Less than what number in the file ht could give 95%?

Attached is column vector ht . I would want to find out below what number in ht would constitute 95% of ht ? For example, numbers in ht that are less than 25 constitute 43.78% of ht.

Answer 1

[Star Strider]

It took me a bit to understand what you wanted, but the result is actually straightforward:

D = load('ht.mat');
ht = D.ans;
ht95idx = fix(0.95*length(ht));
ht95 = ht(ht95idx)

figure(1)
plot(ht, '-r')
hold on
plot(ht95idx, ht95, 'bp')
hold off
grid
xlabel('index')
ylabel('ht')
text(ht95idx-150, ht95, sprintf('95%%‘ht’ at %5.1f \\rightarrow', ht95), 'HorizontalAlignment','right')

The plot of ‘ht’ as a function of its index:

[AbelM Kusemererwa]

Also, what will be the percentage if I need <=10 of elements of ht

[Star Strider]

If I understand correctly what you want, the decimal fraction would be:

DecFrc = 10/length(ht);

for the first 10 elements of ‘ht’. The percentage would be 100 times that:

Pct = 100*DecFrc;

[AbelM Kusemererwa]

What I meant was 95% correspond to 33.7. what percentage will 10 give?

[Star Strider]

I’m still not certain that I understand what you want, but see if this works:

ht10idx = fix(0.10*length(ht));
ht10 = ht(ht10idx)

That’s the same as my previous code, except for the change of constant.

[AbelM Kusemererwa]

What I meant was 95% correspond to 33.7. what percentage of ht will correspond to 10?

[Star Strider]

When I run the code I posted in my last Comment, it evaluates ht10=16.7.

This code will create a plot that may help, since instead of plotting ‘ht’ as a function of the index, it plots it as a function of the percentage:

D = load('ht.mat');
ht = D.ans;
idx = 1:length(ht);                             % Create Index Vector
pct = 100*idx/idx(end);                         % Calculate Index As Percent

ht95idx = fix(0.95*length(ht));                 % Index at 95%
ht95 = ht(ht95idx);                             % Value at 95%

ht10idx = fix(0.10*length(ht));                 % Index at 10%
ht10 = ht(ht10idx);                             % Value at 10%

figure(1)
plot(pct, ht, '-r')
hold on
plot(pct(ht95idx), ht95, 'bp')
plot(pct(ht10idx), ht10, 'bp')
hold off
grid
xlabel('Percentage')
ylabel('ht')
text(pct(ht95idx-150), ht95, sprintf('95%%‘ht’ at %5.1f \\rightarrow', ht95), 'HorizontalAlignment','right')
text(pct(ht10idx+150), ht10, sprintf('\\leftarrow 10%%‘ht’ at %5.1f', ht10), 'HorizontalAlignment','left')

The plot:

__________________________________

This version of the code will return the value of ‘ht’ for any percent you specify (without the plot):

D = load('ht.mat');
ht = D.ans;
idx = 1:length(ht);                             % Create Index Vector
pct = 100*idx/idx(end);                         % Calculate Index As Percent

PrctInp = inputdlg('Enter percent of ‘ht’ desired:', 'Calculate ‘ht’ Percent');
Prct_ht = str2num(PrctInp{:});
ht_Idx = find(pct <= Prct_ht, 1, 'last');
ht_Val  = ht(ht_Idx);
msgbox(sprintf('The value of ‘ht’ at %4.1f%%= %4.1f', Prct_ht, ht_Val), 'Answer');

[AbelM Kusemererwa]

I intend to enter ht desire to obtain the percentage

[Star Strider]

This should work:

D = load('ht.mat');
ht = D.ans;
idx = 1:length(ht);                             % Create Index Vector
pct = 100*idx/idx(end);                         % Calculate Index As Percent

htInp = inputdlg('Enter the ‘ht’ desired:', 'Calculate ‘ht’ Percent');
Des_ht = str2num(htInp{:});
ht_Idx = find(ht <= Des_ht, 1, 'last');
Prct_ht  = ~isempty(ht_Idx)*pct(ht_Idx);
if isempty(ht_Idx)
    Prct_ht = 0;
end
msgbox(sprintf('The value of ‘ht’ at %4.1f%%= %4.1f', Prct_ht, Des_ht), 'Answer');

[AbelM Kusemererwa]

Thanks

[Star Strider]

My pleasure!

Answer 2

[bio lim]

First of all, I think it is bad idea to save your variable as ' ans ', so you should define a unique name. (I have renamed ' ans ' to ' data ' in your ht.mat file.)

This might be a clumsy code but should do the trick.

load('ht.mat')  

x = zeros(size(data)); % Create vector of zeros
  for i = 1 : length(data)
     x(i) = sum(data==data(i)); % Counting the number of occurance
  end

z = [data x]; % Matrix with first column as your initial data, and second as the number of occurence.
h = unique(z, 'rows'); % Only unique elements are selected.
sum = cumsum(h(:,2)); % 95% of 23880 is 22686
g = [h sum];

Thus less than 34.7 could give you 95% (More precisely 95.017%) . Hope it helps.

Answer 3

[Sean de Wolski]

You should be able to use prctile for this.

x = rand(100,1);
xp = prctile(x,[0,95]); % 2nd element

plot(sort(x))
line([1 100],xp([2 2]))
title(['95% is ' num2str(xp(2))])