Check whether an expression is valid using regex in python

Question

I'm trying to check if a string is a valid math expression. Allowed operations are +,-,*,/ and ^. I tried this and don't know why it doesn't work:

a = raw_input("Unesite izraz")
if len( re.findall(r'(\d+(?:.\d+)?(?: [\+\-\/\*\^]\d+(?:.\d+) )* )', a ) ) != 0:

But this regex expression returns [] for valid expressions. Why? Thanks!

Not really an answer, but when you match a number \d+(?:.\d+)?, the dot should be escaped as \., since otherwise '1a3' and '9(10' would be considered valid numbers. — Fernando Matsumoto
– Fernando Matsumoto, Commented Dec 12, 2015 at 21:01

user557597 · Accepted Answer · 2015-12-12 21:04:59Z

2

A validator for a simple symbol math expression could be something like this.
This would be a 1 time match of the entire string.

^\s*[+-]?\s*(?:\d+(?:\.\d*)?|\.\d+)(?:\s*[-+/*^]\s*\s*[+-]?\s*(?:\d+(?:\.\d*)?|\.\d+))*\s*$

Formatted:

 ^                             # BOS
 \s* [+-]? \s*                 # whitespace (opt), sign (opt), whitespace (opt)
 (?:                           # Integer or decimal
      \d+ 
      (?: \. \d* )?
   |  \. \d+ 
 )
 (?:                           # Cluster group
      \s* [-+/*^] \s*               # whitespace (opt), operation symbol (req'd), whitespace (opt)
      \s* [+-]? \s*                 # whitespace (opt), sign (opt), whitespace (opt)
      (?:                           # Integer or decimal
           \d+ 
           (?: \. \d* )?
        |  \. \d+ 
      )
 )*                            # End cluster, do 0 to many times
 \s*                           # optional whitespace
 $                             # EOS

answered Dec 12, 2015 at 21:04

user557597

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2 Comments

Dusan Milosevic Over a year ago

This works, thanks, but shouldn't we use \ in front of +,- etc.

user557597 Over a year ago

The + is inside a character class where its not a quantifier .. no need to escape it. Likewise the * and^.

Tamas Hegedus · Accepted Answer · 2015-12-12 20:52:44Z

1

You have spaces in the wrong places.

# yours / fixed
r'(\d+(?:.\d+)?(?: [\+\-\/\*\^]\d+(?:.\d+) )* )'
r'(\d+(?:.\d+)?(?: [\+\-\/\*\^] \d+(?:.\d+) )*)'

You can try them at pythex.org

answered Dec 12, 2015 at 20:52

Tamas Hegedus

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Comments

macabeus · Accepted Answer · 2015-12-12 20:58:35Z

1

You can simplify your regexp.

There is the operator "except": [^abc] Thus, it will take whatever is not the characters "a", "b" or "c".

import re

e1 = '1 + 2' # correct
e2 = '1 + 3 * 3 / 6 ^ 2' # correct
e3 = '1 + 3 x 3' # wrong

el = [e1, e2, e3]
regexp = re.compile(r'[^+\-*\/^0-9\s]')
for i in el:
    if len(regexp.findall(i)):
        print(i, 'wrong')
    else:
        print(i, 'correct')

You can use this site to learn and test yours regexp: https://regex101.com/

answered Dec 12, 2015 at 20:58

macabeus

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