Find values close to multiples in NumPy array

Question

I have a numpy array as follows:

array([    90.,    180.,    270.,    360.,    450.,    540.,    630.,
      720.,    810.,    900.,    990.,   1080.,   1170.,   1260.,
     1350.,   1440.,   1530.,   1620.,   1710.,   1800.,   1890.,
     1980.,   2070.,   2160.,   2250.,   2340.,   2430.,   2520.,
     2610.,   2700.,   2790.,   2880.,   2970.,   3060.,   3150.,
     3240.,   3330.,   3420.,   3510.,   3600.,   3690.,   3780.,
     3870.,   3960.,   4050.,   4140.,   4230.,   4320.,   4410.,
     4500.,   4590.,   4680.,   4770.,   4860.,   4950.,   5040.,
     5130.,   5220.,   5310.,   5400.,   5490.,   5580.,   5670.,
     5760.,   5850.,   5940.,   6030.,   6120.,   6210.,   6300.,
     6390.,   6480.,   6570.,   6660.,   6750.,   6840.,   6930.,
     7020.,   7110.,   7200.,   7290.,   7380.,   7470.,   7560.,
     7650.,   7740.,   7830.,   7920.,   8010.,   8100.,   8190.,
     8280.,   8370.,   8460.,   8550.,   8640.,   8730.,   8820.,
     8910.,   9000.,   9090.,   9180.,   9270.,   9360.,   9450.,
     9540.,   9630.,   9720.,   9810.,   9900.,   9990.,  10080.,
    10170.,  10260.,  10350.,  10440.,  10530.,  10620.,  10710.,
    10800.])

I am trying to match values that come the closest to 4300 or multiple of 4300. So the condition statement would return TRUE for the value of 4320 and 8640 in the array above and FALSE for all other values. How can I accomplish that?

TIA

Edit 1:

After looking at the answer, I need to clarify. I was looking for the closest value within +-20% of the multiple to return TRUE.

trying to match the closest value within +-20% of the multiple — cndata
– cndata, Commented Dec 12, 2017 at 16:56

jakevdp · Accepted Answer · 2017-12-12 17:23:33Z

3

You can do this using the modulo operator and the floor division operator:

>>> tol = 20
>>> x[x % 4300 <= abs(x // 4300) * tol]
array([ 4320.,  8640.])

Edit:

Given the updated question (finding the closest match within a particular tolerance), what you're looking at is a specialization of a nearest neighbors problem. I'd probably do this with scipy's cKDTree, to be most efficient. For example:

>>> from scipy.spatial import cKDTree

>>> tree = cKDTree(x[:, None])

>>> queries = 4300 * np.arange(np.floor((x / 4300).min()), np.ceil((x / 4300).max()) + 1)

>>> queries
array([     0.,   4300.,   8600.,  12900.])

>>> d, i = tree.query(queries[:, None], k=1)

>>> x[i]
array([    90.,   4320.,   8640.,  10800.])

>>> tol = 0.2

>>> x[i][d < tol * x[i]]
array([  4320.,   8640.,  10800.])

edited Dec 12, 2017 at 17:23

answered Dec 12, 2017 at 16:46

jakevdp

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7 Comments

MaxU - stand with Ukraine Over a year ago

why do we need abs() here?

fuglede Over a year ago

If I'm reading their question correctly, they want at most one result for each multiple; that is, if 4310 were part of the array, only 4310 and not 4320 would get marked.

jakevdp Over a year ago

@fuglede Ah, that wasn't clear to me. But if that's true, why is 10,800 not in the expected output, as it's closest to 3 * 4300?

fuglede Over a year ago

@jakevdp: I was wondering the same: I would imagine the maximum element to be flagged no matter what. cndata: Can you clarify?

cndata Over a year ago

I was only looking for the element to match if it is within a certain range of the multiples of 4300. I realize now that the limits of the range need to be considered as well..so match the closest value within +-20% of the multiple.

|

fuglede · Accepted Answer · 2017-12-12 17:18:48Z

If I'm reading the question correctly, cf. my comment to @jakevdp's answer, only one value should be returned for each multiple. In that case, you can obtain the indices through broadcasting by doing something like

a = np.arange(0, np.max(x)+4300, 4300)[np.newaxis].T
b = np.zeros(x.shape, dtype=np.bool_)
b[np.argmin(np.abs(a - x), axis=1)] = 1

where x is your input array. In the given example, this produces

array([ True, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False,  True, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False,  True, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False,  True], dtype=bool)

Edit: The updated question adds the requirement that the given result moreover be within a certain range of the multiple. We can handle that by adding an extra step to the above:

a = np.arange(0, np.max(x) + 4300, 4300)[np.newaxis].T
idx = np.argmin(np.abs(a - x), axis=1)
b = np.zeros(x.shape, dtype=np.bool_)
b[idx[np.abs(x[idx] - a.ravel()) < 4300*0.2]] = 1

In this case, b becomes

array([ True, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False,  True, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False,  True, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False, False, False, False, False, False, False,
       False, False, False], dtype=bool)

Edit #2 given the comments for @jakevdp's answer: To ignore the first multiple, just use

a = np.arange(4300, np.max(x)+4300, 4300)[np.newaxis].T

instead

You're welcome! If you found the answer helpful, you can accept it.

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