Remove common elements in numpy array

Question

I am trying to do a union of two numpy arrays in the following manner

np.union1d( np.arange(0.1, 0.91, 0.1), np.arange(0.4, 0.81, 0.01)  )

The output reads:

array([ 0.1 ,  0.2 ,  0.3 ,  0.4 ,  0.41,  0.42,  0.43,  0.44,  0.45,
    0.46,  0.47,  0.48,  0.49,  0.5 ,  0.5 ,  0.51,  0.52,  0.53,
    0.54,  0.55,  0.56,  0.57,  0.58,  0.59,  0.6 ,  0.6 ,  0.61,
    0.62,  0.63,  0.64,  0.65,  0.66,  0.67,  0.68,  0.69,  0.7 ,
    0.7 ,  0.71,  0.72,  0.73,  0.74,  0.75,  0.76,  0.77,  0.78,
    0.79,  0.8 ,  0.8 ,  0.9 ])

In the output of this union, the number 0.5 features twice. Even when I use the unique function in numpy, this replication of the number 0.5 doesn't go away. Meaning:

np.unique( np.union1d( np.arange(0.1, 0.91, 0.1), np.arange(0.4, 0.81, 0.01)  ) )

also gives the same output. What am I doing wrong? How can I correct this and get the desired output (i.e. have only one occurrence of the number 0.5 in my array?

It's probably an issue with floating point comparison/precision. Is it possible to use integers for the arange and union operations and then do a division later on to get normalized arrays? — lmNt
– lmNt, Commented Mar 8, 2017 at 8:04
The array that I am intending to use has unequal spacing, i.e. spacing of 0.1 between 0.1 and 0.9 plus a spacing of 0.01 between 0.4 and 0.8. — Commoner
– Commoner, Commented Mar 8, 2017 at 8:10
You should still be able to achieve that using integers. See my answer for example. — lmNt
– lmNt, Commented Mar 8, 2017 at 8:13
Try looking at them separately: a[13] gives 0.5, but a[14] gives 0.50000000000000011 — Eric
– Eric, Commented Mar 8, 2017 at 9:42

Community · Accepted Answer · 2017-05-23 12:34:56Z

Given the input array is sorted, using the same philosophy as in this post -

a[np.r_[True,~np.isclose(a[1:] , a[:-1])]]

Sample run -

In [20]: a = np.union1d( np.arange(0.1, 0.91, 0.1), np.arange(0.4, 0.81, 0.01)  )

In [21]: a
Out[21]: 
array([ 0.1 ,  0.2 ,  0.3 ,  0.4 ,  0.41,  0.42,  0.43,  0.44,  0.45,
        0.46,  0.47,  0.48,  0.49,  0.5 ,  0.5 ,  0.51,  0.52,  0.53,
        0.54,  0.55,  0.56,  0.57,  0.58,  0.59,  0.6 ,  0.6 ,  0.61,
        0.62,  0.63,  0.64,  0.65,  0.66,  0.67,  0.68,  0.69,  0.7 ,
        0.7 ,  0.71,  0.72,  0.73,  0.74,  0.75,  0.76,  0.77,  0.78,
        0.79,  0.8 ,  0.8 ,  0.9 ])

In [22]: a[np.r_[True,~np.isclose(a[1:] , a[:-1])]]
Out[22]: 
array([ 0.1 ,  0.2 ,  0.3 ,  0.4 ,  0.41,  0.42,  0.43,  0.44,  0.45,
        0.46,  0.47,  0.48,  0.49,  0.5 ,  0.51,  0.52,  0.53,  0.54,
        0.55,  0.56,  0.57,  0.58,  0.59,  0.6 ,  0.61,  0.62,  0.63,
        0.64,  0.65,  0.66,  0.67,  0.68,  0.69,  0.7 ,  0.71,  0.72,
        0.73,  0.74,  0.75,  0.76,  0.77,  0.78,  0.79,  0.8 ,  0.9 ])

Community · Accepted Answer · 2017-05-23 10:29:54Z

1

As stated by @ImNt in the comments, this might be due to floating point comparision/precision (probably they are not 0.5 in memory, but 0.500000000001)

You can make a workaround, though. You know your numbers will be at most 2 digits long. Then, you can first np.round the array before applying np.unique.

