CODEWITHSUNDEEP
pythonarraysmatlabfor-loopnumpy
Brandon Ginley
Brandon Ginley

Reputation: 269

for looping through arrays in python- from matlab

Suppose you have the following section of MATLAB code, looping through and modifying index values within the matrix as it iterates:

x = zeros(parts,2);
for i = 1:parts
x(i,1) = (i-1)*L + 1;
x(i,2) = i*L;
end

Now suppose you are a python noob, and have gotten this far:

v = np.zeros((parts,2))
for x in xrange(0,N1/L):

where parts and N1/L are predefined integer values. I've done some searching on indexing and for looping in python, but I'm having difficulty understanding how to reference specific indices and modifying them within the for loop. If someone could direct me in the correct direction to understand how to attack the next section of the code, that would be much appreciated.

Upvotes: 1

Views: 634

Answers (1)

unutbu
unutbu

Reputation: 880897

A literal translation of the Matlab code would be

import numpy as np
x = np.zeros((parts, 2))
for i in range(parts):
    x[i,0] = i*L + 1
    x[i,1] = (i+1)*L

Note that Matlab uses 1-based indexing while Python uses 0-based indexing. This accounts for the differences in where the 1's show up.

However, when using NumPy you'll get much better performance if you avoid element-by-element modification of an array. Instead, you should seek to express the calculation in terms of as few NumPy operators or function calls as you can which affect whole arrays at once. By do this, you off-load as much work as possible to NumPy's underlying fast C/Fortran-compiled function calls and reduce the amount of time spent executing slower Python code.

This usually means you want to avoid Python for-loops, since a loop implies there will be lots of Python statements to be executed.

So, for example, a better way to express the above calculation would be

x = np.zeros((parts, 2))
x[:, 0] = np.arange(1, parts*L, L)
x[:, 1] = x[:, 0] + L - 1

Notice that the values in x are filled in using just 2 assignments. Each assignment affects a whole column of x "all at once".

To give a sense of what a difference array-based operations make, here is an (IPython) timeit test using parts = 10000, L = 3 :

In [16]: %%timeit
   ....: x = np.zeros((parts, 2))
         x[:, 0] = np.arange(1, parts*L, L)
         x[:, 1] = x[:, 0] + L - 1
10000 loops, best of 3: 51.9 µs per loop

In [17]: %%timeit
   ....: x = np.zeros((parts, 2))
         for i in range(parts):
             x[i,0] = i*L + 1
             x[i,1] = (i+1)*L
100 loops, best of 3: 3.58 ms per loop

Upvotes: 2

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