Delete similar values in an array python numpy

Question

Is there an easy possibility to delete similar values in an array (with condition) without using a for loop? For example lets say I have an array

np.array([1.2, 3.4, 3.5, 8.9, 10.9])

In this case, i would set the condition for example difference < 0.3 and get as an output

np.array([1.2, 3.4, 8.9, 10.9])

I haven't seen anything on the internet similar to this question. Of course there is the function .unique() but that is only for exactly the same values.

Can you specify your problem a little bit more? Which values should be deleted if an array contains 3.4, 3.5, 3.6? And what is with the case where you have a row of similar values, for example: [3.4, 3.5, 3.6, 3.7, 3.8, 3.9]? Depending on this you need to define your own function. — Marcello Zago
– Marcello Zago, Commented Nov 28, 2022 at 8:03
What would be the output if you have [1.2, 1.5, 1.7]? Here you can either keep [1.5] or keep [1.2, 1.7] — Dani Mesejo
– Dani Mesejo, Commented Nov 28, 2022 at 8:03
For example for the array [1.2 1.5 1.7] i want to only have the first element. I need this for a data set which is in my case: [1266.20287121 1269.10287778 1420.00321996 1428.10323833 1432.30324785 1804.80409253 1808.50410092] and i want to keep only [1266.20287121 1420.00321996 1804.80409253] — Hrvoje Krizic
– Hrvoje Krizic, Commented Nov 28, 2022 at 8:05
What's the criteria by which to keep values? Should the first, average, or some other value be kept? What if a removed value decides whether to remove another – which values to keep out of 3.4, 3.6, 3.8? Are the values always sorted – which values to keep out of 3.4, 1.2, 3.5? — MisterMiyagi
– MisterMiyagi, Commented Nov 28, 2022 at 8:13

mozway · Accepted Answer · 2022-11-28 08:06:08Z

1

If you want to delete the successive values, you can compute the successive differences and perform boolean indexing:

a = np.array([1.2, 3.4, 3.5, 8.9, 10.9])

out = a[np.r_[True, np.diff(a)>=0.3]]

Or, if you want the absolute difference:

out = a[np.r_[True, np.abs(np.diff(a))>=0.3]]

Output:

array([ 1.2,  3.4,  8.9, 10.9])

Intermediates:

np.diff(a)
# array([2.2, 0.1, 5.4, 2. ])

np.diff(a)>=0.3
# array([ True, False,  True,  True])

np.r_[True, np.diff(a)>=0.3]
# array([ True,  True, False,  True,  True])

answered Nov 28, 2022 at 8:06

mozway

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1 Comment

Hrvoje Krizic Over a year ago

Thank you very much! This is exactly what I searched for :)

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