campare a row value with the next row

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Boby S
Boby S il 20 Feb 2020
Commentato: Boby S il 21 Feb 2020
Hi
I have the following column:
1
1
0
0
0
1
0
1
0
1
I want to campare rows value one by one with the next row ( first with second, second with third ... and ninth with tenth) and check if it changes for 1 to 0, 0 to 0, 0 to 1 and 1 to 1. For each of these conditions, I want to count them. I tried using loop and diff or sign and equations but I could not work out because the results will be similar for two conditions.

Risposta accettata

Rik
Rik il 20 Feb 2020
You were close when using diff. You need to think what characterizes all four combinations. The code below should be what you need.
v=[1;1;0;0;0;1;0;1;0;1];
d=diff(v);
u=v(1:(end-1));%shrink by 1 to make it the same size as d
clc
%if [0;0]
%then diff==0, v==0
L= d==0 & u==0;
fprintf('[0 0]: %d\n',sum(L))
%if [0;1]
%then diff==-1
L= d==-1;
fprintf('[0 1]: %d\n',sum(L))
%if [1;0]
%then diff==1
L= d==1;
fprintf('[1 0]: %d\n',sum(L))
%if [1;1]
%then diff==0, v==1
L= d==0 & u==1;
fprintf('[1 1]: %d\n',sum(L))

Più risposte (1)

Alex Mcaulley
Alex Mcaulley il 20 Feb 2020
Modificato: Alex Mcaulley il 20 Feb 2020
Another option:
a = [1;1;0;0;0;1;0;1;0;1];
b = diff(a);
b(~b) = 2*a(~b);
sol = splitapply(@numel,b,b+2) %Ordered as [1,0],[0,0],[0,1],[1,1]
sol =
3 2 3 1
  4 Commenti
Alex Mcaulley
Alex Mcaulley il 21 Feb 2020
With your second column it should work fine, because you have all the possible combinations.
Boby S
Boby S il 21 Feb 2020
I think this is the reason:
I think if one condition result is 0 then I will get the error.

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