4

I have a structured numpy array:

dtype = [('price', float), ('counter', int)]
values = [(35, 1), (36, 2),
          (36, 3)]
a = np.array(values, dtype=dtype)

I want to sort for price and then for counter if price is equal:

a_sorted = np.sort(a, order=['price', 'counter'])[::-1]

I need the price in a descending order and when prices are equal consider counter in ASCENDING order. In the example above both the price and the counter are in descending order.

What I get is:

a_sorted: [(36., 3), (36., 2), (35., 1)]

what I need is:

 a_sorted: [(36., 2), (36., 3), (35., 1)]
jpp
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Andrew
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1 Answers1

4

You can use np.lexsort:

a_sorted = a[np.lexsort((a['counter'], -a['price']))]

Result:

array([(36.0, 2), (36.0, 3), (35.0, 1)], 
      dtype=[('price', '<f8'), ('counter', '<i4')])

Just remember the order is reversed, i.e. sorting is performed first by -a['price']. Negation takes care of the "descending" aspect.

jpp
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