replace certain not known words in a string

Question

I'm looking for a more elegant solution to replace some upfront not known words in a string, except not,and and or:

Only as an example below, but could be anything but will always be evaluable with eval())

input: (DEFINE_A or not(DEFINE_B and not (DEFINE_C))) and DEFINE_A

output: (self.DEFINE_A or not(self.DEFINE_B and not (self.DEFINE_C))) and self.DEFINE_A

I created a solution, but it looks kind a strange. Is there a more clean way?

s = '(DEFINE_A or not(DEFINE_B and not (DEFINE_C))) and DEFINE_A'
words = re.findall(r'[\w]+|[()]*|[ ]*', s)
for index, word in enumerate(words):
    w = re.findall('^[a-zA-Z_]+$', word)
    if w and w[0] not in ['and','or','not']:
        z = 'self.' + w[0]
        words[index] = z
new = ''.join(str(x) for x in words)
print(new)

Will print correctly:

(self.DEFINE_A or not(self.DEFINE_B and not (self.DEFINE_C))) and self.DEFINE_A

Answer 1

First of all, you can match only words by using a simple \w+. Then, Using a negative lookahead you can exclude the ones you don't want. Now all that's left to do is use re.sub directly with that pattern:

s = '(DEFINE_A or not(DEFINE_B and not (DEFINE_C))) and DEFINE_A'

new = re.sub(r"(?!and|not|or)\b(\w+)", r"self.\1", s)

print(new)

Which will give:

(self.DEFINE_A or not(self.DEFINE_B and not (self.DEFINE_C))) and self.DEFINE_A

You can test-out and see how this regex works here.

---

If the names of your "variables" will always be capitalized, this simplifies the pattern a bit and making it much more efficient. Simply use:

new = re.sub(r"([A-Z\d_]+)", r"self.\1", s)

This is not only a simpler pattern (for readability), but also much more efficient. On this example, it only takes 70 steps compared to 196 of the original (can be seen in the top-right corner in the links).

You can see the new pattern in action here.