labelled variables in Python

Question

I would like to define mutliple variables that are labelled by some coefficients, and these variables are not related to each other. The prime example would be the coefficients of an arbitrary multivariate function:

\sum_{i,j,k} c_{i,j,k} G(i, j, k)

In mathematica I simply would write

c[1,1,1] = 1
...
c[1,2,1] = x
...
c[2,3,2] = x^100-x
...
c[10,10,10] = 0

(* Example of use *)
expr = Sum[ c[i,j,k] G(i,j,k), {i,0,10}, {j,0,10}, {k,0,10}}]

where I defined all the coefficients by hand. Now, I am trying to do the same in Python. The way I currently implement it is via a function:

import sympy
x = sympy.Symbol("x")
def coeff(a,b,c):
    if (a,b,c) == (1,1,1): return 1
    ...
    elif (a,b,c) == (1,2,1): return x
    ...
    elif (a,b,c) == (2,3,2): return x**100-x
    ...
    else:
        return None

It seems it will not scale well if there is a large number of coefficients. My first thought to do it in Python was to use lists, but it only works if the labels are sequential, e.g. i=0,1,2,... which is not my case and does not work if the labels are not integers. Dictionaries also seem tempting but also seem quite messy to define and add coefficients after the fact. Is there a best practice to do such a thing in Python?

Devin Burke · Accepted Answer · 2020-03-29 22:29:30Z

1

Without knowing more about SymPy, it seems like you want a matrix / multi-dimensional array. That could be nested dictionaries -- but in your case I think it's just a dictionary using a tuple of dimensions as the key.

The syntax is virtually identical to your example above.

c = {}
c[1,1,1] = 1
c[1,2,1] = x
c[2,3,2] = x^100-x
c[10,10,10] = 0

The part I'm unsure about is whether the value of x changes over time and is ran as if it were a function. If so, this won't work as the value of x gets staged into the dictionary using whatever its value was at time of being set.

answered Mar 29, 2020 at 22:29

Devin Burke

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1 Comment

Bulkilol Over a year ago

In this case the value of x does change over time so your answer is perfect, thanks!

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