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List Manipulation : {{{a,b},{c,d}},{{e,f},{g,h}}} into {{a,b},{c,d},{e,f},{g,h}}

List manipulation: conditional result based on variable-length sublists

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List manipulation: conditional result based on variable-length sublists

Select rows of matrix that are higher than a given rowSplitting a list into increasing sublistsGroup list by rules in sequence using DateDifference and TotalPartition list into $n$ sublistsListing all distinct exhaustive combinations of sublists of a certain lengthRemove the right sublistlist manipulation by ruleSort a list in a descending orderSpecific list manipulationSelect list elements based on other list

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

2

$begingroup$

Suppose I have the following lists:

eqexp={a,b,c,d}

eqval={e,f,g,h}

signval={{},{1},{-1},{1,-1}}

I want to create a new list of inequalities, where the signval list signals whether to use greater than or less than. I need it to handle empty sublists, and multi-component sublists. My desired output is:

{{},{b>f},{c<g},{d>h,d<h}}

All lists have the same number of top-level elements. eqexp and eqval are one-dimensional. signval is two-dimensional, but the length of the sub-lists is variable. The only numerical values allowed in signval are 1 and -1.

How do I do this?

list-manipulation

share|improve this question

edited 10 hours ago

user64494

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asked 10 hours ago

Kevin AusmanKevin Ausman

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$endgroup$

add a comment | 

2

$begingroup$

Suppose I have the following lists:

eqexp={a,b,c,d}

eqval={e,f,g,h}

signval={{},{1},{-1},{1,-1}}

I want to create a new list of inequalities, where the signval list signals whether to use greater than or less than. I need it to handle empty sublists, and multi-component sublists. My desired output is:

{{},{b>f},{c<g},{d>h,d<h}}

All lists have the same number of top-level elements. eqexp and eqval are one-dimensional. signval is two-dimensional, but the length of the sub-lists is variable. The only numerical values allowed in signval are 1 and -1.

How do I do this?

list-manipulation

share|improve this question

edited 10 hours ago

user64494

4,12422 gold badges1414 silver badges2323 bronze badges

asked 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

$endgroup$

add a comment | 

$begingroup$

Suppose I have the following lists:

eqexp={a,b,c,d}

eqval={e,f,g,h}

signval={{},{1},{-1},{1,-1}}

I want to create a new list of inequalities, where the signval list signals whether to use greater than or less than. I need it to handle empty sublists, and multi-component sublists. My desired output is:

{{},{b>f},{c<g},{d>h,d<h}}

All lists have the same number of top-level elements. eqexp and eqval are one-dimensional. signval is two-dimensional, but the length of the sub-lists is variable. The only numerical values allowed in signval are 1 and -1.

How do I do this?

list-manipulation

share|improve this question

edited 10 hours ago

user64494

4,12422 gold badges1414 silver badges2323 bronze badges

asked 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

$endgroup$

eqexp={a,b,c,d}

eqval={e,f,g,h}

signval={{},{1},{-1},{1,-1}}

{{},{b>f},{c<g},{d>h,d<h}}

share|improve this question

edited 10 hours ago

user64494

4,12422 gold badges1414 silver badges2323 bronze badges

asked 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

share|improve this question

edited 10 hours ago

user64494

4,12422 gold badges1414 silver badges2323 bronze badges

asked 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

edited 10 hours ago

user64494

4,12422 gold badges1414 silver badges2323 bronze badges

edited 10 hours ago

user64494

4,12422 gold badges1414 silver badges2323 bronze badges

asked 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

asked 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

4 Answers
4

active

oldest

votes

4

$begingroup$

MapThread[#3 /. {1 -> #1 > #2, -1 -> #1 < #2} &, {eqexp, eqval, signval}]

(*    {{}, {b > f}, {c < g}, {d > h, d < h}}    *)

share|improve this answer

answered 10 hours ago

RomanRoman

12k11 gold badge1919 silver badges4646 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! I'm embarrassed that I didn't come up with that. Thank you!
  $endgroup$
  – Kevin Ausman
  10 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

And the power of posting... I have found a way to do it:

