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How do make a dominance using python in set?
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I have problem in mycode. I want to compare value which dominance and not dominance is pruned.
mycode is:
def dominan(a,b):
return any([x < y for (x, y) in zip(a, b)])
p1 = ('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'D', 'T'): (8, 12, 35, 6, 7)
p2 = ('S', 'B', 'T'): (7, 3, 30, 4, 4)
pk = list(p2.keys())
pv = list(p2.values())
pks = list(p1.keys())
pvs = list(p1.values())
for x in range(0, len(pk)):
if (len(pks) == 0):
pks.append(pk[x])
pvs.append(pv[x])
for l in range(0, len(pks)):
if dominan(pv[x], pvs[l]):
pks.append(pk[x])
pvs.append(pv[x])
res = dict(zip(pks, pvs))
print("result dominance p2 and p1 =", res)
The logical is p2 is not dominated by ('S', 'B', 'D', 'T'): (8, 12, 35, 6, 7)
because every value are smaller.
Any can help my problem? Thank you very much..
python
|
show 2 more comments
I have problem in mycode. I want to compare value which dominance and not dominance is pruned.
mycode is:
def dominan(a,b):
return any([x < y for (x, y) in zip(a, b)])
p1 = ('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'D', 'T'): (8, 12, 35, 6, 7)
p2 = ('S', 'B', 'T'): (7, 3, 30, 4, 4)
pk = list(p2.keys())
pv = list(p2.values())
pks = list(p1.keys())
pvs = list(p1.values())
for x in range(0, len(pk)):
if (len(pks) == 0):
pks.append(pk[x])
pvs.append(pv[x])
for l in range(0, len(pks)):
if dominan(pv[x], pvs[l]):
pks.append(pk[x])
pvs.append(pv[x])
res = dict(zip(pks, pvs))
print("result dominance p2 and p1 =", res)
The logical is p2 is not dominated by ('S', 'B', 'D', 'T'): (8, 12, 35, 6, 7)
because every value are smaller.
Any can help my problem? Thank you very much..
python
what are you trying to do with p1 and p2 dicts and what is your expected result?
– naivepredictor
Mar 7 at 14:18
my expected output is:('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'T'): (7, 3, 30, 4, 4)
– lilis gumilang
Mar 7 at 14:22
why SBT included and not SBDT?
– naivepredictor
Mar 7 at 14:31
becauseSBDT
is dominated bySBT
.SBDT
is pruned.
– lilis gumilang
Mar 7 at 14:32
what do you mean by dominated? explain your logic how do you assume SBDT dominates SBT
– naivepredictor
Mar 7 at 14:35
|
show 2 more comments
I have problem in mycode. I want to compare value which dominance and not dominance is pruned.
mycode is:
def dominan(a,b):
return any([x < y for (x, y) in zip(a, b)])
p1 = ('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'D', 'T'): (8, 12, 35, 6, 7)
p2 = ('S', 'B', 'T'): (7, 3, 30, 4, 4)
pk = list(p2.keys())
pv = list(p2.values())
pks = list(p1.keys())
pvs = list(p1.values())
for x in range(0, len(pk)):
if (len(pks) == 0):
pks.append(pk[x])
pvs.append(pv[x])
for l in range(0, len(pks)):
if dominan(pv[x], pvs[l]):
pks.append(pk[x])
pvs.append(pv[x])
res = dict(zip(pks, pvs))
print("result dominance p2 and p1 =", res)
The logical is p2 is not dominated by ('S', 'B', 'D', 'T'): (8, 12, 35, 6, 7)
because every value are smaller.
Any can help my problem? Thank you very much..
python
I have problem in mycode. I want to compare value which dominance and not dominance is pruned.
mycode is:
def dominan(a,b):
return any([x < y for (x, y) in zip(a, b)])
p1 = ('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'D', 'T'): (8, 12, 35, 6, 7)
p2 = ('S', 'B', 'T'): (7, 3, 30, 4, 4)
pk = list(p2.keys())
pv = list(p2.values())
pks = list(p1.keys())
pvs = list(p1.values())
for x in range(0, len(pk)):
if (len(pks) == 0):
pks.append(pk[x])
pvs.append(pv[x])
for l in range(0, len(pks)):
if dominan(pv[x], pvs[l]):
pks.append(pk[x])
pvs.append(pv[x])
res = dict(zip(pks, pvs))
print("result dominance p2 and p1 =", res)
The logical is p2 is not dominated by ('S', 'B', 'D', 'T'): (8, 12, 35, 6, 7)
because every value are smaller.
Any can help my problem? Thank you very much..
python
python
asked Mar 7 at 14:08
lilis gumilanglilis gumilang
93
93
what are you trying to do with p1 and p2 dicts and what is your expected result?
– naivepredictor
Mar 7 at 14:18
my expected output is:('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'T'): (7, 3, 30, 4, 4)
– lilis gumilang
Mar 7 at 14:22
why SBT included and not SBDT?
– naivepredictor
Mar 7 at 14:31
becauseSBDT
is dominated bySBT
.SBDT
is pruned.
– lilis gumilang
Mar 7 at 14:32
what do you mean by dominated? explain your logic how do you assume SBDT dominates SBT
– naivepredictor
Mar 7 at 14:35
|
show 2 more comments
what are you trying to do with p1 and p2 dicts and what is your expected result?
– naivepredictor
Mar 7 at 14:18
my expected output is:('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'T'): (7, 3, 30, 4, 4)
– lilis gumilang
Mar 7 at 14:22
why SBT included and not SBDT?
– naivepredictor
Mar 7 at 14:31
becauseSBDT
is dominated bySBT
.SBDT
is pruned.
