Cumulative sum with conditions

I am counting a number of 1's in an array.

But I want to count the cumulative sums only when the number of 1's are followed by a sequence of zeros.

 % For instance,
h1=[1 0 0]
h2=[1 1 0]
h3=[1 1 1]
h4=[0 0 1]
h5=[0 1 0].

I want to count the cumulative sums (S) of those arrays as follows.

 % The ideal outputs are:
S1= 1
S2= 2
S3= 0
S4= 0
S5= 0

since h2: h4 do not meet the condition of 1's followed by 0's.

I would truly appreciate if any one could advise me an effective way of programming the conditional sums.

[Rik]

So you have these as numbered variables? Then you should really have them in a single array. Also, I don't understand the rules here. Are we meant to look row by row? If so, doesn't the 5th row count as well (as there is a 0 after the 1)?

Answer 1

[Bruno Luong]

Change the last line to make it can handle with single row:

    m = size(h,1); 
    d = diff(h,1,2);
    [r,c] = find(d==-1);
    b = sum(abs(d),2)==1;
    s = accumarray(r(:),c(:),[m 1]).*double(b)

This because find() rotates the result for row input array.

[yp78]

It worked perfectly! Many thanks for your patience and kind help. It was also a good opportunity to lean new functions (for me) such as accumarray.

Answer 2

[Sean de Wolski]

h = [1 0 0;
   1 1 0;
   0 0 1;
   1 1 1;
   0 1 0];
s = sum(h, 2).*~h(:,end)

requiring the last element to be zero. This wouldn't work for [1 0 1] but I don't know what you'd want for that.

Answer 3

[yp78]

Hi both, Thank you for your immediate responses.

> Sean, actually your comment of:

 "This wouldn't work for [1 0 1] "

is where I am struggling with. Explaining a bit in detail, I am doing a statistical test described in Test for Cointegration Using the Johansen Test. In one trial (and after some extra operation) what I get is an array of binary values 1's and 0's. e.g.

 h=[1 1 0] .

In order to reach the conclusion of "there are at most 2 cointegrating vectors or less" (hence in this case the number that I want to see is the cumulative sum of 2). But if it were

h= [1 0 1]

I want to count the sum as zero as I cannot make any inference from the output.

Since I have to repeat this test 1000 times, I want to create a function, calculating the cumulative sum of an array when it meets the condition of starting from 1, 1's followed by 0's.

[Sean de Wolski]

In that case what I have above should work. It returns the sum and requires that the last element be a zero. If the last element is non-zero, it returns a zero.

With [1 0 1] it would say zero because the last element is not a zero.

[yp78]

Thank you Sean. But the problem still remains with h= [0 1 0] as it doesn't meet the condition of the array starting from 1(sorry this point was not clear in my previous comment).

When h= [0 1 0] I can't make any statistical inference, hence I need to have a return value of "0" whereas your current code gives "1".

Answer 4

[Bruno Luong]

h = [1 0 0;
1 1 0;
0 0 1;
1 1 1;
0 1 0];

[a,c] = max(1-h,[],2);
s = (c-1).*a.*h(:,1)

[yp78]

Thanks Bruno. It works well with 3 elements, but dosn't work when I increase the number of elements. For instance, with

h = [1 0 1 0 ]

the function returns "1" although I want to have "0" in this case. I am trying to understand your code line by line, but not quite there yet to adjust it by myself. Would you be able to help me to modify your code for an array having more than 3 elements?

[Bruno Luong]

Can you explain why it returns 0? Still not clear the rule...

Do you only count only if the row is composing a sequence of 1s followed by a sequence of 0s (of at least one 0)?

[yp78]

1 and 0 corresponds to the rejection (1) and non-rejection (0) of a statistical test. The hypothesis is tested sequentially. The null hypothesis is:

(trial 1) There are 0 relationship between variables

(trial 2) There are at most 1 or less statistical relationship between variables

(trial 3) There are at most 2 or less statistical relationship between variables

(trial 4) There are at most 3 or less statistical relationship between variables ...etc.

In order obtain a useful inference, I need to know when the first 0 appears after a sequence of 1's. For instance, in the case of;

h=[1 1 0 0 ]

I can make the inference as "there are 2 or less relationships found". But if it were;

h=[1 1 0 1 ]

the result becomes no longer reliable as the trial 3 says;

"there are 2 or less relationships in the data"

whereas the following trial rejects

"there are 3 or less relationships in the data" which is contradictory to the previous trial.

Answer 5

[Bruno Luong]

m = size(h,1); 
d = diff(h,1,2);
[r,c] = find(d==-1);
b = sum(abs(d),2)==1;
s = accumarray(r(:),c(:),[m 1]).*double(b)

[yp78]

Thank you Bruno, The function works fine up to the 3rd line. But I get an error with the last line. If the input was an 1 by 4 array, which number do I need to put into "m" ? Say, could you show me how I need to adjust the function with the array h = [1 0 1 0]?

[Bruno Luong]

sorry, code is now corrected, m is the number of rows

[Torsten]

m = size(h,1);

[yp78]

Very sorry for asking you again. But I still got an error saying:

Error using accumarray Second input VAL must be a vector with one element for each row in SUBS, or a scalar.

Where the code is written as follows.

h = [1 0 1 0 ];
m = size(h,1);
d = diff(h,1,2);
[r,c] = find(d==-1);
b = sum(abs(d),2)==1;
s = accumarray(r,c,[m 1]).*double(b)