Extract All Variables From A String Of Python Code (regex Or Ast)
Solution 1:
I find your question an interesting challenge, so here is a code that do what you want, doing this using Regex
alone it's impossible because there is nested expression, this is a solution using a combination of Regex
and string manipulations to handle nested expressions:
# -*- coding: utf-8 -*-import re
RE_IDENTIFIER = r'\b[a-z]\w*\b(?!\s*[\[\("\'])'
RE_INDEX_ONLY = re.compile(r'(##)(\d+)(##)')
RE_INDEX = re.compile('##\d+##')
defextract_expression(string):
""" extract all identifier and getitem expression in the given order."""defremove_brackets(text):
# 1. handle `[...]` expression replace them with #{#...#}## so we don't confuse them with word[...]
pattern = '(?<!\w)(\s*)(\[)([^\[]+?)(\])'# keep extracting expression until there is no expressionwhile re.search(pattern, text):
text = re.sub(pattern, r'\1#{#\3#}#', string)
return text
defget_ordered_subexp(exp):
""" get index of nested expression."""
index = int(exp.replace('#', ''))
subexp = RE_INDEX.findall(expressions[index])
ifnot subexp:
return exp
return exp + ''.join(get_ordered_subexp(i) for i in subexp)
defreplace_expression(match):
""" save the expression in the list, replace it with special key and it's index in the list."""
match_exp = match.group(0)
current_index = len(expressions)
expressions.append(None) # just to make sure the expression is inserted before it's inner identifier# if the expression contains identifier extract too.if re.search(RE_IDENTIFIER, match_exp) and'['in match_exp:
match_exp = re.sub(RE_IDENTIFIER, replace_expression, match_exp)
expressions[current_index] = match_exp
return'##{}##'.format(current_index)
deffix_expression(match):
""" replace the match by the corresponding expression using the index"""return expressions[int(match.group(2))]
# result that will contains
expressions = []
string = remove_brackets(string)
# 2. extract all expression and keep track of there place in the original code
pattern = r'\w+\s*\[[^\[]+?\]|{}'.format(RE_IDENTIFIER)
# keep extracting expression until there is no expressionwhile re.search(pattern, string):
# every exression that is extracted is replaced by a special key
string = re.sub(pattern, replace_expression, string)
# some times inside brackets can contains getitem expression# so when we extract that expression we handle the brackets
string = remove_brackets(string)
# 3. build the correct result with extracted expressions
result = [None] * len(expressions)
for index, exp inenumerate(expressions):
# keep replacing special keys with the correct expressionwhile RE_INDEX_ONLY.search(exp):
exp = RE_INDEX_ONLY.sub(fix_expression, exp)
# finally we don't forget about the brackets
result[index] = exp.replace('#{#', '[').replace('#}#', ']')
# 4. Order the index that where extracted
ordered_index = ''.join(get_ordered_subexp(exp) for exp in RE_INDEX.findall(string))
# convert it to integer
ordered_index = [int(index[1]) for index in RE_INDEX_ONLY.findall(ordered_index)]
# 5. fix the order of expressions using the ordered indexes
final_result = []
for exp_index in ordered_index:
final_result.append(result[exp_index])
# for debug:# print('final string:', string)# print('expression :', expressions)# print('order_of_expresion: ', ordered_index)return final_result
code = 'foo + bar[1] + baz[1:10:var1[2+1]] + qux[[1,2,int(var2)]] + bob[len("foobar")] + func() + func2 (var3[0])'
code2 = 'baz[1:10:var1[2+1]]'
code3 = 'baz[[1]:10:var1[2+1]:[var3[3+1*x]]]'print(extract_expression(code))
print(extract_expression(code2))
print(extract_expression(code3))
OUTPU:
['foo', 'bar[1]', 'baz[1:10:var1[2+1]]', 'var1[2+1]', 'qux[[1,2,int(var2)]]', 'var2', 'bob[len("foobar")]', 'var3[0]']['baz[1:10:var1[2+1]]', 'var1[2+1]']['baz[[1]:10:var1[2+1]:[var3[3+1*x]]]', 'var1[2+1]', 'var3[3+1*x]', 'x']
I tested this code for very complicated examples and it worked perfectly. and notice that the order if extraction is the same as you wanted, Hope that this is what you needed.
Solution 2:
This answer might be too later. But it is possible to do it using python regex package.
import regex
code= '''foo + bar[1] + baz[1:10:var1[2+1]] +
qux[[1,2,int(var2)]] + bob[len("foobar")] + func() + func2
(var3[0])'''
p=r'(\b[a-z]\w*\b(?!\s*[\(\"])(\[(?:[^\[\]]|(?2))*\])?)'
result=regex.findall(p,code,overlapped=True) #overlapped=True is needed to capture something inside a group like 'var1[2+1]'
[x[0] for x in result] #result variable is list of tuple of two,as each pattern capture two groups ,see below.
output: ['foo','bar[1]','baz[1:10:var1[2+1]]','var1[2+1]','qux[[1,2,int(var2)]]','var2','bob[len("foobar")]','var3[0]']
pattern explaination:
- ( # 1st capturing group start
- \b[a-z]\w*\b #variable name ,eg 'bar'
- (?!\s*[\(\"]) #negative lookahead. so to ignore something like foobar"
- (\[(?:[^\[\]]|(?2))*\]) #2nd capture group,capture nested groups in '[ ]' #eg '[1:10:var1[2+1]]'. #'?2' refer to 2nd capturing group recursively
- ? #2nd capturing group is optional so to capture something like 'foo'
- ) #end of 1st group.
Solution 3:
Regex is not a powerful enough tool to do this. If there is a finite depth of your nesting there is some hacky work around that would allow you to make complicate regex to do what you are looking for but I would not recommend it.
If you really must parse a string for code an AST would technically work but I am not aware of a library to help with this. You would be best off trying to build a recursive function to do the parsing.
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