# campare a row value with the next row

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Boby S on 20 Feb 2020
Commented: Boby S on 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.

Rik on 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))
Boby S on 20 Feb 2020
Thanks

### More Answers (1)

Alex Mcaulley on 20 Feb 2020
Edited: Alex Mcaulley on 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
Alex Mcaulley on 21 Feb 2020
With your second column it should work fine, because you have all the possible combinations.
Boby S on 21 Feb 2020
I think this is the reason:
I think if one condition result is 0 then I will get the error.