Question

import random
import re

requestResponseList = [
[r'.*to Paris ?(.*)',
 ["We apologise, but all our services to Paris {0} have been cancelled!"],
 ["We are sorry, but all our flights on {0} to Paris are fully booked!"],
 ["There are no available flights {0} to Paris! Please accept our apologies."]
 ]
]


def analyze(statement):
 for pattern, responses in requestResponseList:
    match = re.match(pattern, statement.rstrip(".!?"))
    if match:
        response = random.choice(responses)
        if match.group(1):
            return response.format(*[(match.group(1))])
        return response.format(*[""])


def main():
statement = "What would be the first available flight to Paris next year?"
print(analyze(statement))

if __name__ == "__main__":
main()

这会返回太多值以解包（预期2）错误。谷歌搜索解决方案后，我更换了

for pattern, responses in requestResponseList :

用这个：

for pattern, responses in enumerate(requestResponseList):

这仍会产生错误：TypeError：第一个参数必须是字符串或编译模式。

奇怪的是，这个代码（枚举省略）一周前。这可能是Python版本的问题吗？

Answer 1

列表构造不正确 - 您有三个内部包含单个字符串的子列表，但它必须是一个包含三个字符串的子列表

requestResponseList = [
  [r'.*to Paris ?(.*)',
    [
      "We apologise, but all our services to Paris {0} have been cancelled!",
      "We are sorry, but all our flights on {0} to Paris are fully booked!",
      "There are no available flights {0} to Paris! Please accept our apologies."
    ]
  ]
]

现在它适用于我。

如果没有*[...]

，我不知道你使用*[...]的原因

        if match.group(1):
            return response.format( match.group(1) )
        return response.format( "" )

解压缩列表列表的值太多

1 个答案: