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3

$begingroup$

I think my problem is pretty simple, and in SQL this would be trivial. I have two tables

TableOne = a, x1, b, x2, c, x3;
TableTwo = a, y1, c, y2 , a, y3, a, y4, b, y5, c,y6, c, y7

I want to abe able to join these two tables where the values of column 1 in both tables match such that:

DesiredResult = a, x1, a, y1, c, x3 , c, y2 , a, x1, a, y3, a, x1, a, y4, b, x2, b, y5, c, x3, c, y6, c, x3, c, y7

I've tried with both Select[] statements inside a Table[] structure and also looked into JoinAcross[] but haven't been able to acheive the desire effect. In SQL it would be simple, something like:

SELECT Col1.Table1, Col2.Table1, Col1.Table2, Col2.Table2, FROM table2 INNER JOIN table1 ON Col1.Table1 = Col1.Table2

Or something similar.

list-manipulation table

share|improve this question

asked 8 hours ago

QuantumPenguinQuantumPenguin

56922 silver badges1818 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

I think my problem is pretty simple, and in SQL this would be trivial. I have two tables

TableOne = a, x1, b, x2, c, x3;
TableTwo = a, y1, c, y2 , a, y3, a, y4, b, y5, c,y6, c, y7

I want to abe able to join these two tables where the values of column 1 in both tables match such that:

DesiredResult = a, x1, a, y1, c, x3 , c, y2 , a, x1, a, y3, a, x1, a, y4, b, x2, b, y5, c, x3, c, y6, c, x3, c, y7

I've tried with both Select[] statements inside a Table[] structure and also looked into JoinAcross[] but haven't been able to acheive the desire effect. In SQL it would be simple, something like:

SELECT Col1.Table1, Col2.Table1, Col1.Table2, Col2.Table2, FROM table2 INNER JOIN table1 ON Col1.Table1 = Col1.Table2

Or something similar.

list-manipulation table

share|improve this question

asked 8 hours ago

QuantumPenguinQuantumPenguin

56922 silver badges1818 bronze badges

$endgroup$

add a comment | 

$begingroup$

I think my problem is pretty simple, and in SQL this would be trivial. I have two tables

TableOne = a, x1, b, x2, c, x3;
TableTwo = a, y1, c, y2 , a, y3, a, y4, b, y5, c,y6, c, y7

I want to abe able to join these two tables where the values of column 1 in both tables match such that:

DesiredResult = a, x1, a, y1, c, x3 , c, y2 , a, x1, a, y3, a, x1, a, y4, b, x2, b, y5, c, x3, c, y6, c, x3, c, y7

I've tried with both Select[] statements inside a Table[] structure and also looked into JoinAcross[] but haven't been able to acheive the desire effect. In SQL it would be simple, something like:

SELECT Col1.Table1, Col2.Table1, Col1.Table2, Col2.Table2, FROM table2 INNER JOIN table1 ON Col1.Table1 = Col1.Table2

Or something similar.

list-manipulation table

share|improve this question

asked 8 hours ago

QuantumPenguinQuantumPenguin

56922 silver badges1818 bronze badges

$endgroup$

TableOne = a, x1, b, x2, c, x3;
TableTwo = a, y1, c, y2 , a, y3, a, y4, b, y5, c,y6, c, y7

DesiredResult = a, x1, a, y1, c, x3 , c, y2 , a, x1, a, y3, a, x1, a, y4, b, x2, b, y5, c, x3, c, y6, c, x3, c, y7

SELECT Col1.Table1, Col2.Table1, Col1.Table2, Col2.Table2, FROM table2 INNER JOIN table1 ON Col1.Table1 = Col1.Table2

share|improve this question

asked 8 hours ago

QuantumPenguinQuantumPenguin

56922 silver badges1818 bronze badges

share|improve this question

asked 8 hours ago

QuantumPenguinQuantumPenguin

56922 silver badges1818 bronze badges

asked 8 hours ago

QuantumPenguinQuantumPenguin

56922 silver badges1818 bronze badges

asked 8 hours ago

QuantumPenguinQuantumPenguin

56922 silver badges1818 bronze badges

2 Answers
2

active

oldest

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4

$begingroup$

I think you should stay away from associations because your keys aren't unique. Maybe just go over all tuples of elements and pick those that match your criterion:

Reap[Outer[If[#1[[1]] == #2[[1]], Sow@Join@##] &, TableOne, TableTwo, 1]][[2, 1]]

(* a, x1, a, y1, a, x1, a, y3, a, x1, a, y4, b, x2, b, y5, c, x3, c, y2,
 c, x3, c, y6, c, x3, c, y7 *)

share|improve this answer

answered 8 hours ago

RomanRoman

15.6k11 gold badge2121 silver badges5252 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That'll do it, thanks again Roman.
  $endgroup$
  – QuantumPenguin
  8 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

assocOne = AssociationThread[First /@ #, #] & @ TableOne;

f = Join[assocOne[First @ #], #]&;

