优雅的Python函数将CamelCase转换为snake_case？

Question

示例：

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

Answer 1

驼箱到蛇箱

import re

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

如果您多次执行此操作，并且上述操作速度较慢，请提前编译regex：

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

要特别处理更高级的情况(这不再是可逆的)：

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

要同时添加具有两个或更多下划线的案例：

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()

蛇箱到骆驼箱

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

Answer 2

在包索引中有一个inflection library，它可以为您处理这些事情。在本例中，您将查找inflection.underscore()

>>> inflection.underscore('CamelCase')
'camel_case'

Answer 3

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]+)就可以做到这一点

>>> 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，只需在(?!^)后面添加look

>>> 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'

如果您想将ALLCaps与all_caps分开，并希望字符串中包含数字，那么您仍然不需要执行两次单独的运行，只需使用|表达式((?<=[a-z0-9])[A-Z]|(?!^)[A-Z](?=[a-z]))就可以处理本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'

这完全取决于您想要什么，所以请使用最适合您的需求的解决方案，因为它不应该过于复杂。

nJoy！