Numpy: find index of the elements within range

Question

I have a numpy array of numbers, for example,

a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])  

I would like to find all the indexes of the

Answer 1

As in @deinonychusaur's reply, but even more compact:

In [7]: np.where((a >= 6) & (a <=10))
Out[7]: (array([3, 4, 5]),)

Answer 2

I thought I would add this because the a in the example you gave is sorted:

import numpy as np
a = [1, 3, 5, 6, 9, 10, 14, 15, 56] 
start = np.searchsorted(a, 6, 'left')
end = np.searchsorted(a, 10, 'right')
rng = np.arange(start, end)
rng
# array([3, 4, 5])

Answer 3

a = np.array([1,2,3,4,5,6,7,8,9])
b = a[(a>2) & (a<8)]

Answer 4

a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
np.argwhere((a>=6) & (a<=10))

Answer 5

Summary of the answers

For understanding what is the best answer we can do some timing using the different solution. Unfortunately, the question was not well-posed so there are answers to different questions, here I try to point the answer to the same question. Given the array:

a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])

The answer should be the indexes of the elements between a certain range, we assume inclusive, in this case, 6 and 10.

answer = (3, 4, 5)

Corresponding to the values 6,9,10.

To test the best answer we can use this code.

import timeit
setup = """
import numpy as np
import numexpr as ne

a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
# we define the left and right limit
ll = 6
rl = 10

def sorted_slice(a,l,r):
    start = np.searchsorted(a, l, 'left')
    end = np.searchsorted(a, r, 'right')
    return np.arange(start,end)
"""

functions = ['sorted_slice(a,ll,rl)', # works only for sorted values
'np.where(np.logical_and(a>=ll, a<=rl))[0]',
'np.where((a >= ll) & (a <=rl))[0]',
'np.where((a>=ll)*(a<=rl))[0]',
'np.where(np.vectorize(lambda x: ll <= x <= rl)(a))[0]',
'np.argwhere((a>=ll) & (a<=rl)).T[0]', # we traspose for getting a single row
'np.where(ne.evaluate("(ll <= a) & (a <= rl)"))[0]',]

functions2 = [
   'a[np.logical_and(a>=ll, a<=rl)]',
   'a[(a>=ll) & (a<=rl)]',
   'a[(a>=ll)*(a<=rl)]',
   'a[np.vectorize(lambda x: ll <= x <= rl)(a)]',
   'a[ne.evaluate("(ll <= a) & (a <= rl)")]',
]

Results

The results are reported in the following plot. On the top the fastest solutions. If instead of the indexes you want to extract the values you can perform the tests using functions2 but the results are almost the same.

Answer 6

You can use np.clip() to achieve the same:

a = [1, 3, 5, 6, 9, 10, 14, 15, 56]  
np.clip(a,6,10)

However, it holds the values less than and greater than 6 and 10 respectively.