Detecting values in a vector that are different but very close to each other

Question

Paramonte 2020-2-3

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https://ww2.mathworks.cn/matlabcentral/answers/503351-detecting-values-in-a-vector-that-are-different-but-very-close-to-each-other

评论： Paramonte 2020-2-3

采纳的回答： Steven Lord

Hello there:

I have a t vector (time, increasing values) like this:

t=[ 1 1.1 2 3 3.1 4.1 5 6 7.1 7.2]

to which corresponds y values

y=[ 10 12 10 9 1 12 12 4 9 12 ]

I would like to remove in x the values whose difference to the next one is <= 0.1, so I get a

t_new=[ 1 2 3 4.1 5 6 7.1] and then to make a corrspondenece to the new y, in a way that the y values correspondent to the similar x values are added, so:

y_new=[10+12 10 9+1 12 12 4 9+12]

Thanks in advance!!

Regards

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J. Alex Lee 2020-2-3

Then what is the rationale for rounding the 7.1 down to 7?

Might want to move your comment up to this thread

Paramonte 2020-2-3

sorry I did a mistake: i want to keep 7.1. I edited to correct this. Cheeers

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Answer 1

Steven Lord 2020-2-3

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https://ww2.mathworks.cn/matlabcentral/answers/503351-detecting-values-in-a-vector-that-are-different-but-very-close-to-each-other#answer_413513

在 MATLAB Online 中打开

Use uniquetol to unique-ify the data with a tolerance. uniquetol can return a vector of indices that indicate to which of the unique values each original value corresponds. Then use accumarray to accumulate the corresponding values of the second vector together.

t=[ 1     1.1     2      3     3.1    4.1   5     6    7.1    7.2];
y=[ 10  12  10  9   1  12    12    4    9    12 ];
[t2, ~, ind2] = uniquetol(t, 0.11);
ynew = accumarray(ind2, y);
uniqueItemsWithValues = [t2.', ynew]

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J. Alex Lee 2020-2-3

Cool, never knew about uniquetol, but on looking at the doc, isn't there an ambiguity about which value (within a tolerance) is returned? If this can be used as answer to the original question, it suggests that uniquetol will always return the lowest value in the tolerance-group as "the unique" value...running the example, this appears true, and it does not appear to be a side effect of the original order of t. This decision doesn't seem like a unique obvious choice to me...I could imagine wanting the average of the tolerance-group, or the max...or in general wanting to apply some custom function.

Paramonte 2020-2-3

I must thak Image Analist and Steven Lord for your time and effot.

Steven reply worked perfectely.

many thanks indeed

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Answer 2

Paramonte 2020-2-3

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https://ww2.mathworks.cn/matlabcentral/answers/503351-detecting-values-in-a-vector-that-are-different-but-very-close-to-each-other#answer_413495

Thakn you for your reply.

Yes I want to keep the 4.1 since the next valu is 5 so, 5-4.1=0.9 which is over 0.1

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Answer 3

Image Analyst 2020-2-3

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https://ww2.mathworks.cn/matlabcentral/answers/503351-detecting-values-in-a-vector-that-are-different-but-very-close-to-each-other#answer_413504

在 MATLAB Online 中打开

This works:

t=[ 1     1.1     2      3     3.1    4.1   5     6    7.1    7.2]
y=[ 10  12  10  9   1  12    12    4    9    12 ]
dt = diff(t)
bigDiff = dt >= 0.11 % Change according to what you think is a big enough difference.
badIndexes = find(~bigDiff) + 1
goodIndexes = [1, find(bigDiff) + 1]
yCopy = y;
yCopy(badIndexes - 1) = yCopy(badIndexes - 1) + yCopy(badIndexes)
t_new = t(goodIndexes)
y_new = yCopy(goodIndexes)

Adapt as needed.

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Image Analyst 2020-2-3

Not until you give me the t and y you used. Because for the original ones, as in my code, it works beautifully.

Paramonte 2020-2-3

you are right, only needed to transpode a vector. Cheers!!

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