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Establish relationship between two sets of data

1

$begingroup$

I have two data sets - Product to Features and Products to Parts

A = (P1, F1, F2, F3), (P2, F2, F4, F6), (P3, F1, F6, F8)...
B = (P1, M1, M2, M3), (P2, M4, M7), (P3, M1, M5, M7, M9, M10)..

where:

P1, P2, P3... are products

F1, F2, F3... are features

M1, M2, M3... are parts used in building products.

Is it possible to come up with relationship amongst features and parts?

e.g. (F1, M1, M2), (F2, M1, M4, M6)... and so on?

correlation

share|improve this question

edited Apr 2 at 6:15

KT12

8811

asked Apr 1 at 21:45

Viral PatelViral Patel

61

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you looking to join two datasets on one aspect (product id I'm guessing)?
  $endgroup$
  – S van Balen
  Apr 1 at 22:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is the idea. Is there any data science method that can take such metrics and come up with a possible association/correlation?
  $endgroup$
  – Viral Patel
  Apr 2 at 15:23
  

  
  
  
  

add a comment | 

1

$begingroup$

I have two data sets - Product to Features and Products to Parts

A = (P1, F1, F2, F3), (P2, F2, F4, F6), (P3, F1, F6, F8)...
B = (P1, M1, M2, M3), (P2, M4, M7), (P3, M1, M5, M7, M9, M10)..

where:

P1, P2, P3... are products

F1, F2, F3... are features

M1, M2, M3... are parts used in building products.

Is it possible to come up with relationship amongst features and parts?

e.g. (F1, M1, M2), (F2, M1, M4, M6)... and so on?

correlation

share|improve this question

edited Apr 2 at 6:15

KT12

8811

asked Apr 1 at 21:45

Viral PatelViral Patel

61

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you looking to join two datasets on one aspect (product id I'm guessing)?
  $endgroup$
  – S van Balen
  Apr 1 at 22:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is the idea. Is there any data science method that can take such metrics and come up with a possible association/correlation?
  $endgroup$
  – Viral Patel
  Apr 2 at 15:23
  

  
  
  
  

add a comment | 

$begingroup$

I have two data sets - Product to Features and Products to Parts

A = (P1, F1, F2, F3), (P2, F2, F4, F6), (P3, F1, F6, F8)...
B = (P1, M1, M2, M3), (P2, M4, M7), (P3, M1, M5, M7, M9, M10)..

where:

P1, P2, P3... are products

F1, F2, F3... are features

M1, M2, M3... are parts used in building products.

Is it possible to come up with relationship amongst features and parts?

e.g. (F1, M1, M2), (F2, M1, M4, M6)... and so on?

correlation

share|improve this question

edited Apr 2 at 6:15

KT12

8811

asked Apr 1 at 21:45

Viral PatelViral Patel

61

$endgroup$

A = (P1, F1, F2, F3), (P2, F2, F4, F6), (P3, F1, F6, F8)...
B = (P1, M1, M2, M3), (P2, M4, M7), (P3, M1, M5, M7, M9, M10)..

share|improve this question

edited Apr 2 at 6:15

KT12

8811

asked Apr 1 at 21:45

Viral PatelViral Patel

61

share|improve this question

edited Apr 2 at 6:15

KT12

8811

asked Apr 1 at 21:45

Viral PatelViral Patel

61

edited Apr 2 at 6:15

KT12

8811

edited Apr 2 at 6:15

KT12

8811

asked Apr 1 at 21:45

Viral PatelViral Patel

61

asked Apr 1 at 21:45

Viral PatelViral Patel

61

2 Answers
2

active

oldest

votes

1

$begingroup$

I think you may be able to shed light on relationships between features and parts using association rule learning. You can treat the parts and features similarly to items in a market basket.

