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

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

[J. Alex Lee]

To clarify, you want to keep 4.1 as 4.1, but you want to combine 7.1 and 7.2 into 7?

I think the problem is simpler if you declare you want to aggregate all into round number times, so that your new t will be

t_new = [1 2 3 4 5 6 7]

[J. Alex Lee]

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

Might want to move your comment up to this thread

[Paramonte]

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

Answer 1

[Steven Lord]

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]

[J. Alex Lee]

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]

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

Steven reply worked perfectely.

many thanks indeed

Answer 2

[Paramonte]

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

Answer 3

[Image Analyst]

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.

[Image Analyst]

Note: I didn't check it for a run of 3 or more that are close together so you might need to adapt it for that if this code doesn't work in that situation.

[Paramonte]

Thank you for your reply.

I am getting "Index exceeds the number of array elements (10)."

in the line

yCopy(badIndexes - 1) = yCopy(badIndexes - 1) + yCopy(badIndexes)

any idea what it could be?

Many thanks

[Image Analyst]

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

[Paramonte]

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