python: count and replace pattern in string without regex

Question

What is the non-regex equivalent of subn? I would like to replace a pattern in a string and count how many times was the pattern replaced. I ended up with:

def replacen(pat, repl, txt):
    txt2 = txt.replace(pat, repl)
    if (len(pat) != len(repl)):
        return (txt2, (len(txt) - len(txt2)) / (len(pat) - len(repl)))
    else:
        return (txt2, txt.count(pat))

Is there a more elegant solution?

This would be a more appropriate question for codereview.stackexchange.com — heinst
– heinst, Commented Jul 12, 2015 at 18:38

Padraic Cunningham · Accepted Answer · 2015-07-12 18:56:55Z

3

Your own code can simply return txt2 and the count of the pattern:

def replacen(pat, repl, txt):
    txt2 = txt.replace(pat, repl)
    return txt2, txt.count(pat)

If nothing gets replaced then it is because you found no matching substring so count will return 0 if you replaced 5 substring/pats then txt.count(pat) is going to give you 5

answered Jul 12, 2015 at 18:56

Padraic Cunningham

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4 Comments

Kasravnd Over a year ago

Its very funny rally some times we missed the answers in our life because of those simplicity :-D

dimid Over a year ago

right, but I don't want to use count unless I have to (for efficiency considerations)

Burhan Khalid Over a year ago

I doubt you'll get more efficient than count.

dimid Over a year ago

count is at least O(n) while checking the lengths is O(1)

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