Map[f] @ TableTwo 

a, x1, a, y1, c, x3, c, y2, a, x1, a, y3, a, x1, a, y4, b,
x2, b, y5, c, x3, c, y6, c, x3, c, y7

share|improve this answer

answered 5 hours ago

kglrkglr

212k1010 gold badges242242 silver badges485485 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also a very nice solution, +1!
  $endgroup$
  – QuantumPenguin
  3 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

I think you should stay away from associations because your keys aren't unique. Maybe just go over all tuples of elements and pick those that match your criterion:

Reap[Outer[If[#1[[1]] == #2[[1]], Sow@Join@##] &, TableOne, TableTwo, 1]][[2, 1]]

(* a, x1, a, y1, a, x1, a, y3, a, x1, a, y4, b, x2, b, y5, c, x3, c, y2,
 c, x3, c, y6, c, x3, c, y7 *)

share|improve this answer

answered 8 hours ago

RomanRoman

15.6k11 gold badge2121 silver badges5252 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That'll do it, thanks again Roman.
  $endgroup$
  – QuantumPenguin
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

I think you should stay away from associations because your keys aren't unique. Maybe just go over all tuples of elements and pick those that match your criterion:

Reap[Outer[If[#1[[1]] == #2[[1]], Sow@Join@##] &, TableOne, TableTwo, 1]][[2, 1]]

(* a, x1, a, y1, a, x1, a, y3, a, x1, a, y4, b, x2, b, y5, c, x3, c, y2,
 c, x3, c, y6, c, x3, c, y7 *)

share|improve this answer

answered 8 hours ago

RomanRoman

15.6k11 gold badge2121 silver badges5252 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That'll do it, thanks again Roman.
  $endgroup$
  – QuantumPenguin
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I think you should stay away from associations because your keys aren't unique. Maybe just go over all tuples of elements and pick those that match your criterion:

Reap[Outer[If[#1[[1]] == #2[[1]], Sow@Join@##] &, TableOne, TableTwo, 1]][[2, 1]]

(* a, x1, a, y1, a, x1, a, y3, a, x1, a, y4, b, x2, b, y5, c, x3, c, y2,
 c, x3, c, y6, c, x3, c, y7 *)

share|improve this answer

answered 8 hours ago

RomanRoman

15.6k11 gold badge2121 silver badges5252 bronze badges

$endgroup$

Reap[Outer[If[#1[[1]] == #2[[1]], Sow@Join@##] &, TableOne, TableTwo, 1]][[2, 1]]

(* a, x1, a, y1, a, x1, a, y3, a, x1, a, y4, b, x2, b, y5, c, x3, c, y2,
 c, x3, c, y6, c, x3, c, y7 *)

share|improve this answer

answered 8 hours ago

RomanRoman

15.6k11 gold badge2121 silver badges5252 bronze badges

answered 8 hours ago

RomanRoman

15.6k11 gold badge2121 silver badges5252 bronze badges

answered 8 hours ago

RomanRoman

15.6k11 gold badge2121 silver badges5252 bronze badges

2

$begingroup$

assocOne = AssociationThread[First /@ #, #] & @ TableOne;

f = Join[assocOne[First @ #], #]&;

Map[f] @ TableTwo 

a, x1, a, y1, c, x3, c, y2, a, x1, a, y3, a, x1, a, y4, b,
x2, b, y5, c, x3, c, y6, c, x3, c, y7

share|improve this answer

answered 5 hours ago

kglrkglr

212k1010 gold badges242242 silver badges485485 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also a very nice solution, +1!
  $endgroup$
  – QuantumPenguin
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

assocOne = AssociationThread[First /@ #, #] & @ TableOne;

f = Join[assocOne[First @ #], #]&;

Map[f] @ TableTwo 

a, x1, a, y1, c, x3, c, y2, a, x1, a, y3, a, x1, a, y4, b,
x2, b, y5, c, x3, c, y6, c, x3, c, y7

share|improve this answer

answered 5 hours ago

kglrkglr

212k1010 gold badges242242 silver badges485485 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also a very nice solution, +1!
  $endgroup$
  – QuantumPenguin
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

assocOne = AssociationThread[First /@ #, #] & @ TableOne;

f = Join[assocOne[First @ #], #]&;

Map[f] @ TableTwo 

a, x1, a, y1, c, x3, c, y2, a, x1, a, y3, a, x1, a, y4, b,
x2, b, y5, c, x3, c, y6, c, x3, c, y7

share|improve this answer

answered 5 hours ago

kglrkglr

212k1010 gold badges242242 silver badges485485 bronze badges

$endgroup$

assocOne = AssociationThread[First /@ #, #] & @ TableOne;

f = Join[assocOne[First @ #], #]&;

Map[f] @ TableTwo 

share|improve this answer

answered 5 hours ago

kglrkglr

212k1010 gold badges242242 silver badges485485 bronze badges

answered 5 hours ago

kglrkglr

212k1010 gold badges242242 silver badges485485 bronze badges

answered 5 hours ago

kglrkglr

212k1010 gold badges242242 silver badges485485 bronze badges