Pythonic way to rewrite the following C++ string processing code

Question

Previous, I am having a C++ string processing code which is able to do this.

input -> Hello 12
output-> Hello

input -> Hello 12 World
output-> Hello World

input -> Hello12 World
output-> Hello World

input -> Hello12World
output-> HelloWorld

The following is the C++ code.

std::string Utils::toStringWithoutNumerical(const std::string& str) {
    std::string result;

    bool alreadyAppendSpace = false;
    for (int i = 0, length = str.length(); i < length; i++) {
        const char c = str.at(i);
        if (isdigit(c)) {
            continue;
        }
        if (isspace(c)) {
            if (false == alreadyAppendSpace) {
                result.append(1, c);
                alreadyAppendSpace = true;
            }
            continue;
        }
        result.append(1, c);
        alreadyAppendSpace = false;
    }

    return trim(result);
}

May I know in Python, what is the Pythonic way for implementing such functionality? Is regular expression able to achieve so?

Thanks.

Yes, this would be an ideal application of regular expressions. Have you tried it? — jathanism
– jathanism, Commented Nov 11, 2010 at 2:13

Sven Marnach · Accepted Answer · 2010-11-11 02:33:07Z

7

Edit: This reproduces more accurately what the C++ code does than the previous version.

s = re.sub(r"\d+", "", s)
s = re.sub(r"(\s)\s*", "\1", s)

In particular, if the first whitespace in a run of several whitespaces is a tab, it will preserve the tab.

Further Edit: To replace by a space anyway, this works:

s = re.sub(r"\d+", "", s)
s = re.sub(r"\s+", " ", s)

edited Nov 11, 2010 at 2:33

answered Nov 11, 2010 at 2:08

Sven Marnach

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7 Comments

Steven Rumbalski Over a year ago

+1. Was going to post a competing answer before your edit, but now you are correct!

Cheok Yan Cheng Over a year ago

Sorry. I retested my C++ code. Your previous solution is correct. Can you please repost your previous solution? I will edit back my test case.

Sven Marnach Over a year ago

Well, I reproduced what your example code does. The code will collapse any run of consecutive whitespace characters, just leving the first one. My Python code does the same.

Sven Marnach Over a year ago

My second version does essentially the same as the first one, except for the "first whitespace is not a space" issue.

Cheok Yan Cheng Over a year ago

But I tested run your current code, I saw a "smile icon" at the end of string. My guess is caused by "\1"? I think replace "\1" with " " will just fine

|

Mark Ransom · Accepted Answer · 2010-11-11 03:44:36Z

5

Python has a lot of built-in functions that can be very powerful when used together.

def RemoveNumeric(str):
    return ' '.join(str.translate(None, '0123456789').split())

>>> RemoveNumeric('Hello 12')
'Hello'
>>> RemoveNumeric('Hello 12 World')
'Hello World'
>>> RemoveNumeric('Hello12 World')
'Hello World'
>>> RemoveNumeric('Hello12World')
'HelloWorld'

edited Nov 11, 2010 at 3:44

answered Nov 11, 2010 at 2:35

Mark Ransom

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Comments

Rafe Kettler · Accepted Answer · 2010-11-11 02:05:14Z

1

import re
re.sub(r'[0-9]+', "", string)

answered Nov 11, 2010 at 2:05

Rafe Kettler

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2 Comments

Cheok Yan Cheng Over a year ago

No. It will fail at 2nd case. "Hello 12 World" will have 2 spaces in between, instead of the expected single space.

Cheers and hth. - Alf Over a year ago

" ".join( re.sub( r'[0-9]+', "", s ).split() ) Cheers & hth.

John Kugelman · Accepted Answer · 2010-11-11 02:51:13Z

0

import re
re.sub(r"(\s*)\d+(\s*)", lambda m: m.group(1) or m.group(2), string)

Breakdown:

\s* matches zero or more whitespace.
\d+ matches one or more digits.
The parentheses are used to capture the whitespace.
The replacement parameter is normally a string, but it can alternatively be a function which constructs the replacement dynamically.
lambda is used to create an inline function which returns whichever of the two capture groups is non-empty. This preserves a space if there was whitespace and returns an empty string if there wasn't any.

edited Nov 11, 2010 at 2:51

answered Nov 11, 2010 at 2:07

John Kugelman

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1 Comment

Rafe Kettler Over a year ago

Doesn't this have to be a raw string?

Omnifarious · Accepted Answer · 2010-11-11 07:09:50Z

0

The regular expression answers are clearly the right way to do this. But if you're interested in a way to do if you didn't have a regex engine, here's how:

class filterstate(object):
    def __init__(self):
        self.seenspace = False
    def include(self, c):
        isspace = c.isspace()
        if (not c.isdigit()) and (not (self.seenspace and isspace)):
            self.seenspace = isspace
            return True
        else:
            return False

def toStringWithoutNumerical(s):
    fs = filterstate()
    return ''.join((c for c in s if fs.include(c)))

answered Nov 11, 2010 at 7:09

Omnifarious

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