Find group number from a range of data that is not fixed

Hello,

I need to solve a problem like below. I have a matrix with 1 column, with unknown number of rows. I have a random number for number of row, and I need to find out that which group is this row in.

Imagine I have this data and I need to found out that row number=10 is in which group ?

My main concern is since I do not know how many groups do I have (inittially unkown) , how should I modify my code?

number_row=randperm(16,1)
if number_row<8
    disp("group 1")
else 
    disp ("group 2")
end

Thanks in advance for help

[Steven Lord]

What are the rules that you want to follow to determine the boundaries of the groups?

• Is group 1 always the numbers less than 9 and group 2 is everything else?
• Is group 1 always the first half of the data and group 2 the second half?
• Do you divide the data into groups that each contain 8 elements?
• Do you divide the data into groups based on the smallest multiple of 8 that they are less than or equal to?

Each of those four rules would divide the sample data set you showed into the groups you showed but they could behave very differently on other data sets. You tell us how to define the bins and we likely can show you how to figure out into which bin each element belongs.

[Zarak kh]

Hi Steven,

Thanks for your comment, My question lies in the your third category. I have a set of data that the size in unkown, I divide them into groups of 8 elements. finally I need to find out that a random row-number is in which group number. I hope this could help

[Jon]

I think my answer addresses what you describe

Answer 1

[Jon]

Are your groups always as you show in your example? That is, group 1 = 1,2,..8 group 2 = 9, 10,..16, group 3 = 17,18, ...24 ?

In this case you can just use

n = 8;  % number of elements in group

maxVal = 290 % put your value here
x = randperm(maxVal,1)

groupNo = floor(x/n) + 1

[Zarak kh]

Thanks Jon for your answer,

I have a question though,

Imagine two groups which has 6 members. Row 12 is in group 2, but your code generates group number=3. The same problem happen for row 6.

How can we fix that?

n = 6  % number of elements in group
% maxVal = 12 % put your value here
% x = randperm(maxVal,1)
x=12
groupNo = floor(x/n) + 1

Can we use:

groupNo=ceil(x/n)

Thanks

[Jon]

Good catch, you're right. You should use

groupNo = ceil(x/n)

[Zarak kh]

Thanks for your help.