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

这不是一个优雅的方法,是一个非常“低级”的实现一个简单的状态机(位域状态机),可能是最反python模式来解决这个问题,然而re模块也实现了一个太复杂的状态机来解决这个简单的任务,所以我认为这是一个很好的解决方案。

def splitSymbol(s):
    si, ci, state = 0, 0, 0 # start_index, current_index 
    '''
        state bits:
        0: no yields
        1: lower yields
        2: lower yields - 1
        4: upper yields
        8: digit yields
        16: other yields
        32 : upper sequence mark
    '''
    for c in s:

        if c.islower():
            if state & 1:
                yield s[si:ci]
                si = ci
            elif state & 2:
                yield s[si:ci - 1]
                si = ci - 1
            state = 4 | 8 | 16
            ci += 1

        elif c.isupper():
            if state & 4:
                yield s[si:ci]
                si = ci
            if state & 32:
                state = 2 | 8 | 16 | 32
            else:
                state = 8 | 16 | 32

            ci += 1

        elif c.isdigit():
            if state & 8:
                yield s[si:ci]
                si = ci
            state = 1 | 4 | 16
            ci += 1

        else:
            if state & 16:
                yield s[si:ci]
            state = 0
            ci += 1  # eat ci
            si = ci   
        print(' : ', c, bin(state))
    if state:
        yield s[si:ci] 


def camelcaseToUnderscore(s):
    return '_'.join(splitSymbol(s)) 

splitsymbol可以解析所有的大小写类型:UpperSEQUENCEInterleaved, under_score, BIG_SYMBOLS和cammelCasedMethods

我希望它是有用的

一个使用正则表达式的可怕例子(你可以很容易地清理:)):

def f(s):
    return s.group(1).lower() + "_" + s.group(2).lower()

p = re.compile("([A-Z]+[a-z]+)([A-Z]?)")
print p.sub(f, "CamelCase")
print p.sub(f, "getHTTPResponseCode")

但适用于getHTTPResponseCode !

或者,使用lambda:

p = re.compile("([A-Z]+[a-z]+)([A-Z]?)")
print p.sub(lambda x: x.group(1).lower() + "_" + x.group(2).lower(), "CamelCase")
print p.sub(lambda x: x.group(1).lower() + "_" + x.group(2).lower(), "getHTTPResponseCode")

编辑:对于像“Test”这样的情况,应该也很容易看到有改进的空间,因为下划线是无条件插入的。

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()

使用:str.capitalize()将字符串(包含在变量str中)的首字母转换为大写字母,并返回整个字符串。

例子: 命令:“你好”.capitalize () 输出:你好

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。