Indexing through a cell array

I am looking to populate cell array named 'age_list' based on the values i have in two columns; named 'day' and 'pop'

• For each specific 'day', i am looking at the corresponding 'pop' value and then update the 'age_list' based on it.
• Example: On day 4, the population is 3. Therefore the cell array 'age_list' will have 'three' values with the {2,3,4}.

Can anyone suggest me on how to perform this? Any help will be appreciated.

[David Hill]

I have no idea how you are obtaining the values in age_list from day and pop. A better explanation is needed for me.

[Hari krishnan]

I will try to explain using an example. Please let me know if its unclear.

Example : If we look at the day 2, the population is 2, so it means in the 'age_list' there will be two values. For the specific case on day 2, the values in the age_list will be {1,2} . This is because there is a population increment in day 2, so the age list will have a value 1. It will also have a value 2, from the population on the previous day.

[Jan]

The question is not clear yet. The inputs are:

day = 1:7;
pop = [1,2,3,3,3,4,6];

How do you get the output age_list{6} = [1,1,2,5,6,7] ? Why is age_list{6} = [1,4,5,6] ?

[Hari krishnan]

In the age_list{6} = [1,4,5,6], the population is 4 and the corresponding day is 6.

In the age_list{7}, the corresponding day is 7 and the population becomes 6. So there is an increment in population by 2, on day 7. Therefore in the age_list{7}, there will be two more new individuals with age {1,1} along side the updated age of the other individuals {2,5,6,7}. so age_list{7} = [1,1,2,5,6,7]

[Jan]

Is day in all cases 1:numel(day) ? Can the population size shrink? Is {1} a fixed startpoint?

[Hari krishnan]

Yes, day in all cases is 1:numel(day).

The population size cannot shrink.

{1} is a fixed starting point.

Answer 1

[Jan]

A bold try:

day = 1:7;
pop = [1,2,3,3,3,4,6];
age_list = cell(1, numel(day));
age_list{1} = 1;  % Is this given?!
for k = 2:numel(day)
    new         = pop(k) - pop(k - 1);   
    age_list{k} = cat(2, repmat(age_list{1}, 1, new), age_list{k-1} + 1);
end
age_list

age_list = 1×7 cell array
    {[1]}    {[1 2]}    {[1 2 3]}    {[2 3 4]}    {[3 4 5]}    {[1 4 5 6]}    {[1 1 2 5 6 7]}

[Hari krishnan]

Thank you very much for the suggestion.

I would like to clarify an issue, if the length of days and pop is not equal, how will this work.

For example;

day = 1:178;
pop = [2,30,30,37,39,55,69,70,72,77,78,78,89,90,96,98,98,103,103,106,107,121,178];

[Jan]

I cannot guess, what you expect as output in this case.