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python sequence order with optional values
Looking for some help generating some code to create all the possible sequences based on permanent and optional values. Example...
The sequences are saved in a MySQL Db and the each value can be optional
So there is a sequence for example that returns the following order
1,2,3,4
1 & 2 are required
3 & 4 are optional
So the outcome would return the following sequence possibilities:
[1,2], [1,2,3,4], [1,2,3], [1,2,4]
Then I could compare a sequence to those to ensure the sequence received matches the order it should.
Any help is appreciated.
python order sequence
asked Nov 19 '18 at 19:18
tsc1217tsc1217
32
add a comment |
Looking for some help generating some code to create all the possible sequences based on permanent and optional values. Example...
The sequences are saved in a MySQL Db and the each value can be optional
So there is a sequence for example that returns the following order
1,2,3,4
1 & 2 are required
3 & 4 are optional
So the outcome would return the following sequence possibilities:
[1,2], [1,2,3,4], [1,2,3], [1,2,4]
Then I could compare a sequence to those to ensure the sequence received matches the order it should.
Any help is appreciated.
python order sequence
asked Nov 19 '18 at 19:18
tsc1217tsc1217
32
add a comment |
Looking for some help generating some code to create all the possible sequences based on permanent and optional values. Example...
The sequences are saved in a MySQL Db and the each value can be optional
So there is a sequence for example that returns the following order
1,2,3,4
1 & 2 are required
3 & 4 are optional
So the outcome would return the following sequence possibilities:
[1,2], [1,2,3,4], [1,2,3], [1,2,4]
Then I could compare a sequence to those to ensure the sequence received matches the order it should.
Any help is appreciated.
python order sequence
asked Nov 19 '18 at 19:18
tsc1217tsc1217
32
Looking for some help generating some code to create all the possible sequences based on permanent and optional values. Example...
The sequences are saved in a MySQL Db and the each value can be optional
So there is a sequence for example that returns the following order
1,2,3,4
1 & 2 are required
3 & 4 are optional
So the outcome would return the following sequence possibilities:
[1,2], [1,2,3,4], [1,2,3], [1,2,4]
Then I could compare a sequence to those to ensure the sequence received matches the order it should.
Any help is appreciated.
python order sequence
python order sequence
asked Nov 19 '18 at 19:18
tsc1217tsc1217
32
asked Nov 19 '18 at 19:18
tsc1217tsc1217
32
asked Nov 19 '18 at 19:18
tsc1217tsc1217
32
asked Nov 19 '18 at 19:18
tsc1217tsc1217
32
asked Nov 19 '18 at 19:18
tsc1217tsc1217
32
32
add a comment |
add a comment |
1 Answer
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votes
You are actually only interested in the superset of the "optional" sequence. You can get your output by finding this superset and append each of its members to the "required" list.
from itertools import combinations
required = [1, 2]
optional = [3, 4]
superset_optional = [combinations(optional, i) for i in range(len(optional) + 1)]
for combinations in superset_optional:
for comb in combinations :
print(required + list(comb))
# [1, 2]
# [1, 2, 3]
# [1, 2, 4]
# [1, 2, 3, 4]
answered Nov 19 '18 at 19:29
DeepSpaceDeepSpace
37.3k44169
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are actually only interested in the superset of the "optional" sequence. You can get your output by finding this superset and append each of its members to the "required" list.
from itertools import combinations
required = [1, 2]
optional = [3, 4]
superset_optional = [combinations(optional, i) for i in range(len(optional) + 1)]
for combinations in superset_optional:
for comb in combinations :
print(required + list(comb))
# [1, 2]
# [1, 2, 3]
# [1, 2, 4]
# [1, 2, 3, 4]
answered Nov 19 '18 at 19:29
DeepSpaceDeepSpace
37.3k44169
add a comment |
You are actually only interested in the superset of the "optional" sequence. You can get your output by finding this superset and append each of its members to the "required" list.
from itertools import combinations
required = [1, 2]
optional = [3, 4]
superset_optional = [combinations(optional, i) for i in range(len(optional) + 1)]
for combinations in superset_optional:
for comb in combinations :
print(required + list(comb))
# [1, 2]
# [1, 2, 3]
# [1, 2, 4]
# [1, 2, 3, 4]
answered Nov 19 '18 at 19:29
DeepSpaceDeepSpace
37.3k44169
add a comment |
You are actually only interested in the superset of the "optional" sequence. You can get your output by finding this superset and append each of its members to the "required" list.
from itertools import combinations
required = [1, 2]
optional = [3, 4]
superset_optional = [combinations(optional, i) for i in range(len(optional) + 1)]
for combinations in superset_optional:
for comb in combinations :
print(required + list(comb))
# [1, 2]
# [1, 2, 3]
# [1, 2, 4]
# [1, 2, 3, 4]
answered Nov 19 '18 at 19:29
DeepSpaceDeepSpace
37.3k44169
You are actually only interested in the superset of the "optional" sequence. You can get your output by finding this superset and append each of its members to the "required" list.
from itertools import combinations
required = [1, 2]
optional = [3, 4]
superset_optional = [combinations(optional, i) for i in range(len(optional) + 1)]
for combinations in superset_optional:
for comb in combinations :
print(required + list(comb))
# [1, 2]
# [1, 2, 3]
# [1, 2, 4]
# [1, 2, 3, 4]
answered Nov 19 '18 at 19:29
DeepSpaceDeepSpace
37.3k44169
edited Nov 19 '18 at 19:38
answered Nov 19 '18 at 19:29
DeepSpaceDeepSpace
37.3k44169
answered Nov 19 '18 at 19:29
DeepSpaceDeepSpace
37.3k44169
answered Nov 19 '18 at 19:29
DeepSpaceDeepSpace
37.3k44169
37.3k44169
add a comment |
add a comment |
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