CODEWITHSUNDEEP
pythonregexeval
padmanabh pande
padmanabh pande

Reputation: 427

Python: How to replace numeric values in a string with their corresponding data in a dictionary

I have a logical expression like : 

    expression = '(1 & 2) & 3'
    and a dictionary as, x = {1: '1', 2: '0', 3: '0'} 
    --- keys can be int or string, manageable

I need to evaluate this string as,

    '(1 & 2) & 3' 
    => '(x[1] & x[2]) & x[3]'
    => '(1 & 0) & 0'
    => eval('(1 & 0) & 0') = 0

I'm currently doing it like this:

    re.sub('(\d+)', r'x[\1]', expression)
    => '(x[1] & x[2]) & x[3]'
    but can't run eval on this

or

    re.sub('(\d+)', eval(r'x[\1]'), expression)
    => eval(x[\1]) --> logical error

Please share an effective solution to achieve this. The application can define different expressions and one expression is used to evaluate 1000s of dictionaries like x.

Upvotes: 2

Views: 132

Answers (2)

user6882757
user6882757

Reputation:

Incase it helps. without regex

def complement_n(expression):
    expression_dict = {'1': '1','2': '0', '3': '0', '&':'&','(':'(',')':')',' ': ' '
    #expression_dict = dict(a='t', c='g', t='a', g='c'),
                      }
    comp_string = ''
    for base in expression:
        comp_string += expression_dict[base]
        ##comp_string += dna_complement.get(base, 'some_value')
    return comp_string

expression = '(1 & 2 ) & 3'
comp_string = complement_n(expression)
print ("Complement is:", comp_string)

Upvotes: 0

yatu
yatu

Reputation: 88276

What you could do, is define a lambda function as replacement, and lookup in the dictionary for each match:

import re

expression = '(1 & 2) & 3'
d = {1: '1', 2: '0', 3: '0'} 

res = re.sub('(\d+)', lambda x: d[int(x.group(1))], expression)
#'(1 & 0) & 0'

eval(res)
# 0

If the expression could contain integers not present in the dictionary, you could adapt something like:

res = re.sub('(\d+)', lambda x: d.get(int(x.group(1)), '0'), expression)

Upvotes: 4

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