Updated:

A = ('P1', 'F1', 'F2', 'F3'),
 ('P2', 'F2', 'F4', 'F6'),
 ('P3', 'F1', 'F6', 'F8')

B = ('P1', 'M1', 'M2', 'M3'),
 ('P2', 'M4', 'M7'),
 ('P3', 'M1', 'M5', 'M7', 'M9', 'M10')

parts = 'P1','P2','P3'

basket = 
for k in parts:
 temp = []
 for a in A:
 if k in a:
 temp += list(a)
 for b in B:
 if k in b:
 temp += list(b)
 temp.remove(k)
 basket[k] = temp

Then use each value in the basket dictionary as a basket.

share|improve this answer

edited Apr 3 at 1:53

answered Apr 2 at 2:37

KT12KT12

8811

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  would association rule apply across sets or only within a set (like the example they use - items in a given sales transaction and their relationship)
  $endgroup$
  – Viral Patel
  Apr 2 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would combine the features and parts that correspond to each product, and treat each product's features and parts as one basket.
  $endgroup$
  – KT12
  Apr 3 at 1:47
  

  
  
  
  

add a comment | 

0

$begingroup$

It totally depends on how the features are related to the parts.

If they related, you can form a bi-partite graph of (features-parts) for every product. Bi-Partitie graph can be formed if there are no relations between Features and Products among themselves.

If the relation for you is not correct, you can always form a tupled relation.

p1: (f1, m1), (f2, m1)

this finally depends on your dataset and the relation it has among themselves.

share|improve this answer

edited Apr 2 at 1:34

Stephen Rauch♦

1,52551330

answered Apr 1 at 23:27

William ScottWilliam Scott

1063

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is definitely a relationship "HAS" between product and features. There is no explicit relationship between features and corresponding parts to implement the feature. Goal is to derive the same using above 2 sets.
  $endgroup$
  – Viral Patel
  Apr 2 at 15:07
  

  
  
  
  

add a comment | 

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1

$begingroup$

I think you may be able to shed light on relationships between features and parts using association rule learning. You can treat the parts and features similarly to items in a market basket.

Updated:

A = ('P1', 'F1', 'F2', 'F3'),
 ('P2', 'F2', 'F4', 'F6'),
 ('P3', 'F1', 'F6', 'F8')

B = ('P1', 'M1', 'M2', 'M3'),
 ('P2', 'M4', 'M7'),
 ('P3', 'M1', 'M5', 'M7', 'M9', 'M10')

parts = 'P1','P2','P3'

basket = 
for k in parts:
 temp = []
 for a in A:
 if k in a:
 temp += list(a)
 for b in B:
 if k in b:
 temp += list(b)
 temp.remove(k)
 basket[k] = temp

Then use each value in the basket dictionary as a basket.

share|improve this answer

edited Apr 3 at 1:53

answered Apr 2 at 2:37

KT12KT12

8811

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  would association rule apply across sets or only within a set (like the example they use - items in a given sales transaction and their relationship)
  $endgroup$
  – Viral Patel
  Apr 2 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would combine the features and parts that correspond to each product, and treat each product's features and parts as one basket.
  $endgroup$
  – KT12
  Apr 3 at 1:47
  

  
  
  
  

add a comment | 

1

$begingroup$

I think you may be able to shed light on relationships between features and parts using association rule learning. You can treat the parts and features similarly to items in a market basket.

Updated:

A = ('P1', 'F1', 'F2', 'F3'),
 ('P2', 'F2', 'F4', 'F6'),
 ('P3', 'F1', 'F6', 'F8')

B = ('P1', 'M1', 'M2', 'M3'),
 ('P2', 'M4', 'M7'),
 ('P3', 'M1', 'M5', 'M7', 'M9', 'M10')

parts = 'P1','P2','P3'

basket = 
for k in parts:
 temp = []
 for a in A:
 if k in a:
 temp += list(a)
 for b in B:
 if k in b:
 temp += list(b)
 temp.remove(k)
 basket[k] = temp

Then use each value in the basket dictionary as a basket.

share|improve this answer

edited Apr 3 at 1:53

answered Apr 2 at 2:37

KT12KT12

8811

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  would association rule apply across sets or only within a set (like the example they use - items in a given sales transaction and their relationship)
  $endgroup$
  – Viral Patel
  Apr 2 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would combine the features and parts that correspond to each product, and treat each product's features and parts as one basket.
  $endgroup$
  – KT12
  Apr 3 at 1:47
  

  
  
  
  

add a comment | 

$begingroup$

I think you may be able to shed light on relationships between features and parts using association rule learning. You can treat the parts and features similarly to items in a market basket.

