Extract Values That Satisfy A Condition From Numpy Array

Say I have the following arrays: a = np.array([1,1,1,2,2,2]) b = np.array([4,6,1,8,2,1]) Is it possible to do the following: a[np.where(b>3)[0]] #array([1, 1, 2]) Thus select

Solution 1:

Yes, there is a function: numpy.extract(condition, array) returns all values from array that satifsy the condition.

There is not much benefit in using this function over np.where or boolean indexing. All of these approaches create a temporary boolean array that stores the result of b>3. np.where creates an additional index array, while a[b>3]and np.extract use the boolean array directly.

Personally, I would use a[b>3] because it is the tersest form.

Solution 2:

Just use boolean indexing.

>>>a = np.array([1,1,1,2,2,2])                                                                                                   >>>b = np.array([4,6,1,8,2,1])                                                                                                   >>>>>>a[b > 3]                                                                                                                      
array([1, 1, 2])

b > 3 will give you array([True, True, False, True, False, False]) and with a[b > 3] you select all elements from a where the indexing array is True.

Solution 3:

Let's use list comprehension to solve this -

a = np.array([1,1,1,2,2,2])
b = np.array([4,6,1,8,2,1])
indices = [i for i in range(len(b)) if b[i]>3]  # Returns indexes of b where b > 3 - [0, 1, 3]
a[indices]
    array([1, 1, 2])