– lilis gumilang
Mar 7 at 14:32
what do you mean by dominated? explain your logic how do you assume SBDT dominates SBT
– naivepredictor
Mar 7 at 14:35
what are you trying to do with p1 and p2 dicts and what is your expected result?
– naivepredictor
Mar 7 at 14:18
what are you trying to do with p1 and p2 dicts and what is your expected result?
– naivepredictor
Mar 7 at 14:18
my expected output is:
('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'T'): (7, 3, 30, 4, 4)
– lilis gumilang
Mar 7 at 14:22
my expected output is:
('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'T'): (7, 3, 30, 4, 4)
– lilis gumilang
Mar 7 at 14:22
why SBT included and not SBDT?
– naivepredictor
Mar 7 at 14:31
why SBT included and not SBDT?
– naivepredictor
Mar 7 at 14:31
because
SBDT
is dominated by SBT
. SBDT
is pruned.– lilis gumilang
Mar 7 at 14:32
because
SBDT
is dominated by SBT
. SBDT
is pruned.– lilis gumilang
Mar 7 at 14:32
what do you mean by dominated? explain your logic how do you assume SBDT dominates SBT
– naivepredictor
Mar 7 at 14:35
what do you mean by dominated? explain your logic how do you assume SBDT dominates SBT
– naivepredictor
Mar 7 at 14:35
|
show 2 more comments
1 Answer
1
active
oldest
votes
Question: compare value which dominance and not dominance
I have changed any(...
to all(...
.
for p2_key, p2_value in p2.items():
for p1_key, p1_value in p1.items():
d = [x < y for (x, y) in zip(p2_value, p1_value)]
print(''.format(d))
if all(d):
res.append(p2_key: p2_value)
else:
res.append(p1_key:p1_value)
print('dominance:'.format(res))
Output:
[True, True, True, True, False]
[True, True, True, False, False]
[True, True, True, True, True]
dominance:[('S', 'C', 'T'): (10, 12, 44, 5, 3),
('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4),
('S', 'B', 'T'): (7, 3, 30, 4, 4)]
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Question: compare value which dominance and not dominance
I have changed any(...
to all(...
.
for p2_key, p2_value in p2.items():
for p1_key, p1_value in p1.items():
d = [x < y for (x, y) in zip(p2_value, p1_value)]
print(''.format(d))
if all(d):
res.append(p2_key: p2_value)
else:
res.append(p1_key:p1_value)
print('dominance:'.format(res))
Output:
[True, True, True, True, False]
[True, True, True, False, False]
[True, True, True, True, True]
dominance:[('S', 'C', 'T'): (10, 12, 44, 5, 3),
('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4),
('S', 'B', 'T'): (7, 3, 30, 4, 4)]
add a comment |
Question: compare value which dominance and not dominance
I have changed any(...
to all(...
.
for p2_key, p2_value in p2.items():
for p1_key, p1_value in p1.items():
d = [x < y for (x, y) in zip(p2_value, p1_value)]
print(''.format(d))
if all(d):
res.append(p2_key: p2_value)
else:
res.append(p1_key:p1_value)
print('dominance:'.format(res))
Output:
[True, True, True, True, False]
[True, True, True, False, False]
[True, True, True, True, True]
dominance:[('S', 'C', 'T'): (10, 12, 44, 5, 3),
('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4),
('S', 'B', 'T'): (7, 3, 30, 4, 4)]
add a comment |
Question: compare value which dominance and not dominance
I have changed any(...
to all(...
.
for p2_key, p2_value in p2.items():
for p1_key, p1_value in p1.items():
d = [x < y for (x, y) in zip(p2_value, p1_value)]
print(''.format(d))
if all(d):
res.append(p2_key: p2_value)
else:
res.append(p1_key:p1_value)
print('dominance:'.format(res))
Output:
[True, True, True, True, False]
[True, True, True, False, False]
[True, True, True, True, True]
dominance:[('S', 'C', 'T'): (10, 12, 44, 5, 3),
('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4),
('S', 'B', 'T'): (7, 3, 30, 4, 4)]
Question: compare value which dominance and not dominance
I have changed any(...
to all(...
.
for p2_key, p2_value in p2.items():
for p1_key, p1_value in p1.items():
d = [x < y for (x, y) in zip(p2_value, p1_value)]
print(''.format(d))
if all(d):
res.append(p2_key: p2_value)
else:
res.append(p1_key:p1_value)
print('dominance:'.format(res))
Output:
[True, True, True, True, False]
[True, True, True, False, False]
[True, True, True, True, True]
dominance:[('S', 'C', 'T'): (10, 12, 44, 5, 3),
('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4),
('S', 'B', 'T'): (7, 3, 30, 4, 4)]
answered Mar 7 at 17:16
stovflstovfl
8,32341133
8,32341133
add a comment |
add a comment |
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what are you trying to do with p1 and p2 dicts and what is your expected result?
– naivepredictor
Mar 7 at 14:18
my expected output is:
('S', 'A', 'C', 'T'): (12, 14, 55, 1, 4), ('S', 'C', 'T'): (10, 12, 44, 5, 3), ('S', 'B', 'T'): (7, 3, 30, 4, 4)
– lilis gumilang
Mar 7 at 14:22
why SBT included and not SBDT?
– naivepredictor
Mar 7 at 14:31
because
SBDT
is dominated bySBT
.SBDT
is pruned.– lilis gumilang
Mar 7 at 14:32
what do you mean by dominated? explain your logic how do you assume SBDT dominates SBT
– naivepredictor
Mar 7 at 14:35