conflicting cases in python lists

I have two lists of sets -

attribute = [0, 1, 2, 3, 6, 7, 4, 5]

and

decision = [0, 1, 2, 3, 4, 5, 6, 7]

I want -

3, 4

Here, 3, 4 is conflicting, as it is neither a subset of 0, 1, 2, 3, 6, 7, nor of 4, 5.

My code -

 check = 
 for i in attribute:
 for j in decision:
 if j.issubset(i):
 check.append(j)
 print(check)

 for x in decision:
 if not x in check:
 temp = x

 print(temp)

This gives me 3, 4, but is there any easier (and/ or) faster way to do this?

edited Nov 13 '18 at 0:31

asked Nov 13 '18 at 0:18

Ruturaj

66110

add a comment |

I have two lists of sets -

attribute = [0, 1, 2, 3, 6, 7, 4, 5]

and

decision = [0, 1, 2, 3, 4, 5, 6, 7]

I want -

3, 4

Here, 3, 4 is conflicting, as it is neither a subset of 0, 1, 2, 3, 6, 7, nor of 4, 5.

My code -

 check = 
 for i in attribute:
 for j in decision:
 if j.issubset(i):
 check.append(j)
 print(check)

 for x in decision:
 if not x in check:
 temp = x

 print(temp)

This gives me 3, 4, but is there any easier (and/ or) faster way to do this?

edited Nov 13 '18 at 0:31

asked Nov 13 '18 at 0:18

Ruturaj

66110

add a comment |

I have two lists of sets -

attribute = [0, 1, 2, 3, 6, 7, 4, 5]

and

decision = [0, 1, 2, 3, 4, 5, 6, 7]

I want -

3, 4

Here, 3, 4 is conflicting, as it is neither a subset of 0, 1, 2, 3, 6, 7, nor of 4, 5.

My code -

 check = 
 for i in attribute:
 for j in decision:
 if j.issubset(i):
 check.append(j)
 print(check)

 for x in decision:
 if not x in check:
 temp = x

 print(temp)

This gives me 3, 4, but is there any easier (and/ or) faster way to do this?

edited Nov 13 '18 at 0:31

asked Nov 13 '18 at 0:18

Ruturaj

66110

I have two lists of sets -

attribute = [0, 1, 2, 3, 6, 7, 4, 5]

and

decision = [0, 1, 2, 3, 4, 5, 6, 7]

I want -

3, 4

Here, 3, 4 is conflicting, as it is neither a subset of 0, 1, 2, 3, 6, 7, nor of 4, 5.

My code -

 check = 
 for i in attribute:
 for j in decision:
 if j.issubset(i):
 check.append(j)
 print(check)

 for x in decision:
 if not x in check:
 temp = x

 print(temp)

This gives me 3, 4, but is there any easier (and/ or) faster way to do this?

python-3.x nested-lists

edited Nov 13 '18 at 0:31

asked Nov 13 '18 at 0:18

Ruturaj

66110

edited Nov 13 '18 at 0:31

asked Nov 13 '18 at 0:18

Ruturaj

66110

edited Nov 13 '18 at 0:31

asked Nov 13 '18 at 0:18

Ruturaj

66110

asked Nov 13 '18 at 0:18

Ruturaj

66110

asked Nov 13 '18 at 0:18

Ruturaj

66110

add a comment |

2 Answers
2

active

oldest

votes

You can use the following list comprehension:

[d for d in decision if not any(d <= a for a in attribute)]

This returns:

[3, 4]

If you want just the first set that satisfies the criteria, you can use next with a generator expression instead:

next(d for d in decision if not any(d <= a for a in attribute))

This returns:

3, 4

edited Nov 13 '18 at 0:45

answered Nov 13 '18 at 0:34

blhsing

30k41336

thanks @blhsing. But I assume it will take some more time to convert it from [3, 4] to 3, 4

– Ruturaj
Nov 13 '18 at 0:36

1

But there could be more than one set in the decision list that is not a subset of any of of the sets in the attribute list. Returning just 3, 4 makes little sense unless you are guaranteed that there is always exactly one such set that satisfies your criteria. That said, I've updated my answer with a solution that would return what you want, as long as you are sure that you only want the first matching set, or there is always only one matching set.

– blhsing
Nov 13 '18 at 0:42

1

@blhsing This makes more sense to me, thanks for your explanation.

– Ruturaj
Nov 13 '18 at 0:44

add a comment |

result = [i for i in decision if not [j for j in attribute if i.issubset(j)]]

result is the list of all set they are not a subset of attribute. :)

This is the compact version of :

result = 
for i in decision:
 tmp_list = 
 for j in attribute:
 if i.issubset(j):
 tmp_list.append(j)
 if not tmp_list:
 result.append(i)

answered Nov 13 '18 at 0:25

iElden

680417

Thanks, I tried result = [i for i in decision if not [j for j in attribute if i.issubset(j)]] which gives me [3, 4], but not 3,4 (I know the conversion is possible). But is it faster than my solution (with additional conversion time)?

