I need to make groups from array

I have this array, I need to make groups of these 14 elements with the rule that if value of a new element is 1 to 5 percent higher than the value of previous elements then the new element will be part of previous group, otherwise new group will start if it is higher than 5%. Thanks

5 Comments

1 to 5 percent higher than what measurement? The running maximum of the group? If so then an implication would be that if the first value was 0.1 then the next value should form its own group if it is greater than 1.01*0.1 ?
What about the case where the new value is lower than the previous? Or if the value is greater by less than 1%?
Like first element of this arrays is 0.06 that is lower than next element (0.1800),
So if 0.1800 is higher than [ 5% of 0.06+0.06], then new group will start from 0.1800. and previous group will attain only one element 0.06 (1st group).
In this way, third element is 0.15, now if this element 0.15 is higher than [ 5% of 0.1800+0.1800](previous elemnt+5% of previous element), then new group will start from 0.1300 (3rd group). otherwise if it is lower, then 0.15 will be the part of previous group (2nd group with element 0.1800) . and so on.
I suggest writing a loop.
Is the 5% always to be relative to the largest value in the current group? Or to the first value in the group? Or relative to the last value in the group (so in other words you would always be ending up comparing adjacent elements) ?
Yes I need to compare all adjacent elements to form groups. Like firstly I compare element one and two and then two and three and then three and four, and so on.
can you give an example based on our discussion, . then i will try to adjust the loop according to my need.

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 Accepted Answer

A = [.06, .18, .15, .14, .13, .12, .125, .15, .105, .115, .34, .32, .03 .14]
A = 1×14
0.0600 0.1800 0.1500 0.1400 0.1300 0.1200 0.1250 0.1500 0.1050 0.1150 0.3400 0.3200 0.0300 0.1400
mask = A(2:end) > 1.05 * A(1:end-1);
group_lengths = diff([0, find(mask), length(A)])
group_lengths = 1×6
1 6 2 1 3 1
mat2cell(A, 1, group_lengths)
ans = 1×6 cell array
{[0.0600]} {[0.1800 0.1500 0.1400 0.1300 0.1200 0.1250]} {[0.1500 0.1050]} {[0.1150]} {[0.3400 0.3200 0.0300]} {[0.1400]}

6 Comments

thanks. got it. one thing is confusing me, why you used 1.05 ?
x + 5% of x is 100% of x plus 5% of x, collect the x to get (100+5) = 105% of x. Expressed as a fraction, that is 1.05 * x
Thanks for the help. I appreciate it.
Hi walter, if i add another condition x + - 5% of x (mean now x plus/minus 5%), so in this way i will set a threhold a like 95% to 105%, how can i set this in the existing code line.
A = [.06, .18, .15, .14, .13, .12, .125, .15, .105, .115, .34, .32, .03 .14]
A = 1×14
0.0600 0.1800 0.1500 0.1400 0.1300 0.1200 0.1250 0.1500 0.1050 0.1150 0.3400 0.3200 0.0300 0.1400
mask = A(2:end) < 0.95 * A(1:end-1) | A(2:end) > 1.05 * A(1:end-1);
group_lengths = diff([0, find(mask), length(A)])
group_lengths = 1×13
1 1 1 1 1 2 1 1 1 1 1 1 1
mat2cell(A, 1, group_lengths)
ans = 1×13 cell array
{[0.0600]} {[0.1800]} {[0.1500]} {[0.1400]} {[0.1300]} {[0.1200 0.1250]} {[0.1500]} {[0.1050]} {[0.1150]} {[0.3400]} {[0.3200]} {[0.0300]} {[0.1400]}
zhoug zho
zhoug zho on 1 Jul 2021
Edited: zhoug zho on 1 Jul 2021
got it... Thank you so much.

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More Answers (1)

Some guessing. You only said it stays in group is greater by 1-5%, I assumed a new group otherwise. Anyways, you modify the below to meet your needs.
n{1}=yourArray(1);
c=1;
for k=2:length(yourArray)
if yourArray(k)<=1.05*yourArray(k-1)&&yourArray(k)>=1.01*yourArray(k-1)
n{c}=[n{c},yourArray(k)];
else
c=c+1;
n{c}=yourArray(k);
end
end

2 Comments

Thanks David,
However, if it is lower it needs to stay with the group being built according to https://www.mathworks.com/matlabcentral/answers/868868-i-need-to-make-groups-from-array#comment_1613113 and that means that you should not be testing against a lower bound. The 1% is a red herring; the only question is more than 5% increase (new group) otherwise continue the same group.

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