How to bin a simple number array?

1 visualización (últimos 30 días)
That One
That One el 25 de Jul. de 2016
Comentada: Steven Lord el 26 de Jul. de 2016
Hello,
I'm reading a simple number array from a .txt file (all integers) using:
fileID = fopen('control.txt','r');
formatSpec = '%d';
A = fscanf(fileID,formatSpec);
Now I want to bin these integers into a several groups:
Group 1: integers from 21 to 100;
Group 2: integers from 101 to 500;
Group 3: integers from 501 to 1000;
Group 4: integers from 1001 to infinity (max number unknown).
I'm also interested in finding the range (how many numbers) of each group.
Thank you!

Respuesta aceptada

Azzi Abdelmalek
Azzi Abdelmalek el 25 de Jul. de 2016
A=1:120
[b,c]=histc(A,[21 50 60 100 inf])
out=accumarray(c'+1,(1:numel(c))',[],@(x) {A(x)})
out=out(2:end)
celldisp(out)
  1 comentario
That One
That One el 25 de Jul. de 2016
Hi Azzi,
Thanks for the quick answer. I'm actually trying to bin my number array into 4 groups and write the 4 groups into 4 .txt files as:
fileID = fopen('Group_1.txt','w');
fprintf(fileID, '%d\r\n',B);
fclose(fileID);
All I'm missing is the code to bin these numbers. Can you please show me how to do that? Thanks a lot!

Iniciar sesión para comentar.

Más respuestas (1)

Steven Lord
Steven Lord el 25 de Jul. de 2016
Use the histcounts function and specify a vector of edges.
  4 comentarios
Star Strider
Star Strider el 26 de Jul. de 2016
You have already come close to defining them:
Group 1: integers from 21 to 100;
Group 2: integers from 101 to 500;
Group 3: integers from 501 to 1000;
Group 4: integers from 1001 to infinity (max number unknown).
Use the max function to help you define the upper edge.
Experiment with to get the result you want.
Steven Lord
Steven Lord el 26 de Jul. de 2016
Or just specify Inf as the right endpoint of the last bin.

Iniciar sesión para comentar.

Categorías

Más información sobre Matrices and Arrays en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by