How to create a numpy array of N numbers of the same value? [duplicate]

Use numpy.full():

import numpy as np

np.full(
  shape=10,
  fill_value=3,
  dtype=np.int
)
    
> array([3, 3, 3, 3, 3, 3, 3, 3, 3, 3])

Very easy 1)we use arange function :-

arr3=np.arange(0,10)
output=array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

2)we use ones function whose provide 1 and after we will do multiply 3 :-

arr4=np.ones(10)*3
output=array([3., 3., 3., 3., 3., 3., 3., 3., 3., 3.])

An alternative (faster) way to do this would be with np.empty() and np.fill():

import numpy as np

shape = 10
value = 3


myarray = np.empty(shape, dtype=np.int)
myarray.fill(value)

Time comparison

The above approach on my machine executes for:

951 ns ± 14 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

vs using np.full(shape=shape, fill_value=value, dtype=np.int) executes for:

1.66 µs ± 24.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

vs using np.repeat(value, shape) executes for:

2.77 µs ± 41.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

vs using np.ones(shape) * value executes for:

2.71 µs ± 56.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

I find it to be consistently a little bit quicker.

I realize it's not using numpy, but this is very easy with base python.

data = [3]*10
print(data)
> [3, 3, 3, 3, 3, 3, 3, 3, 3, 3]

Try this. It is worked well in my python version (Python 3.9.6).

import numpy as np 
np.array([3]*10)

output: 

In my opinion, this is the easiest way:

import numpy as np

arr = np.arange(0, 11)
arr[:] = 0

output:
[0 0 0 0 0 0 0 0 0 0 0]