例子:
>>> convert('CamelCase')
'camel_case'
例子:
>>> convert('CamelCase')
'camel_case'
当前回答
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
其他回答
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("(?<!^)(?=[A-Z])");
Stringcase是我的首选库;例如:
>>> from stringcase import pascalcase, snakecase
>>> snakecase('FooBarBaz')
'foo_bar_baz'
>>> pascalcase('foo_bar_baz')
'FooBarBaz'
我不知道为什么使用两个.sub()调用?:)我不是正则表达式大师,但我简化了这个函数,这是适合我的特定需求,我只是需要一个解决方案,将camelCasedVars从POST请求转换为vars_with_下划线:
def myFunc(...):
return re.sub('(.)([A-Z]{1})', r'\1_\2', "iTriedToWriteNicely").lower()
它不能使用像getHTTPResponse这样的名字,因为我听说这是一个糟糕的命名惯例(应该像getHTTPResponse一样,很明显,它更容易记住这个形式)。
看一看优秀的Schematics库
https://github.com/schematics/schematics
它允许你创建类型化的数据结构,可以序列化/反序列化从python到Javascript风格,例如:
class MapPrice(Model):
price_before_vat = DecimalType(serialized_name='priceBeforeVat')
vat_rate = DecimalType(serialized_name='vatRate')
vat = DecimalType()
total_price = DecimalType(serialized_name='totalPrice')
轻度改编自https://stackoverflow.com/users/267781/matth 谁使用发电机。
def uncamelize(s):
buff, l = '', []
for ltr in s:
if ltr.isupper():
if buff:
l.append(buff)
buff = ''
buff += ltr
l.append(buff)
return '_'.join(l).lower()