How to avoid nested for loops in NumPy?

Question

I have this code.

n_nodes = len(data_x)
X = np.zeros((n_nodes, n_nodes))

for i in range(n_nodes):
  for j in range(n_nodes):
    X[i, j] = data_x[i] ** j

I want to do the same task with no loops used at all. How can I do that with NumPy functions?

Asked By: sam5922

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Source

Answer 1

If data_x is big, you will be faster using only numpy functions.
You can first repeat the input array and then use np.power with a vector giving the powers. This should be calculated fully vectorised in comparison to the already given list comprehension version.

x = np.arange(10)
X = x[:,np.newaxis].repeat(x.size,axis=1)
X = np.power(X,np.arange(x.size))

If data_x is already a numpy array, you can use it directly, if not you would need to do

x = np.array(data_x)

Answered By: user_na

Answer 2

You can do this in one step with numpy.power.outer:

np.power.outer(data_x, np.arange(len(data_x)))

Answered By: mozway

Answer 3

I’d suggest

data_x[:,None]**np.arange(n_nodes)

A check

In [17]: data_x = np.array((3,5,7,4,6))
    ...: n_nodes = len(data_x)
    ...: X = np.zeros((n_nodes, n_nodes))
    ...: 
    ...: for i in range(n_nodes):
    ...:   for j in range(n_nodes):
    ...:     X[i, j] = data_x[i] ** j
    ...: print(X)
    ...: print('-----------')
    ...: print(data_x[:,None]**np.arange(n_nodes))
[[1.000e+00 3.000e+00 9.000e+00 2.700e+01 8.100e+01]
 [1.000e+00 5.000e+00 2.500e+01 1.250e+02 6.250e+02]
 [1.000e+00 7.000e+00 4.900e+01 3.430e+02 2.401e+03]
 [1.000e+00 4.000e+00 1.600e+01 6.400e+01 2.560e+02]
 [1.000e+00 6.000e+00 3.600e+01 2.160e+02 1.296e+03]]
-----------
[[   1    3    9   27   81]
 [   1    5   25  125  625]
 [   1    7   49  343 2401]
 [   1    4   16   64  256]
 [   1    6   36  216 1296]]

Some timing

In [18]: %timeit data_x[:,None]**np.arange(n_nodes)
2.18 µs ± 7.49 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

In [19]: %%timeit
    ...: for i in range(n_nodes):
    ...:     for j in range(n_nodes):
    ...:         X[i, j] = data_x[i] ** j
10.9 µs ± 107 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

Answered By: gboffi

How to avoid nested for loops in NumPy?

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