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deleting numbers in cell array

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I have a 32x5 cell array.
I want to delete all numbers (+/- 3 either side of each number) that do not appear in every column.
Is there a function that can help me?
Thanks
Sam

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sam van Bohemen
sam van Bohemen on 25 Feb 2020
Yes, I mean that values should be deleted if it is not the case that every column has some entry that is with the range of value-3 to value+3
Sorry I confused myself in the last post:
if the array was:
10 11 10 28 9
39 40 41 40 39
58 47 58 57 59
68 58 68 69 70
I would want to delete:
10 11 10 28 9
47
68 68 69 70
leaving:
39 40 41 40 39
58 58 68 69 70
when values are deleted I want them to be removed so the below entries moves up 'to fill the hole'.
So all values within value-3 to value+3 are in the same row.
I thought this may be easier to do with a cell array, however if a numeric array is best I can use that instead.
Thanks
Sam
Robert U
Robert U on 25 Feb 2020
The description of your problem is ambiguous. It is not clear how your algorithm should work neither the sequential order of commands is clear. I encourage you to describe the desired functionality step-by-step which might help you to write your own code quickly.
From what you wrote (but still not fitting your example values):
  1. Compare the first element of input matrix within its row with all other values.
  2. If all other values of said row are not within first element +/- 3, remove first element (by replacing it with NaN)
  3. Repeat steps 1 & 2 with all other values within input matrix.
  4. Move all remaining values (not NaN) as much rows up keeping the same column as possible without replacing or reordering numerical values.
Example:
dInput = [10 11 10 28 9; ...
39 40 41 40 39; ...
58 47 58 57 59; ...
68 58 68 69 70];
dInMin = dInput - 3;
dInMax = dInput + 3;
for numRow = 1:size(dInput,1)
for numCol = 1:size(dInput,2)
bRange = dInput(numRow,numCol) >= dInMin(numRow,:) & dInput(numRow,numCol) <= dInMax(numRow,:);
bRange(numCol) = 0;
if ~any(bRange)
dInput(numRow,numCol) = NaN;
end
end
end
dInput = [10 11 10 NaN 9
39 40 41 40 39
58 NaN 58 57 59
68 NaN 68 69 70]
Does it really make sense to move remaining values up?
Kind regards,
Robert
sam van Bohemen
sam van Bohemen on 27 Feb 2020
I want to find all values (value-3 to value+3) that are not present in every column and then delete these values.
I will try to describe it step-by-step:
Say the matrix is:
10 11 10 28 9
39 40 41 40 39
58 47 58 57 59
68 58 68 69 70
1) starting with the first element of the matrix (10). I want to search the other 4 columns to determine if they all contain a number between 7 and 13 (10-3 and 10+3).
2) column two contains 11, column three contains 10 and column five contains 9. However there is no number between 7 and 13 in column four.
3) therefore I would delete all of these numbers (10 11 10 9).
4) I would then move on to '28'. none of the other columns contain any number between 25 and 31 (28-3 and 28+3). therefore I would delete 28
5) I would then move on to '39'. all five columns contain a number between 36 and 42 (39-3 and 39+3). therefore I would keep all of these numbers (39 40 41 40 39).
6) moving on to '58' all five columns contain a number between 55 and 61 (58-3 and 58+3). therefore I would keep all of these numbers (58 58 58 57 59).
7) moving onto '47' none of the other columns contain a number between 43 and 50 (47-3 and 47+3). therefore I would delete 47.
8) moving onto '68' all columns except column two contain a number between 65 and 71 (68-3 and 68+3). therefore I would delete all of these numbers (68 68 69 70).
9) This would leave me with:
NaN NaN NaN NaN NaN
39 40 41 40 39
58 NaN 58 57 59
NaN 58 NaN NaN NaN
10) I want to make it so all numbers move up when the number above it is deleted so all numers -3 to +3 are in the same row, as below:
39 40 41 40 39
58 58 68 69 70
Hope this make is more clear. Sorry it is so wordy

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Accepted Answer

Robert U
Robert U on 2 Mar 2020
Hi sam van Bohemen,
Thank you for your description. I guess the following code will do the trick:
dInput = [10 11 10 28 9; ...
39 40 41 40 39; ...
58 47 58 57 59; ...
68 58 68 69 70];
dInMin = dInput - 3;
dInMax = dInput + 3;
for numRow = 1:size(dInput,1)
for numCol = 1:size(dInput,2)
bRange = all(any(dInput(numRow,numCol) >= dInMin(:,[1:numCol-1,numCol+1:end]) & dInput(numRow,numCol) <= dInMax(:,[1:numCol-1,numCol+1:end]),1));
if ~bRange
dInput(numRow,numCol) = NaN;
end
end
end
bRowMoved = true;
while bRowMoved
for numRow = 2:size(dInput,1)
bRowMoved = false;
for numCol = 1:size(dInput,2)
if isnan(dInput(numRow-1,numCol))
dInput(numRow-1,numCol) = dInput(numRow,numCol);
dInput(numRow,numCol) = NaN;
bRowMoved = true;
end
end
end
dInput(all(isnan(dInput),2),:) = [];
end
Kind regards,
Robert

  1 Comment

sam van Bohemen
sam van Bohemen on 4 Mar 2020
Thanks so much Robert, your code worked!

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