simple pattern matching in python

Question

I am trying to write a basic interpreter in python. So, I am at that point where I am trying to declare whether a string entered in command prompt is a method or variable type.

So not trying any fancy stuff..

s="12345" #variable
s ="foo()" method
s = "foo(1234)" method

What is a robust way to do this (for example.. robust for whitespaces ... throw error if syntax is not proper)

My code is pretty straightforward

s = s.strip()

params=   s[s.find("(") + 1:s.find(")")] # find the params..

The above command works in case two and case three but for case 1.. it gives weird results..

A robust way is to write an actual parser, using something like ANTLR or pyparsing or whatever comes up under "python parser library". — millimoose
– millimoose, Commented Feb 9, 2013 at 23:16
the reason why you get the funny result in the first case is because your statement in case of s = 123456 s[s.find("(") + 1:s.find(")")] produces s[-1 + 1 : -1 ] s[0 : -1 ] so you always miss the last character — Redian
– Redian, Commented Feb 9, 2013 at 23:18
Instead of parsing the string directly, you should first break it up into tokens, i.e. strings, bracket, numbers, operators, etc. Then, parse the sequence of tokens. — tobias_k
– tobias_k, Commented Feb 9, 2013 at 23:41

Redian · Accepted Answer · 2013-02-10 11:18:57Z

1

For the scenarios you are asked for i think this could do the trick

have a go

s[ 1+s.find("(") if s.find("(") > 0 else None : -1 if s.find(")") > 0 else None]

edit: making a bit neater as suggested by Paul:

s[ 1+s.find("(") if '(' in s else None : -1 if ')' in s  else None]

edited Feb 10, 2013 at 11:18

answered Feb 9, 2013 at 23:46

Redian

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2 Comments

Ben Mordecai Over a year ago

I think this would be clearer if it weren't crammed on to one line.

PaulMcG Over a year ago

I think this would also be clearer if you used the more up-to-date idiom of if '(' in s instead of if s.find('(') > 0.

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