– Ruturaj
Nov 13 '18 at 0:34

1

EDIT : sorry, my last calcul was wrong, our program has same execution time after convertion on my computer : 0.030 - 0.050 sec

– iElden
Nov 13 '18 at 0:38

Thanks @ iElden i upvoted, but @blhsing's explanation makes more sense to me.

– Ruturaj
Nov 13 '18 at 0:47

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You can use the following list comprehension:

[d for d in decision if not any(d <= a for a in attribute)]

This returns:

[3, 4]

If you want just the first set that satisfies the criteria, you can use next with a generator expression instead:

next(d for d in decision if not any(d <= a for a in attribute))

This returns:

3, 4

edited Nov 13 '18 at 0:45

answered Nov 13 '18 at 0:34

blhsing

30k41336

thanks @blhsing. But I assume it will take some more time to convert it from [3, 4] to 3, 4

– Ruturaj
Nov 13 '18 at 0:36

1

But there could be more than one set in the decision list that is not a subset of any of of the sets in the attribute list. Returning just 3, 4 makes little sense unless you are guaranteed that there is always exactly one such set that satisfies your criteria. That said, I've updated my answer with a solution that would return what you want, as long as you are sure that you only want the first matching set, or there is always only one matching set.

– blhsing
Nov 13 '18 at 0:42

1

@blhsing This makes more sense to me, thanks for your explanation.

– Ruturaj
Nov 13 '18 at 0:44

add a comment |

You can use the following list comprehension:

[d for d in decision if not any(d <= a for a in attribute)]

This returns:

[3, 4]

If you want just the first set that satisfies the criteria, you can use next with a generator expression instead:

next(d for d in decision if not any(d <= a for a in attribute))

This returns:

3, 4

edited Nov 13 '18 at 0:45

answered Nov 13 '18 at 0:34

blhsing

30k41336

thanks @blhsing. But I assume it will take some more time to convert it from [3, 4] to 3, 4

– Ruturaj
Nov 13 '18 at 0:36

1

But there could be more than one set in the decision list that is not a subset of any of of the sets in the attribute list. Returning just 3, 4 makes little sense unless you are guaranteed that there is always exactly one such set that satisfies your criteria. That said, I've updated my answer with a solution that would return what you want, as long as you are sure that you only want the first matching set, or there is always only one matching set.

– blhsing
Nov 13 '18 at 0:42

1

@blhsing This makes more sense to me, thanks for your explanation.

– Ruturaj
Nov 13 '18 at 0:44

add a comment |

You can use the following list comprehension:

[d for d in decision if not any(d <= a for a in attribute)]

This returns:

[3, 4]

If you want just the first set that satisfies the criteria, you can use next with a generator expression instead:

next(d for d in decision if not any(d <= a for a in attribute))

This returns:

3, 4

edited Nov 13 '18 at 0:45

answered Nov 13 '18 at 0:34

blhsing

30k41336

You can use the following list comprehension:

[d for d in decision if not any(d <= a for a in attribute)]

This returns:

[3, 4]

If you want just the first set that satisfies the criteria, you can use next with a generator expression instead:

next(d for d in decision if not any(d <= a for a in attribute))

This returns:

3, 4

edited Nov 13 '18 at 0:45

answered Nov 13 '18 at 0:34

blhsing

30k41336

edited Nov 13 '18 at 0:45

answered Nov 13 '18 at 0:34

blhsing

30k41336

answered Nov 13 '18 at 0:34

blhsing

30k41336

answered Nov 13 '18 at 0:34

blhsing

30k41336

thanks @blhsing. But I assume it will take some more time to convert it from [3, 4] to 3, 4

– Ruturaj
Nov 13 '18 at 0:36

1

But there could be more than one set in the decision list that is not a subset of any of of the sets in the attribute list. Returning just 3, 4 makes little sense unless you are guaranteed that there is always exactly one such set that satisfies your criteria. That said, I've updated my answer with a solution that would return what you want, as long as you are sure that you only want the first matching set, or there is always only one matching set.

– blhsing
Nov 13 '18 at 0:42

1

@blhsing This makes more sense to me, thanks for your explanation.

– Ruturaj
Nov 13 '18 at 0:44

add a comment |

thanks @blhsing. But I assume it will take some more time to convert it from [3, 4] to 3, 4

– Ruturaj
Nov 13 '18 at 0:36

1

But there could be more than one set in the decision list that is not a subset of any of of the sets in the attribute list. Returning just 3, 4 makes little sense unless you are guaranteed that there is always exactly one such set that satisfies your criteria. That said, I've updated my answer with a solution that would return what you want, as long as you are sure that you only want the first matching set, or there is always only one matching set.

– blhsing
Nov 13 '18 at 0:42

1

@blhsing This makes more sense to me, thanks for your explanation.

– Ruturaj
Nov 13 '18 at 0:44

thanks @blhsing. But I assume it will take some more time to convert it from [3, 4] to 3, 4

– Ruturaj
Nov 13 '18 at 0:36

But there could be more than one set in the decision list that is not a subset of any of of the sets in the attribute list. Returning just 3, 4 makes little sense unless you are guaranteed that there is always exactly one such set that satisfies your criteria. That said, I've updated my answer with a solution that would return what you want, as long as you are sure that you only want the first matching set, or there is always only one matching set.

