How to do CamelCase split in python

问题:

What I was trying to achieve, was something like this:

>>> camel_case_split("CamelCaseXYZ")
['Camel', 'Case', 'XYZ']
>>> camel_case_split("XYZCamelCase")
['XYZ', 'Camel', 'Case']

So I searched and found this perfect regular expression:

(?<=[a-z])(?=[A-Z])|(?<=[A-Z])(?=[A-Z][a-z])

As the next logical step I tried:

>>> re.split("(?<=[a-z])(?=[A-Z])|(?<=[A-Z])(?=[A-Z][a-z])", "CamelCaseXYZ")
['CamelCaseXYZ']

Why does this not work, and how do I achieve the result from the linked question in python?

Edit: Solution summary

I tested all provided solutions with a few test cases:

string:                 ''
nfs:                    ['']
casimir_et_hippolyte:   []
two_hundred_success:    []
kalefranz:              string index out of range # with modification: either [] or ['']

string:                 ' '
nfs:                    [' ']
casimir_et_hippolyte:   []
two_hundred_success:    [' ']
kalefranz:              [' ']

string:                 'lower'
all algorithms:         ['lower']

string:                 'UPPER'
all algorithms:         ['UPPER']

string:                 'Initial'
all algorithms:         ['Initial']

string:                 'dromedaryCase'
nfs:                    ['dromedary', 'Case']
casimir_et_hippolyte:   ['dromedary', 'Case']
two_hundred_success:    ['dromedary', 'Case']
kalefranz:              ['Dromedary', 'Case'] # with modification: ['dromedary', 'Case']

string:                 'CamelCase'
all algorithms:         ['Camel', 'Case']

string:                 'ABCWordDEF'
nfs:                    ['ABC', 'Word', 'DEF']
casimir_et_hippolyte:   ['ABC', 'Word', 'DEF']
two_hundred_success:    ['ABC', 'Word', 'DEF']
kalefranz:              ['ABCWord', 'DEF']

In summary you could say the solution by @kalefranz does not match the question (see the last case) and the solution by @casimir et hippolyte eats a single space, and thereby violates the idea that a split should not change the individual parts. The only difference among the remaining two alternatives is that my solution returns a list with the empty string on an empty string input and the solution by @200_success returns an empty list. I don't know how the python community stands on that issue, so I say: I am fine with either one. And since 200_success's solution is simpler, I accepted it as the correct answer.

回答1:

As @nfs has explained, re.split() never splits on an empty pattern match. Therefore, instead of splitting, you should try finding the components you are interested in.

Here is a solution using re.finditer() that emulates splitting:

def camel_case_split(identifier):
    matches = finditer('.+?(?:(?<=[a-z])(?=[A-Z])|(?<=[A-Z])(?=[A-Z][a-z])|$)', identifier)
    return [m.group(0) for m in matches]

回答2:

Use re.sub() and split()

import re

name = 'CamelCaseTest123'
splitted = re.sub('(?!^)([A-Z][a-z]+)', r' \1', name).split()

result

['Camel', 'Case', 'Test123']

回答3:

Most of the time when you don't need to check the format of a string, a global research is more simple than a split (for the same result):

re.findall(r'[A-Z](?:[a-z]+|[A-Z]*(?=[A-Z]|$))', 'CamelCaseXYZ')

returns

['Camel', 'Case', 'XYZ']

To deal with dromedary too, you can use:

re.findall(r'[A-Z]?[a-z]+|[A-Z]+(?=[A-Z]|$)', 'camelCaseXYZ')

Note: (?=[A-Z]|$) can be shorten using a double negation (a negative lookahead with a negated character class): (?![^A-Z])

回答4:

The documentation for python's re.split says:

Note that split will never split a string on an empty pattern match.

When seeing this:

>>> re.findall("(?<=[a-z])(?=[A-Z])|(?<=[A-Z])(?=[A-Z][a-z])", "CamelCaseXYZ")
['', '']

it becomes clear, why the split does not work as expected. The remodule finds empty matches, just as intended by the regular expression.

Since the documentation states that this is not a bug, but rather intended behavior, you have to work around that when trying to create a camel case split:

def camel_case_split(identifier):
    matches = finditer('(?<=[a-z])(?=[A-Z])|(?<=[A-Z])(?=[A-Z][a-z])', identifier)
    split_string = []
    # index of beginning of slice
    previous = 0
    for match in matches:
        # get slice
        split_string.append(identifier[previous:match.start()])
        # advance index
        previous = match.start()
    # get remaining string
    split_string.append(identifier[previous:])
    return split_string

回答5:

I just stumbled upon this case and wrote a regular expression to solve it. It should work for any group of words, actually.

RE_WORDS = re.compile(r'''
    # Find words in a string. Order matters!
    [A-Z]+(?=[A-Z][a-z]) |  # All upper case before a capitalized word
    [A-Z]?[a-z]+ |  # Capitalized words / all lower case
    [A-Z]+ |  # All upper case
    \d+  # Numbers
''', re.VERBOSE)

The key here is the lookahead on the first possible case. It will match (and preserve) uppercase words before capitalized ones:

assert RE_WORDS.findall('FOOBar') == ['FOO', 'Bar']

回答6:

Here's another solution that requires less code and no complicated regular expressions:

def camel_case_split(string):
    bldrs = [[string[0].upper()]]
    for c in string[1:]:
        if bldrs[-1][-1].islower() and c.isupper():
            bldrs.append([c])
        else:
            bldrs[-1].append(c)
    return [''.join(bldr) for bldr in bldrs]

Edit

The above code contains an optimization that avoids rebuilding the entire string with every appended character. Leaving out that optimization, a simpler version (with comments) might look like

def camel_case_split2(string):
    # set the logic for creating a "break"
    def is_transition(c1, c2):
      return c1.islower() and c2.isupper()

    # start the builder list with the first character
    # enforce upper case
    bldr = [string[0].upper()]
    for c in string[1:]:
        # get the last character in the last element in the builder
        # note that strings can be addressed just like lists
        previous_character = bldr[-1][-1]
        if is_transition(previous_character, c):
            # start a new element in the list
            bldr.append(c)
        else:
            # append the character to the last string
            bldr[-1] += c
    return bldr

回答7:

I know that the question added the tag of regex. But still, I always try to stay as far away from regex as possible. So, here is my solution without regex:

def split_camel(text, char):
    if len(text) <= 1: # To avoid adding a wrong space in the beginning
        return text+char
    if char.isupper() and text[-1].islower(): # Regular Camel case
        return text + " " + char
    elif text[-1].isupper() and char.islower(): # Detect Camel case in case of abbreviations
        return text[:-1] + " " + text[-1] + char
    else: # Do nothing part
        return text + char

text = "PathURLFinder"
text = reduce(split_camel, a, "")
print text
# prints "Path URL Finder"
print text.split(" ")
# prints "['Path', 'URL', 'Finder']"

回答8:

I think below is the optimim

Def count_word(): Return(re.findall(‘[A-Z]?[a-z]+’, input(‘please enter your string’))

Print(count_word())