTheSky

Reputation: 1479

import re
def splittext(text, split_by):
    '''the regex will take a string and create groupings of n-characters plus a final grouping of any remainder. if no remainder is desired, the |.+ can be removed''' 
    return re.findall(r".{%d}|.+" % split_by, text)

ret = splittext("HELLOWORLD!", 2)
print "\n".join(ret)

some sample output

>>> re.findall(r".{2}",a)
['HE', 'LL', 'OW', 'OR', 'LD']
>>> re.findall(r".{2}|.{1}",a)
['HE', 'LL', 'OW', 'OR', 'LD', '!']
>>> re.findall(r".{2}|.*",a)
['HE', 'LL', 'OW', 'OR', 'LD', '!', '']
>>> re.findall(r".{2}|.+",a)
['HE', 'LL', 'OW', 'OR', 'LD', '!']
>>> print "\n".join(_)
HE
LL
OW
OR
LD
!
>>>

Upvotes: -1

Darkstarone
Darkstarone

Reputation: 4730

This works: 

string = "abcdefgh"

i = 0
while i < len(string) - 1:
    print(string[i]+string[i+1])
    i += 1

Result:

ab
bc
cd
de
ef
fg
gh

If you don't want gh (it's missing in your example), change the while loop to: while i < len(string) - 2:.

Also another way to do (which hasn't been posted), is via regex:

import re

print("\n".join(re.findall(r'(?=(\w\w))', 'abcdefgh')))

The (?=) (lookahead assertion), allows regex patterns to overlap.

Upvotes: 0

Jon Kiparsky
Jon Kiparsky

Reputation: 7753

if s is the name of your string, this comprehension will do what you want:

[s[i:i+2] for i in range(0, len(s) - 1)]

Using this, you can easily print the strings on separate lines:

for substr in [s[i:i+2] for i in range(0, len(s) -1)]:
  print substr

It can be generalised fairly easily:

def subgroups(s, n):
    return [s[i:i+n] for i in range(0, len(s) - 1)]

(and this function can similarly be used to print the resulting substrings in any fashion you like)

Upvotes: 1

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