String 从字符串python中删除时间戳和url

我有一个字符串，我必须从中删除时间戳和标点符号。我必须删除所有的数字，除了responseCode值 在这种情况下，必须保持原样，例如400。无论400人到哪里，都不应将其移除。我想删除所有的url 文件名以tar.gz结尾

mystr="sun aug 19 13:02:09 2018 I_am.98189:  hello please connect to the local host:8080 
sun aug 19 13:02:10 2018 hey.94289:  hello not able to find the file 
sun aug 19 13:02:10 2018 I_am.94289: Base url for file_transfer is: abc/vd/filename.tar.gz 
mon aug 19 13:02:10 2018 how_94289: $var1={ 
  'responseCode' = '400', 
  'responseDate' = 'Sun, 19 Aug 2018 13:02:08 ET', 
  'responseContent' = 'ABC'  }
mon aug 20 13:02:10 2018 hello!94289: Error performing action, failed with error code [400]
"

预期结果：

"I_am hello please connect to the local host 
hello not able to find the file 
Base url for file_transfer 
var1 
  responseCode = 400 
  responseDate  
  responseContent = ABC 
Error performing action, failed with error code 400
"

我删除标点符号的解决方案：

punctuations = '''!=()-[]{};:'"\,<>.?@#$%^&*_~'''
no_punct = ""
for char in mystr:
   if char not in punctuations:
       no_punct = no_punct + char

# display the unpunctuated string
print(no_punct)

标点符号=“”=()-[]{};:'"\,.?@#$%^&*_~'''
否_punt=“”
对于mystr中的char：
如果字符不在标点符号中：
无点=无点+字符
#显示未定时的字符串
打印（无点）

也许：

patterns = [r"\w{3} \w{3} \d{2} \d{2}:\d{2}:\d{2} \d{4}\s*",    #sun aug 19 13:02:10 2018
        r"\w{3}, \d{2} \w{3} \d{4} \d{2}:\d{2}:\d{2} \w{2}\s*", #Sun, 19 Aug 2018 13:02:08 ET
        r":\s*([\da-zA_Z]+\/)+([a-zA-Z0-9\.]+)",                #URL
        r"([a-zA-Z_!]+)[\.!_]\d+:\s*",                          #word[._!]number:>=0space
        r":\d+",
        "[/':,${}\[\]]"                                         #punctuations
        ]

s = mystr

for p in patterns:
    s = re.sub(p,'', s)

s = s.strip()

print(s)

输出：

hello please connect to the local host
hello not able to find the file
Base url for file_transfer is
var1= 
  responseCode = 400 
  responseDate =  
  responseContent = ABC  
Error performing action failed with error code 400

---

嗨，我更新了问题。98189不总是在我之后。你能编辑你的答案吗？