How to perform the following data splitting?

How to perform the following data splitting?

There are vectors X,Y,T all of these have the same size

I want to split these 3 vectors and store the splitted data in 3 arraies X_array y_array T_array

The splitting will be according to a specific range

for example:

split X and Y according to T

T contains time data from 5.2 to 10 (i,e T=[5.2,5.2002,.........................5.202, ................10])

I want to split these vectors every 0.002 of T

for example include all the data of X in one vector until T reach 5.202 then store in the first cell in the array,then include all the data of X in one vector until T reach 5.204 then store in the 2nd cell in the array,,, and so on

similar to Y and T

Answer 1

[Dyuman Joshi]

%Random data
t = 5.2:0.002:5.3;
x = 1:numel(t);

%Define bins to group data into 
cats = min(t):0.02:max(t);

%Group the data according to the bins defined
idx = discretize(t,cats);

%Split the data according to groups and store in a cell array
%Do the same for y as well
X = splitapply(@(k) {k}, x, idx)

X = 1×5 cell array
    {[1 2 3 4 5 6 7 8 9 10]}    {[11 12 13 14 15 16 17 18 19 20]}    {[21 22 23 24 25 26 27 28 29 30]}    {[31 32 33 34 35 36 37 38 39 40]}    {1×11 double}

X'

ans = 5×1 cell array
    {[            1 2 3 4 5 6 7 8 9 10]}
    {[   11 12 13 14 15 16 17 18 19 20]}
    {[   21 22 23 24 25 26 27 28 29 30]}
    {[   31 32 33 34 35 36 37 38 39 40]}
    {[41 42 43 44 45 46 47 48 49 50 51]}

[Walter Roberson]

We probably should not count on the input T starting from 5.2 so we should probably not use min() for the lower bound. We are told to start from 5.2 so that should be the lower bound.

[M]

@Dyuman Joshi, Thanks, your solution worked with me and the splitting was correct

Answer 2

[Walter Roberson]

idx = round((T(:)-5.2)/0.02) + 1;
Xgrouped = accumarray(idx, X(:), [], @(vals){vals});
Ygrouped = accumarray(idx, Y(:), [], @(vals){vals});
Tgrouped = accumarray(idx, T(:), [], @(vals){vals});

[M]

@Walter Roberson the splitting is not correct

the T shouldnt start from 5.2, we should start from the first index of T until we increase 0.002 and then increase 0.002 and so on

I tried

idx = round((T(:))/0.002) + 1;

the first index in this example start with 5.2 , but it reached to 5.201 not to 5.2019 in the first cell

and the second cell should start from 5.202

even with this same problem

idx = round((T(:)-5.2)/0.002) + 1;

[M]

idx = round((Time(:)-Time(1,1))/0.002) + 1;

Unrecognized function or variable 'Time'.

Xgrouped = accumarray(idx, X(:), [], @(vals){vals});
Ygrouped = accumarray(idx, Y(:), [], @(vals){vals});
Tgrouped = accumarray(idx, Time(:), [], @(vals){vals});

[M]

@Walter Roberson Dyuman solution worked with me,but I am wondered why your solution didn't work