MapThread[

  Function[{a, b, c}, If[c > 0, a > b, a < b], Listable], 

  {eqexp, eqval, signval}]

If there is a better way to do it, I would love to see it, though!

share|improve this answer

answered 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

MapThread[Through @ # @ ##2 &, {signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval}]

{{}, {b > f}, {c < g}, {d > h, d < h}}

Alternatively, define a function f with attribute Listable and use regular function application:

f0 = Function[, # @ ##2, Listable];

f0[signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval]

{{}, {b > f}, {c < g}, {d > h, d < h}}

share|improve this answer

edited 29 mins ago

answered 3 hours ago

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

Which[ 

  #1 === {}, {}, 

  #1 === {1}, {#2 > #3}, 

  #1 === {-1}, {#2 < #3}, 

  #1 === {1, -1}, 

  {#2 > #3, #2 < #3}]&

@@@ Transpose[{signval, eqexp, eqval}]





(*  {{},{b>f},{c<g},{d>h,d<h}}  *)

share|improve this answer

edited 4 hours ago

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

answered 8 hours ago

user1066user1066

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$endgroup$

add a comment | 

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4

$begingroup$

MapThread[#3 /. {1 -> #1 > #2, -1 -> #1 < #2} &, {eqexp, eqval, signval}]

(*    {{}, {b > f}, {c < g}, {d > h, d < h}}    *)

share|improve this answer

answered 10 hours ago

RomanRoman

12k11 gold badge1919 silver badges4646 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! I'm embarrassed that I didn't come up with that. Thank you!
  $endgroup$
  – Kevin Ausman
  10 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

MapThread[#3 /. {1 -> #1 > #2, -1 -> #1 < #2} &, {eqexp, eqval, signval}]

(*    {{}, {b > f}, {c < g}, {d > h, d < h}}    *)

share|improve this answer

answered 10 hours ago

RomanRoman

12k11 gold badge1919 silver badges4646 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! I'm embarrassed that I didn't come up with that. Thank you!
  $endgroup$
  – Kevin Ausman
  10 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

MapThread[#3 /. {1 -> #1 > #2, -1 -> #1 < #2} &, {eqexp, eqval, signval}]

(*    {{}, {b > f}, {c < g}, {d > h, d < h}}    *)

share|improve this answer

answered 10 hours ago

RomanRoman

12k11 gold badge1919 silver badges4646 bronze badges

$endgroup$

MapThread[#3 /. {1 -> #1 > #2, -1 -> #1 < #2} &, {eqexp, eqval, signval}]

(*    {{}, {b > f}, {c < g}, {d > h, d < h}}    *)

share|improve this answer

answered 10 hours ago

RomanRoman

12k11 gold badge1919 silver badges4646 bronze badges

answered 10 hours ago

RomanRoman

12k11 gold badge1919 silver badges4646 bronze badges

answered 10 hours ago

RomanRoman

12k11 gold badge1919 silver badges4646 bronze badges

1

$begingroup$

And the power of posting... I have found a way to do it:

MapThread[

  Function[{a, b, c}, If[c > 0, a > b, a < b], Listable], 

  {eqexp, eqval, signval}]

If there is a better way to do it, I would love to see it, though!

share|improve this answer

answered 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

And the power of posting... I have found a way to do it:

MapThread[

  Function[{a, b, c}, If[c > 0, a > b, a < b], Listable], 

  {eqexp, eqval, signval}]

If there is a better way to do it, I would love to see it, though!

share|improve this answer

answered 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

$endgroup$

add a comment | 

$begingroup$

And the power of posting... I have found a way to do it:

MapThread[

  Function[{a, b, c}, If[c > 0, a > b, a < b], Listable], 

  {eqexp, eqval, signval}]

If there is a better way to do it, I would love to see it, though!

share|improve this answer

answered 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

$endgroup$

MapThread[

  Function[{a, b, c}, If[c > 0, a > b, a < b], Listable], 

  {eqexp, eqval, signval}]

share|improve this answer

answered 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

answered 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

answered 10 hours ago

Kevin AusmanKevin Ausman

81522 silver badges1515 bronze badges

1

$begingroup$

MapThread[Through @ # @ ##2 &, {signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval}]

