How to create vectors out of (histogram) bins and take averages?

I have a vector containing speeds from 0.25m/s to 20m/s. I need to bin the speeds in bins of 0.5m/s widths cenetred at integers (e.g. bin 1: from 0.25m/s to 0.75m/s, bin 2 from 0.75m/s to 1.25m/s, bin 3 from 1.25m/s to 1.75m/s, etc.). How can I make these bins and take the average of each and create a vector containing the mean values of each bin?

I managed to plot it in a histogram:

edges = 0.25:0.5:20;
h = histogram(ws,edges);   % ws = wind speed vector

but I need a vector for each bin and its average.

[KSSV]

See the variable h, it got all your information you wanted.

[Lu Da Silva]

@KSSV Maybe I'm not seeing it... Yes there are some specifics of the histogram in 'h' but my question is how do I get the average of each bin?

Answer 1

[Steven Lord]

You can use groupsummary.

x = rand(10, 1);
y = rand(10, 1);
data = table(x, y) % For display

data = 10×2 table
       x           y    
    ________    ________

      0.2421     0.88207
     0.74513    0.091348
    0.011883     0.88123
     0.67133     0.56919
     0.56695     0.79875
      0.9975     0.27806
     0.74134    0.091931
     0.56092     0.12012
     0.73913     0.64931
     0.30515     0.17256

edges = 0:0.25:1;
% Take the mean of subsets of y corresponding to x values that fall between
% two elements of the edges vector
M = groupsummary(y, x, edges, @mean)

M = 4×1

0.8817 0.1726 0.3868 0.2781

n = 2; % Check bin by manual computation
check = mean(y(edges(n) <= x & x < edges(n+1)))

check = 0.1726

check == M(n) % true

ans = logical
   1

[Lu Da Silva]

I don't have 2 data sets (x and y), I only have the wind speed

[Steven Lord]

That's fine. That means the grouping variable and the data variable will be the same.

x = rand(10, 1)

x = 10×1

0.6509 0.3036 0.2178 0.6356 0.9714 0.3121 0.7992 0.2033 0.9275 0.0883

edges = 0:0.25:1;
% Take the mean of subsets of x corresponding to x values that fall between
% two elements of the edges vector
M = groupsummary(x, x, edges, @mean)

M = 4×1

0.1698 0.3078 0.6432 0.8994

n = 2; % Check bin by manual computation
check = mean(x(edges(n) <= x & x < edges(n+1)))

check = 0.3078

check == M(n) % true

ans = logical
   1