Replace string with multiple reverse patterns in Python

Question

I have a string that need to be replaced.

myString = 'I have a cat, it like food'

I want to replace 'cat' with 'dog', or if there's 'dog' in the string, then replace 'dog' with 'cat'.

meanwhile, I need to replace 'like' with 'love', or 'love' with 'like'.

use OrderedDict can set the conditions' order, but it will skip the rest of the condition if one condition meet

from collections import OrderedDict
def replace_all(text, dic):
    for i, j in dic.items():
        text = text.replace(i, j)
    return text
od = OrderedDict([("cat", "dog"), ("dog", "cat"), ("like ", "love"), ("love", "like")])
myString = "I have a cat, it like food"
replace_all(myString, od)
print(myString)

please advice.

Answer 1

this is another way of doing it

import re
od = OrderedDict([("cat", "dog"), ("dog", "cat"), ("like", "love"), ("love", "like")])
string = "I have a cat, it like food"
for word in re.split(", | ", string):
     if word in od:
             string = string.replace(word, od[word])

Answer 2

Yet a straight forward lambda hash replacement

>>> import re
>>>
>>> negations_dic = {"cat":"dog", "dog":"cat", "like":"love", "love":"like"}
>>> neg_pattern = re.compile(r'(?:cat|dog|like|love)')
>>>
>>> def negate_all(text):
...     return neg_pattern.sub(lambda m: negations_dic[m.group()], text)
...
>>> print ( negate_all("I have a cat, it like food") )
I have a dog, it love food

Answer 3

Use re.sub with lambda func in replacement part. Since the replacement occurs upon every inline match, there won't be any duplicate replacements.

>>> r = [("cat", "dog"), ("dog", "cat"), ("like", "love"), ("love", "like")]
>>> dict(r)
{'like': 'love', 'cat': 'dog', 'love': 'like', 'dog': 'cat'}
>>> d = dict(r)
>>> re.sub(r'\b(?:' + '|'.join(d.keys()) + r')\b', lambda x: d[x.group()], s)
'I have a dog, it love food'
>>>