Dividing numpy array by numpy scalar

Question

I'm trying to divide numpy array by numpy float64 type scalar. Following is my code.

pose_q = np.array(pose_q)
expectecd_q = np.array(expectecd_q)

pose_q = np.squeeze(pose_q)
expectecd_q = np.squeeze(expectecd_q)

q1 = expectecd_q / np.linalg.norm(expectecd_q)
q2 = pose_q / np.linalg.norm(pose_q)

d = abs(np.sum(np.multiply(q1, q2)))

However I'm getting the following error pointing towards expectecd_q / np.linalg.norm(expectecd_q)

TypeError: ufunc 'true_divide' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''

@Mr.RandyTom I think you are printing type(expected_q), I meant expected_q.dtype. — Ehsan
– Ehsan, Commented Aug 28, 2020 at 21:35

Valdi_Bo · Accepted Answer · 2020-08-28 20:43:44Z

4

As you didn't provide your data, I created both arrays as:

a = np.array([12.0, 15.2, 19.3])  # Dividend
b = np.array(3.0)                 # Divider (a Numpy scalar)

If you want to divide a by b run just (no surprise) a / b. The result is:

array([4.        , 5.06666667, 6.43333333])

In your case, maybe you should identify what particular values you have as operands.

I looked in the Web for your error message. I found a suggestion that the cause can be that the array in question has text values (not float). It can happen when you read the array from a database. Check dtype of this array. class 'numpy.ndarray' says only that this is a Numpy array. But what is the type of its elements?

edited Aug 28, 2020 at 20:43

answered Aug 28, 2020 at 20:27

Valdi_Bo

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4 Comments

Mr. Randy Tom Over a year ago

the type of np.linalg.norm(pose_q) is <class 'numpy.float64'> :(

Ehsan Over a year ago

@Mr.RandyTom error is not in pose_q line, it is in expected_q. Please print type of that.

Mr. Randy Tom Over a year ago

and the expected_q is <class 'numpy.ndarray'>

Valdi_Bo Over a year ago

You wrote the name of your variable as expected_q. But in your post there is expectecd_q (note additional "c"). Maybe this is a typo?

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