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Counting certain elements in lists

Splitting up delimited data in listsCounting function, comparing listsTake the next element in a nested listIssue with very large lists in Mathematica“renormalising” a listTiming and memory use is critical:fast partitioning of binary sparse arrayReplicate sublist in new listLess than Nothingrebinning of dataCounting elements in a list

2

$begingroup$

I have the following data set:

n1 = 1000;
data1 = #, RandomInteger[3, 20] & /@ Range[n1];

Dimensions@data1
1000, 2

I want to count how often data1[[All, 2]] == 1.

My solution, which seems to be wrong:

result1 = Table[
 Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
 i, 1, n1
 ];

Dimensions@result1
1000

Now I have 50 appended lists of data1 (= data2).

n3 = 50;

data2 = Array[0 &, n3];

Do[
 data2[[i]] = #, RandomInteger[3, 20] & /@ Range[n1];
 , i, 1, n3
 ];

Dimensions@data2
50, 1000, 2

I want to count how often data2[[j, i, 2]] == 1, where i, 1, n1 and j, 1, n3.

My solution for this more complicated list:

result2 = Array[0 &, n3];

Do[
 result2[[j]] =
 Table[
 Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
 i, 1, n1
 ];
 , j, 1, n3
 ];

Dimensions@result2
50, 1000

How can I replace the Do loops in both cases and improve the performance?

list-manipulation

share|improve this question

edited 51 mins ago

lio

asked 1 hour ago

liolio

1,130217

$endgroup$

add a comment | 

2

$begingroup$

I have the following data set:

n1 = 1000;
data1 = #, RandomInteger[3, 20] & /@ Range[n1];

Dimensions@data1
1000, 2

I want to count how often data1[[All, 2]] == 1.

My solution, which seems to be wrong:

result1 = Table[
 Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
 i, 1, n1
 ];

Dimensions@result1
1000

Now I have 50 appended lists of data1 (= data2).

n3 = 50;

data2 = Array[0 &, n3];

Do[
 data2[[i]] = #, RandomInteger[3, 20] & /@ Range[n1];
 , i, 1, n3
 ];

Dimensions@data2
50, 1000, 2

I want to count how often data2[[j, i, 2]] == 1, where i, 1, n1 and j, 1, n3.

My solution for this more complicated list:

result2 = Array[0 &, n3];

Do[
 result2[[j]] =
 Table[
 Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
 i, 1, n1
 ];
 , j, 1, n3
 ];

Dimensions@result2
50, 1000

How can I replace the Do loops in both cases and improve the performance?

list-manipulation

share|improve this question

edited 51 mins ago

lio

asked 1 hour ago

liolio

1,130217

$endgroup$

add a comment | 

$begingroup$

I have the following data set:

n1 = 1000;
data1 = #, RandomInteger[3, 20] & /@ Range[n1];

Dimensions@data1
1000, 2

I want to count how often data1[[All, 2]] == 1.

My solution, which seems to be wrong:

result1 = Table[
 Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
 i, 1, n1
 ];

Dimensions@result1
1000

Now I have 50 appended lists of data1 (= data2).

n3 = 50;

data2 = Array[0 &, n3];

Do[
 data2[[i]] = #, RandomInteger[3, 20] & /@ Range[n1];
 , i, 1, n3
 ];

Dimensions@data2
50, 1000, 2

I want to count how often data2[[j, i, 2]] == 1, where i, 1, n1 and j, 1, n3.

My solution for this more complicated list:

result2 = Array[0 &, n3];

Do[
 result2[[j]] =
 Table[
 Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
 i, 1, n1
 ];
 , j, 1, n3
 ];

Dimensions@result2
50, 1000

How can I replace the Do loops in both cases and improve the performance?

list-manipulation

share|improve this question

edited 51 mins ago

lio

asked 1 hour ago

liolio

1,130217

$endgroup$

n1 = 1000;
data1 = #, RandomInteger[3, 20] & /@ Range[n1];

Dimensions@data1
1000, 2

result1 = Table[
 Count[Flatten@(data1[[i, 2]]), u_ /; u == 1],
 i, 1, n1
 ];

Dimensions@result1
1000

n3 = 50;

data2 = Array[0 &, n3];

Do[
 data2[[i]] = #, RandomInteger[3, 20] & /@ Range[n1];
 , i, 1, n3
 ];

Dimensions@data2
50, 1000, 2

result2 = Array[0 &, n3];

Do[
 result2[[j]] =
 Table[
 Count[Flatten@(data2[[j, i, 2]]), u_ /; u == 1],
 i, 1, n1
 ];
 , j, 1, n3
 ];

Dimensions@result2
50, 1000

share|improve this question

edited 51 mins ago

lio

asked 1 hour ago

liolio

1,130217

share|improve this question

edited 51 mins ago

lio

asked 1 hour ago

liolio

1,130217

asked 1 hour ago

liolio

1,130217

asked 1 hour ago

liolio

1,130217

2 Answers
2

active

oldest

votes

2

$begingroup$

For your result1:

Count[1] /@ data1[[All, 2, 1]]

For your result2:

Map[Count[1], data2[[All, All, 2, 1]], 2];

share|improve this answer

edited 49 mins ago

answered 1 hour ago

MarcoBMarcoB

37.5k556113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is great ...
  $endgroup$
  – lio
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have an idea for result2 improvement?
  $endgroup$
  – lio
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
  $endgroup$
  – MarcoB
  46 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2] 

Short @ %

5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5

r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions

50, 1000

share|improve this answer

edited 40 mins ago

answered 1 hour ago

kglrkglr

188k10205422

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
  $endgroup$
  – lio
  51 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lio, please see the update.
  $endgroup$
  – kglr
  42 mins ago
  

  
  
  
  

add a comment | 

Your Answer

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2

$begingroup$

For your result1:

Count[1] /@ data1[[All, 2, 1]]

For your result2:

Map[Count[1], data2[[All, All, 2, 1]], 2];

share|improve this answer

edited 49 mins ago

answered 1 hour ago

MarcoBMarcoB

37.5k556113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is great ...
  $endgroup$
  – lio
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have an idea for result2 improvement?
  $endgroup$
  – lio
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
  $endgroup$
  – MarcoB
  46 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

For your result1:

Count[1] /@ data1[[All, 2, 1]]

For your result2:

Map[Count[1], data2[[All, All, 2, 1]], 2];

share|improve this answer

edited 49 mins ago

answered 1 hour ago

MarcoBMarcoB

37.5k556113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is great ...
  $endgroup$
  – lio
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have an idea for result2 improvement?
  $endgroup$
  – lio
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lio Yes, I've added something for result2 as well. You can check that they give the same output as your result1 and result2 respectively.
  $endgroup$
  – MarcoB
  46 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

For your result1:

Count[1] /@ data1[[All, 2, 1]]

For your result2:

Map[Count[1], data2[[All, All, 2, 1]], 2];

share|improve this answer

edited 49 mins ago

answered 1 hour ago

MarcoBMarcoB

37.5k556113

$endgroup$

Count[1] /@ data1[[All, 2, 1]]

Map[Count[1], data2[[All, All, 2, 1]], 2];

share|improve this answer

edited 49 mins ago

answered 1 hour ago

MarcoBMarcoB

37.5k556113

answered 1 hour ago

MarcoBMarcoB

37.5k556113

answered 1 hour ago

MarcoBMarcoB

37.5k556113

1

$begingroup$

r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2] 

Short @ %

5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5

r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions

50, 1000

share|improve this answer

edited 40 mins ago

answered 1 hour ago

kglrkglr

188k10205422

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
  $endgroup$
  – lio
  51 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lio, please see the update.
  $endgroup$
  – kglr
  42 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2] 

Short @ %

5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5

r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions

50, 1000

share|improve this answer

edited 40 mins ago

answered 1 hour ago

kglrkglr

188k10205422

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great ... what do you think about result2. Now the Indices in the result2 double loop are correct.
  $endgroup$
  – lio
  51 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lio, please see the update.
  $endgroup$
  – kglr
  42 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2] 

Short @ %

5,5,4,6,5,8,5,6,5,4,8,6,3,3,2,7,2,9,5,6,4,5,3,8,7,6,8,3,7,5,9,4,7,6,5,4,5,5,<<925>>,7,5,6,6,3,5,4,6,6,9,7,9,5,6,4,3,8,4,4,6,6,9,6,6,3,8,3,4,1,8,4,4,4,5,7,8,5

r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions

50, 1000

share|improve this answer

edited 40 mins ago

answered 1 hour ago

kglrkglr

188k10205422

$endgroup$

r1 = Total[1 - Unitize[data1[[All, 2, 1]] - 1], 2] 

Short @ %

r2 = Join @@@ Total[1 - Unitize[data2[[All, All, 2]] - 1], 4];
r2 // Dimensions

share|improve this answer

edited 40 mins ago

answered 1 hour ago

kglrkglr

188k10205422

answered 1 hour ago

kglrkglr

188k10205422

answered 1 hour ago

kglrkglr

188k10205422