在Python中处理多个键错误
当我两次尝试捕捉一个键错误时,就会出现这个错误。python中是否有任何东西阻止您两次捕获相同的错误在Python中处理多个键错误,python,Python,当我两次尝试捕捉一个键错误时,就会出现这个错误。python中是否有任何东西阻止您两次捕获相同的错误 $ ./scratch.py try getting a can't get a try getting b Traceback (most recent call last): File "./scratch.py", line 13, in <module> print dict['b'] KeyError: 'b' 您需要额外尝试…除了: dict={} dict
$ ./scratch.py
try getting a
can't get a try getting b
Traceback (most recent call last):
File "./scratch.py", line 13, in <module>
print dict['b']
KeyError: 'b'
您需要额外尝试…除了:
dict={}
dict['c'] = '3'
try:
print 'try getting a'
print dict['a']
except KeyError:
print 'can\'t get a try getting b'
try:
print dict['b']
except KeyError as e:
print "Got another exception", e
except:
print 'can\'t get a or b'
我将通过一个简单的for循环来实现这一点:
>>> d = {'c': 1}
>>> keys = ['a', 'b', 'c']
>>> for key in keys:
... try:
... value = d[key]
... break
... except KeyError:
... pass
... else:
... raise KeyError('not any of keys in dict')
...
>>> value
1
>>> key
'c'
如果要在一行中完成此操作:
key,value=next((k,d[k]),如果k在d中,则k在d中)
为什么不使用dict.get
?您也没有尝试使用print dict['b']
。您好,Padriac-dict.get似乎没有抛出异常并以静默方式失败?如果找不到钥匙,我想做点什么。从逻辑上讲,我认为缩进不适合第二次尝试。因为如果dict['a']存在,就不会有错误,而且它从不尝试检查dict['b'],我相信@nev正在使用代码片段来理解python异常处理机制。我想指出的关键点是:如果您想处理异常处理程序中抛出的异常,则需要添加另一个try…except块。操作系统中有一个类似的问题:
>>> d = {'c': 1}
>>> keys = ['a', 'b', 'c']
>>> for key in keys:
... try:
... value = d[key]
... break
... except KeyError:
... pass
... else:
... raise KeyError('not any of keys in dict')
...
>>> value
1
>>> key
'c'