How to identify duplicate elements index values in this array without deleting them

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sakshi chopra 2021-7-25

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/885669-how-to-identify-duplicate-elements-index-values-in-this-array-without-deleting-them

评论： sakshi chopra 2021-7-26

A=[53	54	80	1
43	12	0
71	14	0
55	70	23
69	26	0
66	38	0
78	60	0
3	48	0
7	42	0
17	74	0
30	64	0
35	67	0]

how to identify duplicate elements index values in this array without deleting them like 67 is repeated twice and so as 64

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Answer 1

Image Analyst 2021-7-25

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https://ww2.mathworks.cn/matlabcentral/answers/885669-how-to-identify-duplicate-elements-index-values-in-this-array-without-deleting-them#answer_753624

在 MATLAB Online 中打开

Try this:

A = [53	54	80	1
21	43	12	0
27	71	14	0
52	55	70	23
36	69	26	0
46	66	38	0
68	78	60	0
59	3	48	0
74	7	42	0
42	17	74	0
67	30	64	0
64	35	67	0]
minValue = min(A(:))
maxValue = max(A(:))
[counts, edges] = histcounts(A, minValue : maxValue)
for k = 1 : length(counts)
	if counts(k) >= 2
		fprintf('%f is in there %d times.\n', edges(k), counts(k));
	end
end

You'll see:

0.000000 is in there 10 times.

42.000000 is in there 2 times.

64.000000 is in there 2 times.

67.000000 is in there 2 times.

74.000000 is in there 2 times.

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Image Analyst 2021-7-26

在 MATLAB Online 中打开

@sakshi chopra, you didn't mention any of that originally.

To find out the rows that have duplicate values within the row, do this:

[rows, columns] = size(A)
for row = 1 : rows
	% Get the whole row.
	thisRow = A(row, :);
	% Delete anything that has a value of 0
	thisRow(thisRow == 0) = [];
	% Find the number of unique values
	uniqueValues = unique(thisRow);
	if length(thisRow) > length(uniqueValues)
		fprintf('Row %d has at least one duplicate value.n', row);
	end
end

Note that no row has duplicate values in the example you gave.

If you want to find duplicates anywhere in the array, even if the duplicate lives on a different row or column, and ignore zeros, use this:

A = [53	54	80	1
21	43	12	0
27	71	14	0
52	55	70	23
36	69	26	0
46	66	38	0
68	78	60	0
59	3	48	0
74	7	42	0
42	17	74	0
67	30	64	0
64	35	67	0]
minValue = min(A(:))
maxValue = max(A(:))
[counts, edges] = histcounts(A, minValue : maxValue)
ca = cell(1, length(counts));
for k = 1 : length(counts)
	if counts(k) >= 2  && edges(k) ~= 0
		fprintf('%f is in there %d times at these locations.\n', edges(k), counts(k));
		% Find locations where this value lives.
		[r, c] = find(A == edges(k));
		ca{k} = [r, c];
		for k2 = 1 : length(r)
			fprintf('   at row %d, column %d.\n', r(k2), c(k2));
		end
	end
end

You see

42.000000 is in there 2 times at these locations.

at row 10, column 1.

at row 9, column 3.

64.000000 is in there 2 times at these locations.

at row 12, column 1.

at row 11, column 3.

67.000000 is in there 2 times at these locations.

at row 11, column 1.

at row 12, column 3.

74.000000 is in there 2 times at these locations.

at row 9, column 1.

at row 10, column 3.

and ca is a cell array that contains the rows and columns that the number appears in for each number. It's a cell array because each number could appear a different number of times (otherwise you could have used a nromal double 3-D array).

sakshi chopra 2021-7-26