Python - get variables from file using regular expressions

Question

Problem

Given a file, for example:

#Example
x = 'foo = {}'

bar = {
  'color': 'blue'
}

print('Hello World')

Using regular expressions, the program will find variables names that are corresponding to dictionaries or set literals and print them. The output for the above would just be:

bar

A line like this

yes = { "no={}" }

Should only return yes.

The variables could be found with \w+? expression I'm just not sure where to start.

Current code

for line in open('program.txt'):
   array = line.split(' ')

   for item in array:
      if '{' in item:
         print(array[0])

This only functions properly for basic examples and there are some problems. Multiable variables on the same line don't work either. Also it doesn't use Regular Expressions.

Avinash Raj · Accepted Answer · 2015-08-16 23:51:09Z

1

You may do like this,

>>> s = '''#Example
x = 'foo = {}'

bar = {
  'color': 'blue'
}

print('Hello World')
yes = { "no={}" }'''
>>> re.findall(r'^\s*(\w+)\s*=\s*\{[^}]*\}', s, re.M)
['bar', 'yes']

or

>>> [i for i in re.findall(r'\{(?:\{[^{}]*\}|[^{}])*\}|^\s*(\w+)\s*=\s*(?=\{)', s, re.M) if i]
['bar', 'yes']
>>>

If you want to define it as a function, you may use the below.

def check(s):
    return [i for i in re.findall(r'\{(?:\{[^{}]*\}|[^{}])*\}|^\s*(\w+)\s*=\s*(?=\{)', s, re.M) if i]

edited Aug 16, 2015 at 23:51

answered Aug 16, 2015 at 5:54

Avinash Raj

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1 Comment

sir_triple Over a year ago

How would I modify the pattern so that it can find dictionaries in one line functions. Example: def check(): b = {} would output b.

Kevin · Accepted Answer · 2015-08-16 06:04:22Z

1

Don't use regular expressions for this.

Well, that's probably a bit too strong. Don't just use regular expressions for this.

Start by matching the code against this regex:

(\w+)\s*=\s*(.*)

Then, take the second capturing group, and pass it through ast.literal_eval().

To wit:

results = {}
for line in file:
    match = re.match(r'(\w+)\s*=\s*(.*)')
    if match:
        try:
            results[match.group(1)] = ast.literal_eval(match.group(2))
        except ValueError:
            continue  # It wasn't a valid literal

Note that ast.literal_eval() does not handle set and byte literals in 2.x. This is yet another reason to upgrade to Python 3.

edited Aug 16, 2015 at 6:04

answered Aug 16, 2015 at 5:58

Kevin

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