Generate sequential numbers between 0 and 1 with equal distance

Question

I am generating random variables in my array: np.random.rand(5,3,20)

How can I create the same shape and size but sequentially between 0 and 1?

JkShaw · Accepted Answer · 2017-05-10T13:02:22.173

3

Create evenly spaced numbers over a specified interval using linspace and then reshape to shape required as follows:

np.linspace(0, 1, 300).reshape(5, 3, 20)

Note:

'The new shape should be compatible with the original shape'

So let's say, for np.linspace(0, 1, t).reshape(x, y, z) the condition that should be met is t = x*y*z

edited May 10 '17 at 13:02

answered May 10 '17 at 12:30

JkShaw

when I run `np.linspace(0, 1, 300).reshape(5,5,20)` , i get this error: `ValueError: total size of new array must be unchanged` – jacky May 10 '17 at 12:37
`5*3*20 = 300`, where as `5*5*20 = 500`, you need to provide `shape` whose total elements must be same, that is for `np.linspace(0, 1, t).reshape(x, y, z)` the condition that should be met `t = x*y*z` – JkShaw May 10 '17 at 12:40

score 2 · Answer 2 · edited Mar 04 '22 at 18:33

2

Just use:

np.linspace(0, 1, num=5).reshape((5, 3))

and set the num param to implicitly define how big needs to be your step.

edited Mar 04 '22 at 18:33

Asclepius

answered May 10 '17 at 12:30

FrankBr

i get this error `ValueError: total size of new array must be unchanged` and where is the value `20` gone? – jacky May 10 '17 at 12:41

score 1 · Answer 3 · answered May 10 '17 at 13:00

1

Another option:

#start:stop:stepj inside np.r_ is interpreted as np.linspace(start, stop, step, endpoint=1) inside of the brackets
np.r_[0:1:300j].reshape(5,3,20)

answered May 10 '17 at 13:00

Allen Qin

3 Answers3