Create a n x m array of polynomials using a (n x 1) data through Numpy/Pandas

Question

I created an array using the following:

𝑥𝑖 = np.random.normal(0,1,50), which gave me
array([ 1.92024714, -0.19882742, -0.26836024,  0.32805879, -0.32085809,
       -0.23569939,  0.22310599,  0.5483915 , -0.13106083, -1.03798811,
        0.4586899 , -1.7378367 , -0.49868295,  1.58943447,  0.92153814,
        0.38894787, -1.26605208,  0.44308314,  1.10222734,  0.40031394,
       -1.2126154 ,  0.26871733, -0.85161259,  0.15853002, -0.18531145,
       -0.18069696,  0.19121711,  0.16586507,  0.43668293,  0.38395065,
       -1.02418998,  0.10464186, -0.02777545, -0.30571787,  1.0690931 ,
       -0.67266002,  2.00256049, -0.05156432, -1.03735733,  0.27650841,
       -0.53300549, -0.4301668 ,  1.01371008, -0.70780846,  0.11577668,
        0.19328765, -0.72971236,  1.61804424, -0.69770352, -1.33161613])

For each element of this array how can I do the following to give me a 50x3 matrix something like this – ANY SUGGESTIONS ?

𝑥1^1     𝑥1^2     𝑥1^3
𝑥2^1     𝑥2^2     𝑥2^3
𝑥3^1     𝑥3^2     𝑥3^3
.
.
𝑥50^1     𝑥50^2     𝑥50^3

ie the numbers in 50x1 array above would look this in an 50 x 3 array

1.92024714     3.68734907867818      7.08062152251341
-0.19882742    0.03953234294385     -0.00786011375408 
-0.26836024     0.07201721841285     -0.01932655801740
. 
.
.
.
.
-1.33161613     1.77320151767618     -2.36122374267808

Answer 1

Using np.column_stack

np.column_stack((a, a**2, a**3))

---

array([[ 1.92024714e+00,  3.68734908e+00,  7.08062152e+00],
       [-1.98827420e-01,  3.95323429e-02, -7.86011375e-03],
              ...      ,        ...     ,       ...
       [-6.97703520e-01,  4.86790202e-01, -3.39635237e-01],
       [-1.33161613e+00,  1.77320152e+00, -2.36122374e+00]])

Answer 2

Here's one way leveraging broadcasting :

a = np.random.normal(0,1,50)

out = a[:,None]**np.arange(1,4)

print(out.shape)
# (50, 3)

Answer 3

What you're describing here is called a Vandermonde matrix . numpy has this built in (and more performant than broadcasting on large matrices)

The first column of a Vandermonde matrix is always 1 , so you can filter that out if you wish.

---

a = np.random.normal(0, 1, 50)

np.vander(a, 4, increasing=True)[:, 1:]

array([[ 4.21022633e-01,  1.77260058e-01,  7.46304963e-02],   
       [-9.37208666e-02,  8.78360084e-03, -8.23206683e-04],   
                          ...   
       [-9.02260087e-01,  8.14073265e-01, -7.34505815e-01],   
       [ 1.21125200e+00,  1.46713140e+00,  1.77706584e+00]])  

---

Just for a bit of validation:

>>> np.isclose(np.vander(a, 4, increasing=True)[:, 1:], a[:, None]**np.arange(1, 4)).all()
True

---

On large matricies, vander beats broadcasting:

a = np.random.normal(0, 1, 10_000)

In [99]: %timeit np.vander(a, 100, increasing=True)[:, 1:]
8.37 ms ± 97 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [100]: %timeit a[:, None]**np.arange(1, 100)
51.4 ms ± 904 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

---

If you don't want a strictly increasing matrix, this becomes far less useful, and will calculate unnecessary powers, in which case you should fall back to the broadcasted solution.

Answer 4

All, many thanks for your responses. I'm a beginner at Python and it's great to see three different ways to address this problem. I read and educated myself about all three.

Thanks again !!