{{}, {b > f}, {c < g}, {d > h, d < h}}

Alternatively, define a function f with attribute Listable and use regular function application:

f0 = Function[, # @ ##2, Listable];

f0[signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval]

{{}, {b > f}, {c < g}, {d > h, d < h}}

share|improve this answer

edited 29 mins ago

answered 3 hours ago

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

MapThread[Through @ # @ ##2 &, {signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval}]

{{}, {b > f}, {c < g}, {d > h, d < h}}

Alternatively, define a function f with attribute Listable and use regular function application:

f0 = Function[, # @ ##2, Listable];

f0[signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval]

{{}, {b > f}, {c < g}, {d > h, d < h}}

share|improve this answer

edited 29 mins ago

answered 3 hours ago

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

$endgroup$

add a comment | 

$begingroup$

MapThread[Through @ # @ ##2 &, {signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval}]

{{}, {b > f}, {c < g}, {d > h, d < h}}

Alternatively, define a function f with attribute Listable and use regular function application:

f0 = Function[, # @ ##2, Listable];

f0[signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval]

{{}, {b > f}, {c < g}, {d > h, d < h}}

share|improve this answer

edited 29 mins ago

answered 3 hours ago

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

$endgroup$

MapThread[Through @ # @ ##2 &, {signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval}]

f0 = Function[, # @ ##2, Listable];

f0[signval /. {1 -> Greater, -1 -> Less}, eqexp, eqval]

share|improve this answer

edited 29 mins ago

answered 3 hours ago

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

answered 3 hours ago

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

answered 3 hours ago

kglrkglr

201k1010 gold badges230230 silver badges458458 bronze badges

0

$begingroup$

Which[ 

  #1 === {}, {}, 

  #1 === {1}, {#2 > #3}, 

  #1 === {-1}, {#2 < #3}, 

  #1 === {1, -1}, 

  {#2 > #3, #2 < #3}]&

@@@ Transpose[{signval, eqexp, eqval}]





(*  {{},{b>f},{c<g},{d>h,d<h}}  *)

share|improve this answer

edited 4 hours ago

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

answered 8 hours ago

user1066user1066

6,39822 gold badges2020 silver badges3333 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

Which[ 

  #1 === {}, {}, 

  #1 === {1}, {#2 > #3}, 

  #1 === {-1}, {#2 < #3}, 

  #1 === {1, -1}, 

  {#2 > #3, #2 < #3}]&

@@@ Transpose[{signval, eqexp, eqval}]





(*  {{},{b>f},{c<g},{d>h,d<h}}  *)

share|improve this answer

edited 4 hours ago

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

answered 8 hours ago

user1066user1066

6,39822 gold badges2020 silver badges3333 bronze badges

$endgroup$

add a comment | 

$begingroup$

Which[ 

  #1 === {}, {}, 

  #1 === {1}, {#2 > #3}, 

  #1 === {-1}, {#2 < #3}, 

  #1 === {1, -1}, 

  {#2 > #3, #2 < #3}]&

@@@ Transpose[{signval, eqexp, eqval}]





(*  {{},{b>f},{c<g},{d>h,d<h}}  *)

share|improve this answer

edited 4 hours ago

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

answered 8 hours ago

user1066user1066

6,39822 gold badges2020 silver badges3333 bronze badges

$endgroup$

Which[ 

  #1 === {}, {}, 

  #1 === {1}, {#2 > #3}, 

  #1 === {-1}, {#2 < #3}, 

  #1 === {1, -1}, 

  {#2 > #3, #2 < #3}]&

@@@ Transpose[{signval, eqexp, eqval}]





(*  {{},{b>f},{c<g},{d>h,d<h}}  *)

share|improve this answer

edited 4 hours ago

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

answered 8 hours ago

user1066user1066

6,39822 gold badges2020 silver badges3333 bronze badges

edited 4 hours ago

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

edited 4 hours ago

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

answered 8 hours ago

user1066user1066

6,39822 gold badges2020 silver badges3333 bronze badges

answered 8 hours ago

user1066user1066

6,39822 gold badges2020 silver badges3333 bronze badges