What are the operators == and * being used for in this code?

Question

I need to understand the usage of operators in the following python code snippet. I'm not familiar with the usage of * in the np.zeros() statement. It looks like the pointer dereferencing operator in c, but I'm guessing not.

Also, what is the usage of == in the assignment statement? This looks like an equality test, but True and False are not valid indexes into a numpy array.

new_segs = np.zeros((*raw_segs.shape, 3), dtype=np.uint8)
i = 0
for val in [x['handle'] for x in detections]:
    colors = [(255,0,255), (55,0,55), (34,33,87)]
    new_segs[raw_segs == val] = colors[i % len(colors)]

Sorry for the crappy question. Have tried looking for answers but I get unsatisfactory answers with searches on operator usage.

`*` is not an operator here: https://stackoverflow.com/questions/2921847/what-does-the-star-operator-mean — Xantium, Mar 09 '18 at 00:00

Totoro · Accepted Answer · 2018-03-09T12:28:21.607

4

The star * unpacking was explained. The == in numpy is a boolean mask. You are passing an array that should be the same size of the other numpy array but this tells it which elements of the array to include or not.

edited Mar 09 '18 at 12:28

answered Mar 09 '18 at 00:09

Totoro

867
9
10

score 2 · Answer 2 · edited Mar 09 '18 at 00:23

2

The star * is the unpacking operator; it expands raw_segs.shape to a tuple of values.

You're incorrect about indices: True and False are valid, being interpreted as 1 and 0, respectively. Try it and see:

>>> new_segs = np.array((3.3, 4.4))
>>> new_segs[0]
3.2999999999999998
>>> new_segs[True]
4.4000000000000004
>>>

edited Mar 09 '18 at 00:23

Right leg

16,080
7
48
81

answered Mar 09 '18 at 00:02

Prune

76,765
14
60
81

`raw_segs == val` doesn't yield a boolean. – John Kugelman Mar 09 '18 at 00:16

score 1 · Answer 3 · answered Mar 09 '18 at 01:20

* unpacks the shape tuple, allowing it be joined with the 3 to make a bigger shape tuple:

In [26]: x = np.zeros((2,3))
In [28]: y = np.zeros((*x.shape, 3))
In [29]: y.shape
Out[29]: (2, 3, 3)

another way to do that:

In [30]: y = np.zeros(x.shape+(3,))
In [31]: y.shape
Out[31]: (2, 3, 3)

and

In [32]: i,j = x.shape
In [33]: y = np.zeros((i,j,3))
In [34]: y.shape
Out[34]: (2, 3, 3)

What are the operators == and * being used for in this code?

3 Answers3