在我的假日方案中,我正在发现这一错误:
KeyError: variablename
根据该法典:
path = meta_entry[ path ].strip( / ),
谁能解释为什么发生这种情况?
在我的假日方案中,我正在发现这一错误:
KeyError: variablename
根据该法典:
path = meta_entry[ path ].strip( / ),
谁能解释为什么发生这种情况?
A KeyError
通常是指存在关键吨数。 因此,您是否相信<代码>path 关键?
摘自官方书目:
http://www.ohchr.org。
Raised when a mapping (dictionary) key is not found in the set of existing keys.
例如:
>>> mydict = { a : 1 , b : 2 }
>>> mydict[ a ]
1
>>> mydict[ c ]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
KeyError: c
>>>
因此,试图印刷<代码>meta_enter/code>的内容,并检查是否存在<代码>path<>/code>。
>>> mydict = { a : 1 , b : 2 }
>>> print mydict
{ a : 1 , b : 2 }
或者,你可以做到:
>>> a in mydict
True
>>> c in mydict
False
我完全赞同关键错误评论。 也可使用方法,并避免例外情况。 如下文所示,这也可以用来提供一条缺省道路,而不是<代码>None。
>>> d = {"a":1, "b":2}
>>> x = d.get("A",None)
>>> print x
None
字典,公正使用
在关键清单中不使用搜索
<代码>关键词()
后者将耗费更多的时间。
是的,这很可能是由非重要因素造成的。
In my program, I used setdefault to mute this error, for efficiency concern. depending on how efficient is this line
>>> a in mydict.keys()
I am new to Python too. In fact I have just learned it today. So forgive me on the ignorance of efficiency.
在Adhur3,你也可以使用这一功能。
get(key[, default]) [function doc][1]
据称,这永远不会造成重大错误。
如果你重新使用3枚空洞,那就让我们简单化。
mydict = { a : apple , b : boy , c : cat }
check = c in mydict
if check:
print( c key is present )
如果需要,
mydict = { a : apple , b : boy , c : cat }
if c in mydict:
print( key present )
else:
print( key not found )
为了具有活力的关键价值,你也可以通过审判例外处理。
mydict = { a : apple , b : boy , c : cat }
try:
print(mydict[ c ])
except KeyError:
print( key value not found )
mydict = { a : apple , b : boy , c : cat }
我收到这一错误,当时我对<条码>的编码<>/条码>附后<条码>。
cats = { Tom : { color : white , weight : 8}, Klakier : { color : black , weight : 10}}
cat_attr = {}
for cat in cats:
for attr in cat:
print(cats[cat][attr])
追查:
Traceback (most recent call last):
File "<input>", line 3, in <module>
KeyError: K
由于第二版应为<条码>目录[目录]编码>,而仅限<条码>。 (只有钥匙)
因此:
cats = { Tom : { color : white , weight : 8}, Klakier : { color : black , weight : 10}}
cat_attr = {}
for cat in cats:
for attr in cats[cat]:
print(cats[cat][attr])
Gives
black
10
white
8
这意味着你的独裁者没有你所期待的关键。 我处理这一职能时,如果存在这种价值,或者收回一个违约价值,就能够收回价值。
def keyCheck(key, arr, default):
if key in arr.keys():
return arr[key]
else:
return default
myarray = { key1 :1, key2 :2}
print keyCheck( key1 , myarray, #default )
print keyCheck( key2 , myarray, #default )
print keyCheck( key3 , myarray, #default )
产出:
1
2
#default
例如,如果是:
ouloulou={
1:US,
2:BR,
3:FR
}
ouloulou[1]()
它的工作是完美的,but 如果你使用:
ouloulou[input("select 1 2 or 3"]()
由于你的投入回馈表明1点,因此它不工作。 因此,你需要使用<>int()。
ouloulou[int(input("select 1 2 or 3"))]()
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