Reputation: 3680
I have a string of integers e.g. s = "1234"
I want to split it to the individual sequential combinations of integers split = [ 1234, 1, 2, 3, 4, 12, 123, 23, 234, 34 ]
How can I code this in Python?
What i tried:
for i in range(0,len(number)-1):
x =["" + number[j] for j in range(i, len(number))]
print(x)
Output:
['1', '2', '3', '4', '5']
['2', '3', '4', '5']
['3', '4', '5']
['4', '5']
Upvotes: 1
Views: 1955
Reputation: 2505
Are you looking for somethng like this:
stringA = "1234";
lenA = len(stringA);
# Loop through the number of times stringA is long.
for n in range(lenA):
print("---");
# Loop through string, print part of the string.
for x in range(lenA-n):
print(stringA[n:n + (x+1)])
I suggest having a look at substrings, that is what I also do in my example above.
Link
Upvotes: 0
Reputation: 117636
You can use combinations
from itertools
library combined with list comprehension
:
>>> from itertools import combinations
>>> s = "1234"
>>> [int(''.join(x)) for i in range(len(s)) for x in combinations(s, i + 1)]
[1, 2, 3, 4, 12, 13, 14, 23, 24, 34, 123, 124, 134, 234, 1234]
update As you need only sequential combinations, you can use all substrings from the string (using How To Get All The Contiguous Substrings Of A String In Python?):
>>> l = len(s)
>>> [int(s[i:j+1]) for i in range(l) for j in range(i,l)]
[1, 12, 123, 1234, 2, 23, 234, 3, 34, 4]
Upvotes: 2
Reputation: 107357
You need all the combinations, so you can use itertools.combinations
and a generator expression in order to generate all of them:
In [25]: from itertools import combinations
In [26]: list(''.join(sub) for i in range(1, len(s) + 1) for sub in combinations(s, i))
Out[26]:
['1',
'2',
'3',
'4',
'12',
'13',
'14',
'23',
'24',
'34',
'123',
'124',
'134',
'234',
'1234']
Upvotes: 2