CODEWITHSUNDEEP
pythonarraysnumpy
mel
mel

Reputation: 2790

Numpy: get the index of the top value sorted that are superior to 0

I have a numpy array and I wan to have the index of the top value sorted that are superior to 0 for instance:

[-0.4, 0.6, 0, 0, 0.4, 0.2, 0.7]

And I want to have:

[6, 1, 4, 5]

I can do it using a function I implemented but I guess for this kind of task there is something already implemented in Numpy.

Upvotes: 2

Views: 111

Answers (4)

Abdul Fatir
Abdul Fatir

Reputation: 6357

Use np.where().

d > 0.0 generates a boolean mask and where fetches all the values where the mask is true.

>>> d=np.array([-0.4, 0.6, 0, 0, 0.4, 0.2, 0.7])
>>> r=np.where( d > 0)
>>> s=sorted(r[0].tolist(), key=lambda x:d[x], reverse=True)
>>> s
[6L, 1L, 4L, 5L]

EDIT

Here's what I mean by mask. 

>>> mask = d > 0
>>> mask
array([False,  True, False, False,  True,  True,  True], dtype=bool)

Upvotes: 1

Divakar
Divakar

Reputation: 221754

Here's a vectorized approach -

idx = np.where(A>0)[0]
out = idx[A[idx].argsort()[::-1]]

Sample run -

In [37]: A = np.array([-0.4, 0.6, 0, 0, 0.4, 0.2, 0.7])

In [38]: idx = np.where(A>0)[0]

In [39]: idx[A[idx].argsort()[::-1]]
Out[39]: array([6, 1, 4, 5])

Upvotes: 2

Colonel Beauvel
Colonel Beauvel

Reputation: 31181

You can also do:

L = [-0.4, 0.6, 0, 0, 0.4, 0.2, 0.7]

[L.index(i) for i in sorted(filter(lambda x: x>0, L), reverse=True)]

Out[72]: [6, 1, 4, 5]

Upvotes: 3

Rahul K P
Rahul K P

Reputation: 16081

You can implement with np.where

a = np.array([-0.4, 0.6, 0, 0, 0.4, 0.2, 0.7])
np.where(a > 0)[0].tolist()

Result

[1, 4, 5, 6]

The result of np.where(a > 0) is in the form of tuple of numpy array. So can converted into list with using tolist()

Upvotes: 2

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