Ttgjyuk

Brexit - No Deal Rejection

Why does Deadpool say "You're welcome, Canada," after shooting Ryan Reynolds in the end credits?

What is the greatest age difference between a married couple in Tanach?

Theorems like the Lovász Local Lemma?

Why would a flight no longer considered airworthy be redirected like this?

Be in awe of my brilliance!

How to deal with taxi scam when on vacation?

Make a transparent 448*448 image

How is the Swiss post e-voting system supposed to work, and how was it wrong?

Possible Leak In Concrete

Official degrees of earth’s rotation per day

How do I hide Chekhov's Gun?

Why are there 40 737 Max planes in flight when they have been grounded as not airworthy?

PTIJ: is Mi Yodeya found in the Torah codes?

Instead of Universal Basic Income, why not Universal Basic NEEDS?

Does this AnyDice function accurately calculate the number of ogres you make unconcious with three 4th-level castings of Sleep?

My adviser wants to be the first author

What is a good source for large tables on the properties of water?

Did CPM support custom hardware using device drivers?

How to answer questions about my characters?

I need to drive a 7/16" nut but am unsure how to use the socket I bought for my screwdriver

Do I need life insurance if I can cover my own funeral costs?

Can elves maintain concentration in a trance?

Can the damage from a Talisman of Pure Good (or Ultimate Evil) be non-lethal?

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

4

$begingroup$

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 3 hours ago

lio

asked 3 hours ago

liolio

1,140217

$endgroup$

add a comment | 

4

$begingroup$

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 3 hours ago

lio

asked 3 hours ago

liolio

1,140217

$endgroup$

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 3 hours ago

lio

asked 3 hours ago

liolio

1,140217

$endgroup$

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 3 hours ago

lio

asked 3 hours ago

liolio

1,140217

share|improve this question

edited 3 hours ago

lio

asked 3 hours ago

liolio

1,140217

asked 3 hours ago

liolio

1,140217

asked 3 hours ago

liolio

1,140217

2 Answers
2

active

oldest

votes

4

$begingroup$

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 3 hours ago

answered 3 hours ago

MarcoBMarcoB

37.5k556113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is great ...
  $endgroup$
  – lio
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have an idea for result2 improvement?
  $endgroup$
  – lio
  3 hours 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
  3 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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 2 hours ago

answered 3 hours ago

kglrkglr

189k10205422

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lio, please see the update.
  $endgroup$
  – kglr
  2 hours ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193270%2fcounting-certain-elements-in-lists%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

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 3 hours ago

answered 3 hours ago

MarcoBMarcoB

37.5k556113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is great ...
  $endgroup$
  – lio
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have an idea for result2 improvement?
  $endgroup$
  – lio
  3 hours 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
  3 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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 3 hours ago

answered 3 hours ago

MarcoBMarcoB

37.5k556113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is great ...
  $endgroup$
  – lio
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have an idea for result2 improvement?
  $endgroup$
  – lio
  3 hours 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
  3 hours 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 3 hours ago

answered 3 hours 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 3 hours ago

answered 3 hours ago

MarcoBMarcoB

37.5k556113

answered 3 hours ago

MarcoBMarcoB

37.5k556113

answered 3 hours ago

MarcoBMarcoB

37.5k556113

3

$begingroup$

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 2 hours ago

answered 3 hours ago

kglrkglr

189k10205422

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lio, please see the update.
  $endgroup$
  – kglr
  2 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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 2 hours ago

answered 3 hours ago

kglrkglr

189k10205422

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @lio, please see the update.
  $endgroup$
  – kglr
  2 hours 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 2 hours ago

answered 3 hours ago

kglrkglr

189k10205422

$endgroup$

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 2 hours ago

answered 3 hours ago

kglrkglr

189k10205422

answered 3 hours ago

kglrkglr

189k10205422

answered 3 hours ago

kglrkglr

189k10205422