例子:

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

当前回答

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

其他回答

我认为这个解决方案比之前的答案更直接:

import re

def convert (camel_input):
    words = re.findall(r'[A-Z]?[a-z]+|[A-Z]{2,}(?=[A-Z][a-z]|\d|\W|$)|\d+', camel_input)
    return '_'.join(map(str.lower, words))


# Let's test it
test_strings = [
    'CamelCase',
    'camelCamelCase',
    'Camel2Camel2Case',
    'getHTTPResponseCode',
    'get200HTTPResponseCode',
    'getHTTP200ResponseCode',
    'HTTPResponseCode',
    'ResponseHTTP',
    'ResponseHTTP2',
    'Fun?!awesome',
    'Fun?!Awesome',
    '10CoolDudes',
    '20coolDudes'
]
for test_string in test_strings:
    print(convert(test_string))

输出:

camel_case
camel_camel_case
camel_2_camel_2_case
get_http_response_code
get_200_http_response_code
get_http_200_response_code
http_response_code
response_http
response_http_2
fun_awesome
fun_awesome
10_cool_dudes
20_cool_dudes

正则表达式匹配三种模式:

[a - z]吗?[a-z]+:连续小写字母,可选以大写字母开头。 [a - z] {2,} (? = [a - z] [a - z] | | \ \ d W | $):两个或两个以上的连续大写字母。如果最后一个大写字母后面跟着一个小写字母,它使用一个超前来排除它。 \d+:连续数字。

通过使用re.findall,我们得到了一个单独的“单词”列表,这些单词可以转换为小写字母并用下划线连接。

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 !

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])");