Text data replacement using dictionary

问题:

Dataframe with below structure -

ID text
0  Language processing in python th is great
1  Relace the string 

Dictionary named custom fix

{'Relace': 'Replace', 'th' : 'three'}

Tried the code and the output is coming as - Current output -

ID text
0  Language processing in pythirdon three is great
1  Replace threee string 

Code:

def multiple_replace(dict, text):
  # Create a regular expression  from the dictionary keys
  regex = re.compile("(%s)" % "|".join(map(re.escape, dict.keys())))

  # For each match, look-up corresponding value in dictionary
  return regex.sub(lambda mo: dict[mo.string[mo.start():mo.end()]], text) 

df['col1'] = df.apply(lambda row: multiple_replace(custom_fix, row['text']), axis=1)

Expected Output -

ID text
0  Language processing in python three is great
1  Replace the string

回答1:

I'm not an regex expert, and maybe this is not the best solution, but using word boundaries \b in your regex should fix the problem, here the fixed function:

def multiple_replace(d, text):
    # Create a regular expression  from the dictionary keys
    regex = re.compile("(%s)" % "|".join(["\\b" + x + "\\b" for x in d.keys()]))

    # For each match, look-up corresponding value in dictionary
    return regex.sub(lambda mo: d[mo.string[mo.start():mo.end()]], text)

回答2:

You can also split the string to get all the words and iterate through the list.

    def multiple_replace(d, text):
        splitText=text.split()
        disc=len(set(splitText).intersection(set(d.keys())))
        if disc==0:    
            return ' '.join(splitText)
        else:
            for k in range(len(splitText)):      
                try:        
                    splitText[k]=d[splitText[k]]        
                except KeyError:        
                    pass
            return ' '.join(splitText)

Hope it helps.