例子:
>>> convert('CamelCase')
'camel_case'
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("(?<!^)(?=[A-Z])");