x = np.union1d( np.arange(0.1, 0.91, 0.1), np.arange(0.4, 0.81, 0.01)  )
x = np.round(x, 2) # Round 2 floating points
x = np.unique(x)

Output:

array([ 0.1 ,  0.2 ,  0.3 ,  0.4 ,  0.41,  0.42,  0.43,  0.44,  0.45,
        0.46,  0.47,  0.48,  0.49,  0.5 ,  0.51,  0.52,  0.53,  0.54,
        0.55,  0.56,  0.57,  0.58,  0.59,  0.6 ,  0.61,  0.62,  0.63,
        0.64,  0.65,  0.66,  0.67,  0.68,  0.69,  0.7 ,  0.71,  0.72,
        0.73,  0.74,  0.75,  0.76,  0.77,  0.78,  0.79,  0.8 ,  0.9 ])

edited May 23, 2017 at 10:29

CommunityBot

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answered Mar 8, 2017 at 8:10

Carles Mitjans

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Comments

lmNt · Accepted Answer · 2017-03-08 08:15:31Z

1

As I have written in my comment, it will be an issue due to floating point precision and their comparison. If applicable in your particular case I would suggest working with integers and normalizing later on.

For example

x = np.union1d( np.arange(10, 91, 10), np.arange(40, 81, 1)  )
x = x/100.0

Output:

[ 0.1   0.2   0.3   0.4   0.41  0.42  0.43  0.44  0.45  0.46  0.47  0.48
  0.49  0.5   0.51  0.52  0.53  0.54  0.55  0.56  0.57  0.58  0.59  0.6
  0.61  0.62  0.63  0.64  0.65  0.66  0.67  0.68  0.69  0.7   0.71  0.72
  0.73  0.74  0.75  0.76  0.77  0.78  0.79  0.8   0.9 ]

edited Mar 8, 2017 at 8:15

answered Mar 8, 2017 at 8:09

lmNt

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Comments

Paul Panzer · Accepted Answer · 2017-03-08 08:45:16Z

Or you could use Fractions:

>>> import numpy as np
>>> from fractions import Fraction
>>> np.union1d( np.arange(Fraction(1,10), Fraction(91,100), Fraction(1,10)), np.arange(Fraction(4,10), Fraction(81,100),Fraction(1,100)))
array([Fraction(1, 10), Fraction(1, 5), Fraction(3, 10), Fraction(2, 5),
       Fraction(41, 100), Fraction(21, 50), Fraction(43, 100),
       Fraction(11, 25), Fraction(9, 20), Fraction(23, 50),
       Fraction(47, 100), Fraction(12, 25), Fraction(49, 100),
       Fraction(1, 2), Fraction(51, 100), Fraction(13, 25),
       Fraction(53, 100), Fraction(27, 50), Fraction(11, 20),
       Fraction(14, 25), Fraction(57, 100), Fraction(29, 50),
       Fraction(59, 100), Fraction(3, 5), Fraction(61, 100),
       Fraction(31, 50), Fraction(63, 100), Fraction(16, 25),
       Fraction(13, 20), Fraction(33, 50), Fraction(67, 100),
       Fraction(17, 25), Fraction(69, 100), Fraction(7, 10),
       Fraction(71, 100), Fraction(18, 25), Fraction(73, 100),
       Fraction(37, 50), Fraction(3, 4), Fraction(19, 25),
       Fraction(77, 100), Fraction(39, 50), Fraction(79, 100),
       Fraction(4, 5), Fraction(9, 10)], dtype=object)
>>> _.astype(float)
array([ 0.1 ,  0.2 ,  0.3 ,  0.4 ,  0.41,  0.42,  0.43,  0.44,  0.45,
        0.46,  0.47,  0.48,  0.49,  0.5 ,  0.51,  0.52,  0.53,  0.54,
        0.55,  0.56,  0.57,  0.58,  0.59,  0.6 ,  0.61,  0.62,  0.63,
        0.64,  0.65,  0.66,  0.67,  0.68,  0.69,  0.7 ,  0.71,  0.72,
        0.73,  0.74,  0.75,  0.76,  0.77,  0.78,  0.79,  0.8 ,  0.9 ])

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