Numpy "Where" with Multiple Conditions

Numpy where function multiple conditions

The best way in your particular case would just be to change your two criteria to one criterion:

dists[abs(dists - r - dr/2.) <= dr/2.]

It only creates one boolean array, and in my opinion is easier to read because it says, is dist within a dr or r? (Though I'd redefine r to be the center of your region of interest instead of the beginning, so r = r + dr/2.) But that doesn't answer your question.

The answer to your question:
You don't actually need where if you're just trying to filter out the elements of dists that don't fit your criteria:

dists[(dists >= r) & (dists <= r+dr)]

Because the & will give you an elementwise and (the parentheses are necessary).

Or, if you do want to use where for some reason, you can do:

 dists[(np.where((dists >= r) & (dists <= r + dr)))]

Why:
The reason it doesn't work is because np.where returns a list of indices, not a boolean array. You're trying to get and between two lists of numbers, which of course doesn't have the True/False values that you expect. If a and b are both True values, then a and b returns b. So saying something like [0,1,2] and [2,3,4] will just give you [2,3,4]. Here it is in action:

In [230]: dists = np.arange(0,10,.5)
In [231]: r = 5
In [232]: dr = 1

In [233]: np.where(dists >= r)
Out[233]: (array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19]),)

In [234]: np.where(dists <= r+dr)
Out[234]: (array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12]),)

In [235]: np.where(dists >= r) and np.where(dists <= r+dr)
Out[235]: (array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12]),)

What you were expecting to compare was simply the boolean array, for example

In [236]: dists >= r
Out[236]: 
array([False, False, False, False, False, False, False, False, False,
       False,  True,  True,  True,  True,  True,  True,  True,  True,
        True,  True], dtype=bool)

In [237]: dists <= r + dr
Out[237]: 
array([ True,  True,  True,  True,  True,  True,  True,  True,  True,
        True,  True,  True,  True, False, False, False, False, False,
       False, False], dtype=bool)

In [238]: (dists >= r) & (dists <= r + dr)
Out[238]: 
array([False, False, False, False, False, False, False, False, False,
       False,  True,  True,  True, False, False, False, False, False,
       False, False], dtype=bool)

Now you can call np.where on the combined boolean array:

In [239]: np.where((dists >= r) & (dists <= r + dr))
Out[239]: (array([10, 11, 12]),)

In [240]: dists[np.where((dists >= r) & (dists <= r + dr))]
Out[240]: array([ 5. ,  5.5,  6. ])

Or simply index the original array with the boolean array using fancy indexing

In [241]: dists[(dists >= r) & (dists <= r + dr)]
Out[241]: array([ 5. ,  5.5,  6. ])

Numpy where with multiple conditions

You can use a ternary:

np.where(consumption_energy > 400, 'high', 
         (np.where(consumption_energy < 200, 'low', 'medium')))

How do I use np.where with multiple conditions

You have to use bitwise operators, & for and, | for or, and so on.

With your example,

a = np.array([1, 2, 3, 4, 5, 6])
np.where((a > 2) & (a < 5))

returns

(array([2, 3], dtype=int64),)

np.where with arbitrary number of conditions

You can do this by combining a reordered view of the columns of your array (which is a fancy way of saying "use a list of indexes") with a broadcast comparison, reduced over rows with np.all

>>> arr[np.where(np.all(arr[:,cols] > thds, axis=1))]
array([[9, 8, 7, 6],
       [9, 7, 6, 5]])

As your first link indicates (and as mentioned in the Note at the top of the documentation for np.where), there is actually no need for np.where in this case; it's only slowing things down. You can use a boolean list to slice a Numpy array, so you don't need to change the boolean list to a list of indexes. Since np.all, like the & operator, returns a Numpy array of boolean values, there is also no need for np.asarray or np.nonzero (as suggested in the aforementioned note):

>>> arr[np.all(arr[:,cols] > thds, axis=1)]
array([[9, 8, 7, 6],
       [9, 7, 6, 5]])

How to give multiple conditions to numpy.where()

numpy.any will return True/False, so that won't be suitable to use here.

You can stick to just numpy.where, but with a small tweak in the condition -- you introduce the bitwise OR operator into the condition like this:

np.where((letters == 'A') | (letters == 'C'))

In fact, numpy has this built in as well, as numpy.bitwise_or:

np.where(np.bitwise_or(letters == 'A', letters == 'C'))

Here's a short working example:

import numpy as np

A = 'A'
B = 'B'
C = 'C'

letters = np.array([A, B, C, A, B, C, A, B, C])

# Either this:
letter_indexes = np.where((letters == 'A') | (letters == 'C'))

# Or this:
letter_indexes = np.where(np.bitwise_or(letters == 'A', letters == 'C'))

print(letter_indexes[0])

# [0 2 3 5 6 8]

Using numpy.where function with multiple conditions but getting valueError

You should use bitwise & and parantheses, rather than and.

df['R'] = numpy.where((df['H'] > df['T']) & (df['P'] > 0),
                      df['C'] / df['T'] - 1, 0)

Numpy: Selecting Rows based on Multiple Conditions on Some of its Elements

Elegant is np.equal

Z[np.equal(Z[:, [0,1]], 1).all(axis=1)]

Or:

Z[np.equal(Z[:,0], 1) & np.equal(Z[:,1], 1)]