Updated:

A = ('P1', 'F1', 'F2', 'F3'),
 ('P2', 'F2', 'F4', 'F6'),
 ('P3', 'F1', 'F6', 'F8')

B = ('P1', 'M1', 'M2', 'M3'),
 ('P2', 'M4', 'M7'),
 ('P3', 'M1', 'M5', 'M7', 'M9', 'M10')

parts = 'P1','P2','P3'

basket = 
for k in parts:
 temp = []
 for a in A:
 if k in a:
 temp += list(a)
 for b in B:
 if k in b:
 temp += list(b)
 temp.remove(k)
 basket[k] = temp

Then use each value in the basket dictionary as a basket.

share|improve this answer

edited Apr 3 at 1:53

answered Apr 2 at 2:37

KT12KT12

8811

$endgroup$

A = ('P1', 'F1', 'F2', 'F3'),
 ('P2', 'F2', 'F4', 'F6'),
 ('P3', 'F1', 'F6', 'F8')

B = ('P1', 'M1', 'M2', 'M3'),
 ('P2', 'M4', 'M7'),
 ('P3', 'M1', 'M5', 'M7', 'M9', 'M10')

parts = 'P1','P2','P3'

basket = 
for k in parts:
 temp = []
 for a in A:
 if k in a:
 temp += list(a)
 for b in B:
 if k in b:
 temp += list(b)
 temp.remove(k)
 basket[k] = temp

share|improve this answer

edited Apr 3 at 1:53

answered Apr 2 at 2:37

KT12KT12

8811

answered Apr 2 at 2:37

KT12KT12

8811

answered Apr 2 at 2:37

KT12KT12

8811

0

$begingroup$

It totally depends on how the features are related to the parts.

If they related, you can form a bi-partite graph of (features-parts) for every product. Bi-Partitie graph can be formed if there are no relations between Features and Products among themselves.

If the relation for you is not correct, you can always form a tupled relation.

p1: (f1, m1), (f2, m1)

this finally depends on your dataset and the relation it has among themselves.

share|improve this answer

edited Apr 2 at 1:34

Stephen Rauch♦

1,52551330

answered Apr 1 at 23:27

William ScottWilliam Scott

1063

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is definitely a relationship "HAS" between product and features. There is no explicit relationship between features and corresponding parts to implement the feature. Goal is to derive the same using above 2 sets.
  $endgroup$
  – Viral Patel
  Apr 2 at 15:07
  

  
  
  
  

add a comment | 

0

$begingroup$

It totally depends on how the features are related to the parts.

If they related, you can form a bi-partite graph of (features-parts) for every product. Bi-Partitie graph can be formed if there are no relations between Features and Products among themselves.

If the relation for you is not correct, you can always form a tupled relation.

p1: (f1, m1), (f2, m1)

this finally depends on your dataset and the relation it has among themselves.

share|improve this answer

edited Apr 2 at 1:34

Stephen Rauch♦

1,52551330

answered Apr 1 at 23:27

William ScottWilliam Scott

1063

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is definitely a relationship "HAS" between product and features. There is no explicit relationship between features and corresponding parts to implement the feature. Goal is to derive the same using above 2 sets.
  $endgroup$
  – Viral Patel
  Apr 2 at 15:07
  

  
  
  
  

add a comment | 

$begingroup$

It totally depends on how the features are related to the parts.

If they related, you can form a bi-partite graph of (features-parts) for every product. Bi-Partitie graph can be formed if there are no relations between Features and Products among themselves.

If the relation for you is not correct, you can always form a tupled relation.

p1: (f1, m1), (f2, m1)

this finally depends on your dataset and the relation it has among themselves.

share|improve this answer

edited Apr 2 at 1:34

Stephen Rauch♦

1,52551330

answered Apr 1 at 23:27

William ScottWilliam Scott

1063

$endgroup$

p1: (f1, m1), (f2, m1)

share|improve this answer

edited Apr 2 at 1:34

Stephen Rauch♦

1,52551330

answered Apr 1 at 23:27

William ScottWilliam Scott

1063

edited Apr 2 at 1:34

Stephen Rauch♦

1,52551330

edited Apr 2 at 1:34

Stephen Rauch♦

1,52551330

answered Apr 1 at 23:27

William ScottWilliam Scott

1063

answered Apr 1 at 23:27

William ScottWilliam Scott

1063