如果你的目标是以最少的时间复杂度完成任务，也许利用Python内置方法比依赖正则表达式更有效：

str_1 = "Www.GooGle.com"
print(str_1.swapcase())

使用随机生成的长度为1000个字符的mixalpha字符串进行5000次迭代的效率比较：

import timeit
import re
import random
import string

def swapLowerUpper_regex(source):
    return re.sub(r'([A-Z]+)|([a-z]+)', lambda x: x.group(1).lower() if x.group(1) else x.group(2).upper(), source)

def swapLowerUpper_swapcase(source):
    return source.swapcase()

# Generate a random mixalpha string with length 1000
word = ''.join(random.choices(string.ascii_letters, k=1000))

# Define the number of iterations to run
num_iterations = 5000

# Time the execution of the swapLowerUpper() function using regular expressions
elapsed_times_regex = []
for i in range(num_iterations):
    start_time = timeit.default_timer()
    swapLowerUpper_regex(word)
    elapsed_time_regex = timeit.default_timer() - start_time
    elapsed_times_regex.append(elapsed_time_regex)

# Time the execution of the swapcase() method
elapsed_times_swapcase = []
for i in range(num_iterations):
    start_time = timeit.default_timer()
    swapLowerUpper_swapcase(word)
    elapsed_time_swapcase = timeit.default_timer() - start_time
    elapsed_times_swapcase.append(elapsed_time_swapcase)

# Compute the average elapsed time of each method
avg_elapsed_time_regex = sum(elapsed_times_regex) / len(elapsed_times_regex)
avg_elapsed_time_swapcase = sum(elapsed_times_swapcase) / len(elapsed_times_swapcase)

# Print the average elapsed times
print(f"Average elapsed time using swapLowerUpper_regex(): {avg_elapsed_time_regex:.10f} seconds")
print(f"Average elapsed time using swapLowerUpper_swapcase(): {avg_elapsed_time_swapcase:.10f} seconds")

输出：

Average elapsed time using swapLowerUpper_regex(): 0.0002279637 seconds
Average elapsed time using swapLowerUpper_swapcase(): 0.0000108802 seconds

检查x.group(1)或x.group(2)是否匹配，并返回相应的替换项。

def swapLowerUpper(source):
    return re.sub(r'([A-Z]+)|([a-z]+)', lambda x: x.group(1).lower() if x.group(1) else x.group(2).upper(), source)

尝试：

import re

str_1 = "Www.GooGle.com"

str_1 = re.sub(
    r"([a-z]+)|([A-Z]+)",
    lambda g: g.group(1).upper() if g.group(1) else g.group(2).lower(),
    str_1,
)
print(str_1)

图纸：

wWW.gOOgLE.COM