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How to add values in coloumn with corresponding of logical array?

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Pritam Daundkar
Pritam Daundkar 2023-2-7
编辑： Jan 2023-2-8
I have a matrix which is shown in following snipped table.
I have to add the value in colomn no 7 when the logical value as per in last coloumn is 1 with the previous value of coloumn 7.
For example: In 3rd row of last coloumn I have logical value 1 so I can add corresponding value of coloumn 7 i.e, 7.6079 with previous value 4.1703 and replace the 4.1703 with output 11.7782.
For that wrote following code:
% Event is the Matrix as shown in the image.
St = find(Event(:,10)); % Index of ones in last coloumn
m = length(St); % No of instance
n = length(Event);
for i = 1:n-m
if Event(i,10) == 1
s = Event(i,7) + Event(i-1,7);
Event(i-1,7)=s;
Event(i,:)=[];
end
end
This worked for all the cases where logical 1 in last coloumn is followed by 0. But, problem encountered when there are consecutive 1 in last coloumn. The above code is adding the corresponding value of 7th coloumn with respect to last coloumn.
The output should be 77.8971(5.0724+12.730+10.910+19.562+24.285+5.3377). i.e, all the corresponding values of all consecutive 1 should be added with previous 0.
But I'm geting 17.8024 (5.0724+12.730).

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Jan
Jan 2023-2-7
编辑：Jan 2023-2-7
Start from the end:
s = 0;
for k = size(Event, 1):-1:1
if Event(k, 10) % You can omit the == 1
s = s + Event(k, 7);
elseif s ~= 0
Event(k, 7) = Event(k, 7) + s;
s = 0;
end
end
% Remove the rows with 1 in column 10 afterwards:
Event(Event(:, 10) == 1, :) = [];
With your approach deleting the i.th row causes troubles, because then the former i+1.th row becomes the new i.th row. But in the next iteration i is increased and the former i+1.th row is not considered ever.
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Pritam Daundkar
Pritam Daundkar 2023-2-8
编辑：Pritam Daundkar 2023-2-8
% Input
Event = [ 0 0 0 0 0 0 5.0724 4 305 0
0 0 0 0 0 0 12.7300 9 2 1
0 0 0 0 0 0 10.9100 7 3 1
0 0 0 0 0 0 19.5620 14 2 1
0 0 0 0 0 0 24.2850 16 3 1
0 0 0 0 0 0 5.3377 4 2 1
0 0 0 0 0 0 4.9985 4 1461 0 ]
Output using your code
Event = [ 0 0 0 0 0 0 17.8024 4 305 0
0 0 0 0 0 0 4.9985 4 1461 0 ]
Which is similar answer with the my previous code.
Just adding 1st logical 1 with previous 0.
Jan
Jan 2023-2-8
编辑：Jan 2023-2-8
Ran in:
@Pritam Daundkar: Nope, I definitely get a different result:
Event = [ 0 0 0 0 0 0 5.0724 4 305 0; ...
0 0 0 0 0 0 12.7300 9 2 1; ...
0 0 0 0 0 0 10.9100 7 3 1; ...
0 0 0 0 0 0 19.5620 14 2 1; ...
0 0 0 0 0 0 24.2850 16 3 1; ...
0 0 0 0 0 0 5.3377 4 2 1; ...
0 0 0 0 0 0 4.9985 4 1461 0 ];
s = 0;
for k = size(Event, 1):-1:1
if Event(k, 10) % You can omit the == 1
s = s + Event(k, 7);
elseif s ~= 0
Event(k, 7) = Event(k, 7) + s;
s = 0;
end
end
% Remove the rows with 1 in column 10 afterwards:
Event(Event(:, 10) == 1, :) = [];
Event(:, 7)
ans = 2×1
77.8971 4.9985
So if you get another result, you do not use my code.
A sorter version:
for k = size(Event, 1) - 1:-1:1
Event(k, 7) = Event(k, 7) + Event(k+1, 7) * Event(k+1, 10);
end
Event(Event(:, 10) == 1, :) = [];

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