Reputation: 43
Number of possible combinations for the given Indices
for example the indice[0] of the word BANANA
should give me :
{'B',
'BA',
'BAN',
'BANA',
'BANAN',
'BANANA'}
word='BANANA'
indices=[0,2,4]
def find_combinations(word:str,indices:list):
a=[''.join(l) for i in range(len(word)) for l in combinations(word, i+1)]
b=[x for x in a if x.startswith('B')]
return b
output:
set(b)
{'B',
'BA',
'BAA',
'BAAA',
'BAAN',
'BAANA',
'BAN',
'BANA',
'BANAA',
'BANAN',
'BANANA',
'BANN',
'BANNA',
'BN',
'BNA',
'BNAA',
'BNAN',
'BNANA',
'BNN',
'BNNA'}
desired output:
{'B',
'BA',
'BAN',
'BANA',
'BANAN',
'BANANA'}
Upvotes: 3
Views: 359
Reputation: 3565
You don't need combinations, and you can generate prefixes starting at a specific index pretty easily using slices and range
.
from typing import List
def get_parts(word: str, start: int) -> List[str]:
return [word[start:i] for i in range(start + 1, len(word) + 1)]
(Obviously you can change to return { ... }
if you need it to be a set
.
>>> get_parts("BANANA", 0)
['B', 'BA', 'BAN', 'BANA', 'BANAN', 'BANANA']
>>> get_parts("BANANA", 2)
['N', 'NA', 'NAN', 'NANA']
>>> get_parts("BANANA", 4)
['N', 'NA']
Upvotes: 2
Reputation: 2117
You can create the combinations, based on index on a given word, forward way.
word = "BANANA"
indice = [0,2,4]
def find_comb(word:str, indice:list):
final = []
for i in indice:
local = []
new = ""
for j in word[i:]:
new = new + j
local.append(new)
final.append(local)
return final
print(*find_comb(word, indice), sep='\n')
This will give you list of lists as combinations index wise.
Output:
['B', 'BA', 'BAN', 'BANA', 'BANAN', 'BANANA']
['N', 'NA', 'NAN', 'NANA']
['N', 'NA']
Upvotes: 1