Index of True and False of a numpy array

Question

Code:

import numpy as np    
a = np.arange(5)   
print(a[True])    
print(a[False])

Result:

[[0 1 2 3 4]]
[]

For above code when I am passing True to index of a numpy array it's returning complete array while for False an empty array. What's happening in background ?

Could you please double check the output from your first print statement? Is it [0,1,2,3,4] or [[0,1,2,3,4]]? — Paul Panzer
– Paul Panzer, Commented Sep 27, 2019 at 10:35

Paul Panzer · Accepted Answer · 2019-09-28 11:41:06Z

4

Your example---applying a scalar, or 0D---boolean mask to a 1D array yields

print(a[True])
# [[0 1 2 3 4]]
print(a[False])
# []

For clarity, especially w.r.t. the second (False) case let us query the shapes

print(a[True].shape)
# (1, 5)
print(a[False].shape)
# (0, 5)

Now that is at first sight slightly puzzling. Why the extra dimension?

Let's start from a less edgy case and work out why this is the logical behavior:

x = np.arange(6).reshape(3,2)
m2 = np.array([[True,False],[True,True],[False,True]])
m1 = np.array([True,False,True])
m0 = np.array(True)

We have created a 2D array and matching 2D,1D and 0D masks.

Masking with a 2D mask gives a 1D result

x[m2]
# array([0, 2, 3, 5])

Masking with a 1D mask selects entire rows, hence gives a 2D result

x[m1]
# array([[0, 1],
#        [4, 5]])

We could verify in higher dimesions also, that removing a dimension from the mask adds one to the result.

Therefore it is logical that masking with 0D we shold indeed get one dimension more than we started with. A 2D mask selects individual points (0D -> list of those: 1D), a 1D mask selected entire rows (1D -> list of those: 2D); consequently, a 0D mask should and does select entire sheets (2D -> list of those: 3D); as the base array is 2D a sheet is the whole of it.

x[m0]
# array([[[0, 1],
#         [2, 3],
#         [4, 5]]])

x[m0].ndim
# 3

In general:

x[m2].ndim == x.ndim - m2.ndim + 1
# True 
x[m1].ndim == x.ndim - m1.ndim + 1
# True
x[m0].ndim == x.ndim - m0.ndim + 1
# True

edited Sep 28, 2019 at 11:41

answered Sep 27, 2019 at 10:19

Paul Panzer

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2 Comments

FObersteiner Over a year ago

nicely presented! I see the logic, and it also makes sense to me for dimensions >= 1 (both array and mask) but why should it be useful to increase the dimension if 1D is masked with zero-D?

Paul Panzer Over a year ago

@MrFuppes In my experience this kind of consistent behavior is the easiest to work with. For example, if you are writing some code that you want to work with all kind of dimensionalities you may need to be able to predict the properties of intermediate results. In this case being able to rely on a formula like at the very end of the answer and not having to make an extra branch for 0D masks will come in handy.

Akash Kumar · Accepted Answer · 2019-09-27 08:59:31Z

0

a be your numpy array and i.e [0,1,2,3,4] you suppose to put condition on index if condition is satisfied then it will return that element if not then discard those element. here you are putting True which mean every element satisfied this condition which is True in every case if False then non of the element satisfied. check this section

Boolean or “mask” index arrays

answered Sep 27, 2019 at 8:59

Akash Kumar

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