I am trying to reduce the vector length backward each loop, like the following: I think I should use for loop but i can not figure out how to do it
x=0:.1:1;
x=0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0;
Then I would like the OUTPUT to be
x(1)= 0 .1 .2 .3 .4 .5 .6 .7 .8 .9
x(2)= 0 .1 .2 .3 .4 .5 .6 .7 .8
x(3)= 0 .1 .2 .3 .4 .5 .6 .7
.. ..
and so on.. until I reach to x(11)=0

 Réponse acceptée

madhan ravi
madhan ravi le 4 Juin 2020

0 votes

The end indexing inside the loop would be total number of elements in x minus iterator.

3 commentaires

Ali Tawfik
Ali Tawfik le 4 Juin 2020
I tried the following lines, but the final answer is empty by matrix... so how could I solve that problem ... and reach to the last answer to be zero
clear all;
clc;
x=0:0.1:1;
for i=1:length(x)
x(1:length(x)-i)
end
madhan ravi
madhan ravi le 4 Juin 2020
x = 0:.5:1;
N = numel(x);
Wanted = cell(N,1);
for ii = 1:N
Wanted{ii} = x(1: (N - ii)+1);
end
celldisp(Wanted)
Ali Tawfik
Ali Tawfik le 5 Juin 2020
Modifié(e) : Ali Tawfik le 5 Juin 2020
x=0:0.1:1;
for i=1:length(x)
x(1:(11-i)+1) % so the addition of one here to aviod make 1:0
end
thank you

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