# How to take an average?

2 visualizaciones (últimos 30 días)
Haya Ali el 17 de Mzo. de 2023
Editada: Haya Ali el 17 de Mzo. de 2023
I have this array and according to excel histogram plot the mostly values are located in the range [-0.5,0.5]. I want to sort all values from the array A that lies in this range and take there average. How can I do this?
A= [0.0000 0.4341 -0.0000 -0.5910 -0.0352 2.0350 -0.0000 -0.9597 0.0000 -1.2164 -2.7826 -0.0000 0.3716 -0.0000 -0.0000 -0.0000 1.4557 0.0000 -0.0000 0.5599 -0.0000 -0.2463 -0.7001 0.0000]
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Iniciar sesión para comentar.

Raghvi el 17 de Mzo. de 2023
Hey Haya,
To find the range in which most values are present, you can use the histcount() function: https://www.mathworks.com/help/matlab/ref/histcounts.html
[n,e] = histcounts(A);
This will partitions the A values into bins(ranges), and returns the count in each bin(n), as well as the bin edges(e).
So the maximum values will be present in the range:
[maxn,i] = max(n);
range = [e(i), e(i+1)];
In this case it will be (0,1)
To find the average of only these values:
x = A >= range(1) & A<= range(2); % or > and < instead of >= and <=
result = mean(A(x));
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Iniciar sesión para comentar.

### Más respuestas (1)

HimeshNayak el 17 de Mzo. de 2023
Hi Haya,
From what I understand, you want to find the average of the range of the elements present in the array. Follow the following steps:
1. Find the range of the element in the array.
[minA, maxA] = bounds(A);
2. Store the range values in an array.
M = [minA, maxA];
3. Find the average / mean of the values.
avg = mean(M);
Regards
HimeshNayak
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Iniciar sesión para comentar.

### Categorías

Más información sobre Logical en Help Center y File Exchange.

### Community Treasure Hunt

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

Start Hunting!

Translated by