Code for Counting and Lookup
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Waqas Siddique
el 4 de Oct. de 2020
Comentada: Adam Danz
el 19 de Feb. de 2021
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/370756/image.png)
I have imported a csv file into MATLAB. The number of rows in the table is 81 so there is a long series of '1's and '0's in the single column. I want to check that whenever the consecutive number of rows with '1's is greater than 4, then flag it and note the time stamp corresponding to the first '1' for that consecutive rows of '1'. Do this for every series of 1's appearing successively for more than 4 times i.e obtain the corresponding time. The time stamps are given in the column Var1.
4 comentarios
Rik
el 4 de Oct. de 2020
Did you mean elseif?
You forgot to format the code, and you forgot to account for the length of your variable. What happens when i is equal to the length of the table?
Respuesta aceptada
Adam Danz
el 4 de Oct. de 2020
Editada: Adam Danz
el 4 de Oct. de 2020
Here's a demo.
v is the input vector of 1s and 0s (or Trues and Falses)
r is the output vector of row numbers in v that start 4+ consecutive 1s.
% Demo data: v is a vec of 1s and 0s (or Trues and Falses)
A = [1 0 0 1 1 1 1 1 1 0 0 1 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 1 1 1 1 1]';
% Length of each group of consecutive 1s
B = diff(find([0;A(:);0]==0))-1;
B(B==0) = [];
% Index of 1st '1' in each group of consecutive 1s
firstIdx = find(diff([0;A(:)])==1);
% Row number of the first 1 in groups of 4 or more consecutive 1s
minConsec = 4;
r = firstIdx(B >= minConsec);
Result:
r =
4
20
25
42
48
14 comentarios
Adam Danz
el 19 de Feb. de 2021
>I need help in counting the consecutive number of 1's.
Look at variable B in my answer. A comment in my answer describes B as "Length of each group of consecutive 1s".
Más respuestas (1)
Image Analyst
el 17 de Oct. de 2020
If you have the Image Processing Toolbox, you can also use bwareaopen() and regionprops():
A = [1 0 0 1 1 1 1 1 1 0 0 1 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 1 1 1 1 1]';
A = bwareaopen(logical(A), 4); % Just extract runs of 4 or longer.
props = regionprops(A, 'PixelIdxList') % Find indexes of all runs of 1's.
% Done! Results are in the table.
% Print out all runs where the length is 4 or more
startingIndexes = zeros(height(props), 1);
for k = 1 : height(props)
startingIndexes(k) = props(k).PixelIdxList(1);
fprintf('Region %d starts at index %d.\n', ...
k, startingIndexes(k));
end
Gives the same results as Adam's solution.
Region 1 starts at index 4.
Region 2 starts at index 20.
Region 3 starts at index 25.
Region 4 starts at index 42.
Region 5 starts at index 48.
If you also want the lengths of the runs in addition to their starting index, just ask regionprops() for 'Area':
props = regionprops(A, 'PixelIdxList', 'Area') % Find lengths and indexes of all runs of 1's.
allLengths = [props.Area]
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