Ter*_*row 3 python dictionary list
我有以下代码:
TYPES = {'hotmail':{'type':'hotmail', 'lookup':'mixed', 'dkim': 'no', 'signatures':['|S|Return-Path: postmaster@hotmail.com','|R|^Return-Path:\s*[^@]+@(?:hot|msn)','^Received: from .*hotmail.com$']},
'gmail':{'type':'gmail', 'lookup':'mixed', 'dkim': 'yes', 'signatures':['|S|Subject: unsubscribe','','','']}
}
for type_key, type in TYPES.iteritems():
for sub_type_key, sub_type in type.iteritems():
for sig in sub_type['signatures']:
if ("|S|" in sig):
#String based matching
clean_sig = sig[3:len(sig)]
if (clean_sig in file_contents):
sig_match += 1
elif ("|R|" in sig):
clean_sig = sig[3:len(sig)]
#REGMATCH later
if (sig_match == sig.count):
return sub_type['type']
return None
Run Code Online (Sandbox Code Playgroud)
但是,它会生成错误:
for sig in sub_type['signatures']:
TypeError: string indices must be integers, not str
Run Code Online (Sandbox Code Playgroud)
我假设它会看到列表从字典元素中拉出来,并允许我循环它?
Python新手是一个新手:(
for type_key, type in TYPES.iteritems():
for sub_type_key, sub_type in type.iteritems():
for sig in sub_type['signatures']:
Run Code Online (Sandbox Code Playgroud)
应该:
for type_key, type in TYPES.iteritems():
for sig in type['signatures']:
Run Code Online (Sandbox Code Playgroud)
但是'type'在这种情况下是一个糟糕的名字选择...你不想暗示内置.
基本上,'type_key'具有名称('hotmail'或'gmail'),'type'具有与该键相关联的值的字典.所以输入['signatures']就是你想要的.
此外,您可能不需要在嵌套字典中包含"gmail"; 只返回'type_key'而不是type['type'].
将它们整合在一起,也许这会更好:(警告:未经测试)
providers = {
'hotmail':{
'type':'hotmail',
'lookup':'mixed',
'dkim': 'no',
'signatures':[
'|S|Return-Path: postmaster@hotmail.com',
'|R|^Return-Path:\s*[^@]+@(?:hot|msn)',
'^Received: from .*hotmail.com$']
},
'gmail':{
'type':'gmail',
'lookup':'mixed',
'dkim': 'yes',
'signatures':['|S|Subject: unsubscribe','','','']
}
}
for provider, provider_info in providers.iteritems():
for sig in provicer_info['signatures']:
if ("|S|" in sig):
#String based matching
clean_sig = sig[3:len(sig)]
if (clean_sig in file_contents):
sig_match += 1
elif ("|R|" in sig):
clean_sig = sig[3:len(sig)]
#REGMATCH later
if (sig_match == sig.count):
return provider
return None
Run Code Online (Sandbox Code Playgroud)