How can I (easily) take a string such as "sin(x)*x^2" which might be entered by a user at runtime and produce a Python function that could be evaluated for any value of x?
Answers:
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Method 1
Python’s own internal compiler can parse this, if you use Python notation.
If your change the notation slightly, you’ll be happier.
import compiler eq= "sin(x)*x**2" ast= compiler.parse( eq )
You get an abstract syntax tree that you can work with.
Method 2
You can use Python parser:
import parser from math import sin formula = "sin(x)*x**2" code = parser.expr(formula).compile() x = 10 print(eval(code))
It performs better than pure eval and, of course, avoids code injection!
Method 3
f = parser.parse('sin(x)*x^2').to_pyfunc()
Where parser could be defined using PLY, pyparsing, builtin tokenizer, parser, ast.
Don’t use eval on user input.
Method 4
pyparsing might do what you want (http://pyparsing.wikispaces.com/) especially if the strings are from an untrusted source.
See also http://pyparsing.wikispaces.com/file/view/fourFn.py for a fairly full-featured calculator built with it.
Method 5
To emphasize J.F. Sebastian’s advice, ‘eval’ and even the ‘compiler’ solutions can be open to subtle security holes. How trustworthy is the input? With ‘compiler’ you can at least filter out things like getattr lookups from the AST, but I’ve found it’s easier to use PLY or pyparsing for this sort of thing than it is to secure the result of letting Python help out.
Also, ‘compiler’ is clumsy and hard to use. It’s deprecated and removed in 3.0. You should use the ‘ast’ module (added in 2.6, available in 2.5 as ‘_ast’).
Method 6
Sage is intended as matlab replacement and in intro videos it’s demonstrated how similar to yours cases are handled. They seem to be supporting a wide range of approaches. Since the code is open-source you could browse and see for yourself how the authors handle such cases.
Method 7
In agreement with vartec. I would use SymPy – in particular the lambdify function should do exactly what you want.
See: http://showmedo.com/videotutorials/video?name=7200080&fromSeriesID=720
for a very nice explanation of this.
Best wishes,
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0