Convert Equation from One Language to Another

Question

I have a task where I need to convert equations in one language (MATLAB) to another (C++) using Python. Everything is straightforward except the power operation. The equations are represented by Python strings.

For example, I want to convert

Eq = ((a+b)^2 + (c+d)^(1/2)) * e

to,

Eq = (pow((a+b),2) + pow(c+d, 1/2)) * e

I tried regular expression, but it does not seem to work because of nested parenthesis. Is there any good way to do this?

In C, `pow(c+d, 1/2)` is always 1. So it's not quite as simple as just changing operators. — rici, May 12 '14 at 04:30
Do you want to convert the first string to second string? Your question is not clear. I am close-voting it. — thefourtheye, May 12 '14 at 04:32
It has been covered many times why using regex to parse structured data is a bad idea. Usually people try it with [HTML](http://stackoverflow.com/questions/1732348/regex-match-open-tags-except-xhtml-self-contained-tags/1732454#1732454) so this is slightly novel. But it's still a bad idea. — tripleee, May 12 '14 at 04:40
This is the kind of task that [source-to-source compilers](https://en.wikipedia.org/wiki/Source-to-source_compiler) are designed to handle. — Anderson Green, May 12 '14 at 04:40
@thefourtheye yes I want to convert the first string to the second. I edited my question. — ysakamoto, May 12 '14 at 04:42
To expound on @rici's comment, `1/2` evaluates to `0` in C, you need to convert integer literals to floating point literals in expressions, as `1.0/2.0`. — U2EF1, May 12 '14 at 04:54

score 2 · Accepted Answer · answered May 12 '14 at 04:50

The best way to solve this would be to use Abstract Syntax Tree of Python. We can get that using ast module.

import ast, _ast

ops = {_ast.Mult: "*", _ast.Add: "+", _ast.BitXor: "pow", _ast.Div: "/"}

def rec(n):
    if isinstance(n, _ast.Expr):
        return rec(n.value)
    elif isinstance(n, _ast.BinOp):
        if isinstance(n.op, _ast.BitXor):
            return "{}({}, {})".format(ops[type(n.op)], rec(n.left),rec(n.right))
        else:
            return "({} {} {})".format(rec(n.left), ops[type(n.op)],rec(n.right))
    elif isinstance(n, _ast.Name):
        return n.id
    elif isinstance(n, _ast.Num):
        return n.n

print rec(ast.parse("(((a+b)^2) + ((c+d)^(1/2))) * e").body[0])
# ((pow((a + b), 2) + pow((c + d), (1 / 2))) * e)

Note: Since ^ means Binary XOR operation in Python, you need to enclose that expression with an extra parenthesis, like I have shown in the example.

First, we check if it is an expression, then we process its value. If it is a binary operation, we check of it is Bitwise XOR, in that case we return the result in a different format, otherwise we return the result like left op right. If the current item looks like a name, then we return its id or if it is a number we return the attribute n.

Thanks! I added `_ast.Sub` and `_ast.USub` for `-` operation and sign, and it works great! — ysakamoto, May 12 '14 at 16:21

sshashank124 · Answer 2 · 2014-05-12T04:32:18.107

1

You can do it as follows. Note: This is still a very primitive approach and I'd be careful about relying on it 100%:

import re

s = "Eq = ((a+b)^2 + (c+d)^(1/2)) * e"

>>> print re.sub(r'\(([^()\s]*)\)\^\(?([^()\s]*)\)?', 'pow(\\1,\\2)', s)
Eq = (pow(a+b,2) + pow(c+d,1/2)) * e

edited May 12 '14 at 04:32

answered May 12 '14 at 04:26

sshashank124

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OP wants to convert to C++ using Python, not convert to Python using Python. `**` is not going to work in C++. – rici May 12 '14 at 04:31
@rici, Sorry. My mistake. I have removed that part. – sshashank124 May 12 '14 at 04:33

score 0 · Answer 3 · edited May 23 '17 at 12:05

0

I think I'd do this with a simple tokeniser and parser rather than regular expressions. It's not that hard to write and will be more robust and readable than a regular expression based parser.

edited May 23 '17 at 12:05

Community

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answered May 12 '14 at 04:37

Noufal Ibrahim

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Convert Equation from One Language to Another

3 Answers3