– blhsing
Nov 13 '18 at 0:42

@blhsing This makes more sense to me, thanks for your explanation.

– Ruturaj
Nov 13 '18 at 0:44

add a comment |

result = [i for i in decision if not [j for j in attribute if i.issubset(j)]]

result is the list of all set they are not a subset of attribute. :)

This is the compact version of :

result = 
for i in decision:
 tmp_list = 
 for j in attribute:
 if i.issubset(j):
 tmp_list.append(j)
 if not tmp_list:
 result.append(i)

answered Nov 13 '18 at 0:25

iElden

680417

Thanks, I tried result = [i for i in decision if not [j for j in attribute if i.issubset(j)]] which gives me [3, 4], but not 3,4 (I know the conversion is possible). But is it faster than my solution (with additional conversion time)?

– Ruturaj
Nov 13 '18 at 0:34

1

EDIT : sorry, my last calcul was wrong, our program has same execution time after convertion on my computer : 0.030 - 0.050 sec

– iElden
Nov 13 '18 at 0:38

Thanks @ iElden i upvoted, but @blhsing's explanation makes more sense to me.

– Ruturaj
Nov 13 '18 at 0:47

add a comment |

result = [i for i in decision if not [j for j in attribute if i.issubset(j)]]

result is the list of all set they are not a subset of attribute. :)

This is the compact version of :

result = 
for i in decision:
 tmp_list = 
 for j in attribute:
 if i.issubset(j):
 tmp_list.append(j)
 if not tmp_list:
 result.append(i)

answered Nov 13 '18 at 0:25

iElden

680417

Thanks, I tried result = [i for i in decision if not [j for j in attribute if i.issubset(j)]] which gives me [3, 4], but not 3,4 (I know the conversion is possible). But is it faster than my solution (with additional conversion time)?

– Ruturaj
Nov 13 '18 at 0:34

1

EDIT : sorry, my last calcul was wrong, our program has same execution time after convertion on my computer : 0.030 - 0.050 sec

– iElden
Nov 13 '18 at 0:38

Thanks @ iElden i upvoted, but @blhsing's explanation makes more sense to me.

– Ruturaj
Nov 13 '18 at 0:47

add a comment |

result = [i for i in decision if not [j for j in attribute if i.issubset(j)]]

result is the list of all set they are not a subset of attribute. :)

This is the compact version of :

result = 
for i in decision:
 tmp_list = 
 for j in attribute:
 if i.issubset(j):
 tmp_list.append(j)
 if not tmp_list:
 result.append(i)

answered Nov 13 '18 at 0:25

iElden

680417

result = [i for i in decision if not [j for j in attribute if i.issubset(j)]]

result is the list of all set they are not a subset of attribute. :)

This is the compact version of :

result = 
for i in decision:
 tmp_list = 
 for j in attribute:
 if i.issubset(j):
 tmp_list.append(j)
 if not tmp_list:
 result.append(i)

answered Nov 13 '18 at 0:25

iElden

680417

answered Nov 13 '18 at 0:25

iElden

680417

answered Nov 13 '18 at 0:25

iElden

680417

answered Nov 13 '18 at 0:25

iElden

680417

Thanks, I tried result = [i for i in decision if not [j for j in attribute if i.issubset(j)]] which gives me [3, 4], but not 3,4 (I know the conversion is possible). But is it faster than my solution (with additional conversion time)?

– Ruturaj
Nov 13 '18 at 0:34

1

EDIT : sorry, my last calcul was wrong, our program has same execution time after convertion on my computer : 0.030 - 0.050 sec

– iElden
Nov 13 '18 at 0:38

Thanks @ iElden i upvoted, but @blhsing's explanation makes more sense to me.

– Ruturaj
Nov 13 '18 at 0:47

add a comment |

Thanks, I tried result = [i for i in decision if not [j for j in attribute if i.issubset(j)]] which gives me [3, 4], but not 3,4 (I know the conversion is possible). But is it faster than my solution (with additional conversion time)?

– Ruturaj
Nov 13 '18 at 0:34

1

EDIT : sorry, my last calcul was wrong, our program has same execution time after convertion on my computer : 0.030 - 0.050 sec

– iElden
Nov 13 '18 at 0:38

Thanks @ iElden i upvoted, but @blhsing's explanation makes more sense to me.

– Ruturaj
Nov 13 '18 at 0:47

Thanks, I tried result = [i for i in decision if not [j for j in attribute if i.issubset(j)]] which gives me [3, 4], but not 3,4 (I know the conversion is possible). But is it faster than my solution (with additional conversion time)?

– Ruturaj
Nov 13 '18 at 0:34

EDIT : sorry, my last calcul was wrong, our program has same execution time after convertion on my computer : 0.030 - 0.050 sec

– iElden
Nov 13 '18 at 0:38

Thanks @ iElden i upvoted, but @blhsing's explanation makes more sense to me.

– Ruturaj
Nov 13 '18 at 0:47

add a comment |

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