How do i generate numpy linspace type numpy zeros array for initialization?

Question

While generating a linspace array in Numpy we get an array of the form (len(array), ), i.e. it doesn't have any 2nd dimension. How do I generate a similar array and initialize it using Numpy zeros? Because it takes a 2nd argument, like 1, so I get (len(array), 1) while initializing, which I wanted to avoid if possible.

Eg. np.linspace(0,10,5) = [0, 2.5, 5, 7.5, 10] ; It's array dimension is (5, ).

On the other hand, a zeros array is defined as np.zeros((5,1)) and our output is a vector [0 0 0 0 0] ^ (Transpose). I wanted to be a flat array not like a vector.

Is there a way?

It is not entirely clear to me what you want to achieve. Please updated with an example with input and expected output. — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Mar 13, 2017 at 13:05

Vince W. · Accepted Answer · 2017-03-13 14:25:16Z

2

your first argument (5,1) is defining the shape of the array as a 5x1 explicitly 2d shape. Just pass (5,), or more explicitly as follows:

import numpy as np
z = np.zeros(shape=(5,), dtype=float)
print(z)
print(z.shape)

output is:

[ 0.  0.  0.  0.  0.]
(5,)

edited Mar 13, 2017 at 14:25

answered Mar 13, 2017 at 13:21

Vince W.

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1 Comment

hpaulj Over a year ago

Use shape=5 or shape=(5,). The 2nd is a 1 element tuple.

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