例子:

>>> convert('CamelCase')
'camel_case'

当前回答

这个网站上提出了非常好的RegEx:

(?<!^)(?=[A-Z])

如果python有一个字符串分割方法,它应该工作…

在Java中:

String s = "loremIpsum";
words = s.split("(?&#60;!^)(?=[A-Z])");

其他回答

简洁,没有正则表达式,但HTTPResponseCode=> httpresponse_code:

def from_camel(name):
    """
    ThisIsCamelCase ==> this_is_camel_case
    """
    name = name.replace("_", "")
    _cas = lambda _x : [_i.isupper() for _i in _x]
    seq = zip(_cas(name[1:-1]), _cas(name[2:]))
    ss = [_x + 1 for _x, (_i, _j) in enumerate(seq) if (_i, _j) == (False, True)]
    return "".join([ch + "_" if _x in ss else ch for _x, ch in numerate(name.lower())])
''.join('_'+c.lower() if c.isupper() else c for c in "DeathToCamelCase").strip('_')
re.sub("(.)([A-Z])", r'\1_\2', 'DeathToCamelCase').lower()

哇,我刚从django片段中偷了这个。ref http://djangosnippets.org/snippets/585/

很优雅

camelcase_to_underscore = lambda str: re.sub(r'(?<=[a-z])[A-Z]|[A-Z](?=[^A-Z])', r'_\g<0>', str).lower().strip('_')

例子:

camelcase_to_underscore('ThisUser')

返回:

'this_user'

REGEX演示

骆驼案变蛇案

import re

name = 'CamelCaseName'
name = re.sub(r'(?<!^)(?=[A-Z])', '_', name).lower()
print(name)  # camel_case_name

如果你这样做了很多次,上面的速度很慢,提前编译正则表达式:

pattern = re.compile(r'(?<!^)(?=[A-Z])')
name = pattern.sub('_', name).lower()

为了处理更高级的情况(这是不可逆的了):

def camel_to_snake(name):
    name = re.sub('(.)([A-Z][a-z]+)', r'\1_\2', name)
    return re.sub('([a-z0-9])([A-Z])', r'\1_\2', name).lower()

print(camel_to_snake('camel2_camel2_case'))  # camel2_camel2_case
print(camel_to_snake('getHTTPResponseCode'))  # get_http_response_code
print(camel_to_snake('HTTPResponseCodeXYZ'))  # http_response_code_xyz

添加带有两个或两个以上下划线的also大小写:

def to_snake_case(name):
    name = re.sub('(.)([A-Z][a-z]+)', r'\1_\2', name)
    name = re.sub('__([A-Z])', r'_\1', name)
    name = re.sub('([a-z0-9])([A-Z])', r'\1_\2', name)
    return name.lower()

斯内克案转帕斯卡案

name = 'snake_case_name'
name = ''.join(word.title() for word in name.split('_'))
print(name)  # SnakeCaseName

我不知道为什么这些都这么复杂。

对于大多数情况,简单的表达式([A-Z]+)就可以了

>>> re.sub('([A-Z]+)', r'_\1','CamelCase').lower()
'_camel_case'  
>>> re.sub('([A-Z]+)', r'_\1','camelCase').lower()
'camel_case'
>>> re.sub('([A-Z]+)', r'_\1','camel2Case2').lower()
'camel2_case2'
>>> re.sub('([A-Z]+)', r'_\1','camelCamelCase').lower()
'camel_camel_case'
>>> re.sub('([A-Z]+)', r'_\1','getHTTPResponseCode').lower()
'get_httpresponse_code'

要忽略第一个字符,只需添加look behind (?!^)

>>> re.sub('(?!^)([A-Z]+)', r'_\1','CamelCase').lower()
'camel_case'
>>> re.sub('(?!^)([A-Z]+)', r'_\1','CamelCamelCase').lower()
'camel_camel_case'
>>> re.sub('(?!^)([A-Z]+)', r'_\1','Camel2Camel2Case').lower()
'camel2_camel2_case'
>>> re.sub('(?!^)([A-Z]+)', r'_\1','getHTTPResponseCode').lower()
'get_httpresponse_code'

如果你想分离ALLCaps到all_caps,并期望字符串中的数字,你仍然不需要做两次单独的运行,只需使用|这个表达式((?<=[a-z0-9])[A-Z]|(?!^)[A-Z](?=[A-Z]))可以处理书中的几乎所有场景

>>> a = re.compile('((?<=[a-z0-9])[A-Z]|(?!^)[A-Z](?=[a-z]))')
>>> a.sub(r'_\1', 'getHTTPResponseCode').lower()
'get_http_response_code'
>>> a.sub(r'_\1', 'get2HTTPResponseCode').lower()
'get2_http_response_code'
>>> a.sub(r'_\1', 'get2HTTPResponse123Code').lower()
'get2_http_response123_code'
>>> a.sub(r'_\1', 'HTTPResponseCode').lower()
'http_response_code'
>>> a.sub(r'_\1', 'HTTPResponseCodeXYZ').lower()
'http_response_code_xyz'

这完全取决于你想要什么,所以使用最适合你需要的解决方案,因为它不应该过于复杂。

nJoy !