pythonic way to fill an array with geometric cumulative function-CodePudding

I need to fill an array with the position of the boundary of n_cells (starting from 0 up to n_cells 1) with the size of the cell in geometric progression (with parameter c) and the total size will be thick (thus x[0]=0 and x[n_cells]=thick)

I have this which works:

n_cells=10
c=0.9
thick=2

x = np.zeros(n_cells   1)
x_0=thick * (c - 1) / (c**n_cells - 1)
for i in range(n_cells):
    x[i 1] = x[i]   x_0 * c ** i

Since I'm learning, is there a simpler pythonic comprehension way to do the same?

CodePudding user response：

You could try to use the numba.njit module. It will speed up your code. It is not more pythonic but it will be faster if you use a larger n_cells. See the example below

import numpy as np
from numda import njit
@njit
def fast_goem():
    n_cells=10
    c=0.9
    thick=2
    x = np.zeros(n_cells   1)
    x_0=thick * (c - 1) / (c**n_cells - 1)
    for i in range(n_cells):
        x[i 1] = x[i]   x_0 * c ** i
    return x

CodePudding user response：

Ok I found a way by using numpy stuff:

x_0=thick * (c - 1) / (c**n_cells - 1)
x=np.cumsum(np.concatenate(([0],np.geomspace(x_0, x_0*c**(n_cells-1), num=n_cells))))

however readability is not improved (I don't like the concatenate) so I wouldn't mark this answer as more "pythonic" just shorter