How to add zeros to 2 x vectors where there are no matches, to get the same length ?

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x=[1 2 3 4];
y=[1 2 3 1];
x1=[3 4 5 6 7 8];
y1=[2 6 4 2 5 3];
[x, y, x1, y1] = Tout_Ajuster(x, y, x1, y1)
I want to put zeros where there is no match in x axes to get the same length for (x and x1) so I write a small code (Tout_Ajuster.m) in the output, x1 misses (5 and 6) so y1 miss(4 and 2) What is missing in my function < Tout_Ajuster.m > ???
  2 comentarios
Jon
Jon el 29 de Jul. de 2015
It would help if you gave us the expected outputs for x1 and y1. Otherwise it's not clear what you're trying to do.
Joel Sande
Joel Sande el 29 de Jul. de 2015
% This is the expected output
x = [1 2 3 4 0 0 0 0]
x1 = [0 0 3 4 5 6 7 8]
y = [1 2 3 1 0 0 0 0]
y1 = [0 0 2 6 4 2 5 3]
%% Jon is very close to the answer, %% this is an other example :
ax = [3 5 9];
ay = [1 5 6];
bx = [1 2 10 15];
by = [3 4 5 2];
from that I want :
ax = [0 0 3 5 9 0 0];
ay = [0 0 1 5 6 0 0];
bx = [1 2 0 0 0 10 15];
by = [3 4 0 0 0 5 2];

Iniciar sesión para comentar.

Respuesta aceptada

Joel Sande
Joel Sande el 29 de Jul. de 2015
In this example, Test_Ajuster.m is calling Tout_Ajuster.m

Más respuestas (3)

Jon
Jon el 29 de Jul. de 2015
Your question seems ambiguous to me, but based off my interpretation, see if this code gives you the result you're looking for:
% This assumes that none of the matching x values are 0
N = max(numel(x),numel(x1));
xout = zeros(1,N);
yout = zeros(1,N);
x1out = zeros(1,N);
y1out = zeros(1,N);
% Reinsert the known values of x and y
xout(1:numel(x)) = x;
yout(1:numel(y)) = y;
% Values in x1 (and the correponding y1 positions) that do not appear in x are set to zero
for i = 1:numel(x)
matching = find(x(i) == x1);
if isempty(matching) == 1
continue
else
x1out(matching) = x1(matching);
y1out(matching) = y1(matching);
end
end
Gives the following results:
xout =
1 2 3 4 0 0
yout =
1 2 3 1 0 0
x1out =
3 4 0 0 0 0
y1out =
2 6 0 0 0 0

Azzi Abdelmalek
Azzi Abdelmalek el 29 de Jul. de 2015
n=max(numel(x),numel(x1))
x=[x zeros(1,n)]
x1=[x1 zeros(1,n)]
y=[y zeros(1,n)]
y1=[y1 zeros(1,n)]

Joel Sande
Joel Sande el 29 de Jul. de 2015
%% It is working now. Thank
ax = [3 5 9];
ay = [1 5 6];
bx = [1 2 10 15];
by = [3 4 5 2];
>> [a, b, c, d] = Tout_Ajuster(ax,ay,bx,by)
a =
0 0 3 5 9 0 0
b =
0 0 1 5 6 0 0
c =
1 2 0 0 0 10 15
d =
3 4 0 0 0 5 2

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