python - 如何使复杂的 if 条件可维护，因为 Python 不支持 if 条件中的赋值？

Question

我有以下 Python 代码，其中函数 print_processed_username 包含一个 if/else 构造。由于重复的正则表达式，if 行很长。如果需要对正则表达式进行修改，则必须在该正则表达式的每次出现（包括对 process_the_username 的调用）中进行相同的修改，这使得代码难以维护。

import re

def process_the_username(username):
    return 'e' + username[1:]

def print_processed_username(args):
    if len(args) == 1 and type(args[0]) is str and re.compile(r'[\'"]username[\'"]: ?[\'"](\S*)[\'"]', re.IGNORECASE).search(args[0]) and len(re.compile(r'[\'"]username[\'"]: ?[\'"](\S*)[\'"]', re.IGNORECASE).search(args[0]).groups()) == 1 and len(re.compile(r'[\'"]username[\'"]: ?[\'"](\S*)[\'"]', re.IGNORECASE).search(args[0]).groups()[0]) == 7 and re.compile(r'[\'"]username[\'"]: ?[\'"](\S*)[\'"]', re.IGNORECASE).search(args[0]).groups()[0][0] == '_':
        # Here args is a list containing one item which is a string and the string contains 'username': '<user>' only once where <user> is 7 characters long and starts with _.
        print process_the_username(re.compile(r'[\'"]username[\'"]: ?[\'"](\S*)[\'"]', re.IGNORECASE).search(args[0]).groups()[0])
    else:
        print "Missing or correct user name format. Nothing to do."

如果 Python 像许多其他语言一样支持 if 条件中的赋值，那么这个问题将很容易解决。但正如我们所知，Python 不支持这一点。

因此，我正在寻求有关如何以 Pythonic 方式编写 if 条件的建议，其中消除了正则表达式的重复。高度赞赏所有使代码更简单和更易于维护的建议。

以下是一些按预期处理用户名的示例执行。

>>> args = ["'location': 'Frankfurt', 'Phone': '[phone]', 'UserName': '_beka01'"]
>>> print_processed_username(args)
ebeka01
>>> 
>>> args = ["'UserName': '_beka01', 'location': 'Frankfurt', 'Phone': '[phone]'"]
>>> print_processed_username(args)
ebeka01
>>> 
>>> args = ["'UserName': '_beka01'"]
>>> print_processed_username(args)
ebeka01
>>> 
>>> args = ["'USERNAME': '_beka01'"]
>>> print_processed_username(args)
ebeka01
>>> 
>>> args = ['"location":"Frankfurt", "Phone":"[phone]", "UserName":"_beka01"']
>>> print_processed_username(args)
ebeka01
>>> 
>>> args = ['"location":"Frankfurt","Phone":"[phone]","UserName":"_beka01"']
>>> print_processed_username(args)
ebeka01
>>>

以下是一些未按预期处理用户名的示例执行。

>>> args = ["'location': 'Frankfurt', 'Phone': '[phone]', 'UserName': 'abeka01'"]
>>> print_processed_username(args)
Missing or correct user name format. Nothing to do.
>>> 
>>> args = ["'location': 'Frankfurt', 'Phone': '[phone]'"]
>>> print_processed_username(args)
Missing or correct user name format. Nothing to do.
>>> 
>>> args = ["'UserName': '_beka0132'"]
>>> print_processed_username(args)
Missing or correct user name format. Nothing to do.
>>> 

标签： python

Answer 1

第 1 步：编译一次正则表达式并将其保存在变量中。它没有变化，因此在调用该方法之前提前执行此操作。

username_regex = re.compile(r'[\'"]username[\'"]: ?[\'"](\S*)[\'"]', re.IGNORECASE)

def print_processed_username(args):
    if len(args) == 1 and type(args[0]) is str and username_regex.search(args[0]) and len(username_regex.search(args[0]).groups()) == 1 and len(username_regex.search(args[0]).groups()[0]) == 7 and username_regex.search(args[0]).groups()[0][0] == '_':
        print process_the_username(username_regex.search(args[0]).groups()[0])
    else:
        print "Missing or correct user name format. Nothing to do."

第 2 步：消除对 的重复调用search()。

username_regex = re.compile(r'[\'"]username[\'"]: ?[\'"](\S*)[\'"]', re.IGNORECASE)

def print_processed_username(args):
    if len(args) != 1 or type(args[0]) is not str:
        print "Missing or correct user name format. Nothing to do."
        return

    result = username_regex.search(args[0])

    if result and len(result.groups()) == 1 and len(result.groups()[0]) == 7 and result.groups()[0][0] == '_':
        print process_the_username(result.groups()[0])
    else:
        print "Missing or correct user name format. Nothing to do."

第 3 步：将用户名保存在变量中。

username_regex = re.compile(r'[\'"]username[\'"]: ?[\'"](\S*)[\'"]', re.IGNORECASE)

def print_processed_username(args):
    if len(args) != 1 or type(args[0]) is not str:
        print "Missing or correct user name format. Nothing to do."
        return

    result = username_regex.search(args[0])

    if not result or len(result.groups()) != 1:
        print "Missing or correct user name format. Nothing to do."
        return

    username = result.groups()[0]

    if len(username) == 7 and username[0] == '_':
        print process_the_username(username)
    else:
        print "Missing or correct user name format. Nothing to do."

第 4 步：从处理结果的代码中提取字符串解析。编写一个纯粹解析字符串并将结果留给调用者的解析器。

username_regex = re.compile(r'[\'"]username[\'"]: ?[\'"](\S*)[\'"]', re.IGNORECASE)

def parse_username(args):
    if len(args) != 1 or type(args[0]) is not str: return None

    result = username_regex.search(args[0])
    if not result or len(result.groups()) != 1: return None

    username = result.groups()[0]
    if len(username) != 7 or username[0] != '_': return None

    return username

def print_processed_username(args):
    username = parse_username(args)

    if username:
        print process_the_username(username)
    else:
        print "Missing or correct user name format. Nothing to do."