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How do make a dominance using python in set?



The Next CEO of Stack OverflowHow do I check whether a file exists without exceptions?Calling an external command in PythonWhat are metaclasses in Python?How can I safely create a nested directory in Python?Does Python have a ternary conditional operator?How to get the current time in PythonHow can I make a time delay in Python?How to make a chain of function decorators?How to make a flat list out of list of lists?Does Python have a string 'contains' substring method?










0















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..










share|improve this question






















  • 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 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















0















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..










share|improve this question






















  • 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 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













0












0








0








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..










share|improve this question














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






share|improve this question













share|improve this question











share|improve this question




share|improve this question










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











  • 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 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 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 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












1 Answer
1






active

oldest

votes


















0















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)]






share|improve this answer























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    1 Answer
    1






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    active

    oldest

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    active

    oldest

    votes









    0















    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)]






    share|improve this answer



























      0















      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)]






      share|improve this answer

























        0












        0








        0








        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)]






        share|improve this answer














        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)]







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 7 at 17:16









        stovflstovfl

        8,32341133




        8,32341133





























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