blocks|key|597874|text|我完全同意关键错误注释。您也可以使用dictionary's+get()方法来避免异常。这也可以用来提供默认路径，而不是None，如下所示。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|597875|>>>+d+=+{"a":1,+"b":2}
>>>+x+=+d.get("A",None)
>>>+print+x
None|code-block|syntax|javascript|597876|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/stdtypes.html#dict^0|1O|4|I|I|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

I fully agree with the Key error comments. You could also use the <a href="http://docs.python.org/library/stdtypes.html#dict">dictionary's get()</a> method as well to avoid the exceptions. This could also be used to give a default path rather than <code>None</code> as shown below.

<pre><code>&gt;&gt;&gt; d = {"a":1, "b":2}
&gt;&gt;&gt; x = d.get("A",None)
&gt;&gt;&gt; print x
None
</code></pre>

blocks|key|597938|text|是的，这很可能是由不存在的key引起的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|597939|在我的程序中，出于效率考虑，我使用了setdefault来静音这个错误。取决于这条线路的效率|597940|>>>'a'+in+mydict.keys()++|code-block|syntax|javascript|597941|我也是Python的新手。事实上，我今天才学到的。所以请原谅我对效率的忽视。|597942|在Python+3中，您也可以使用此函数，|597943|get(key[,+default])+[function+doc][1]|597944|据说它永远不会引发键错误。|597945|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|X|8|@]|9|@]|A|$G|H]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Yes, it is most likely caused by non-exsistent key.

In my program, I used setdefault to mute this error, for efficiency concern. 
depending on how efficient is this line 

<pre><code>&gt;&gt;&gt;'a' in mydict.keys() 
</code></pre>

I am new to Python too. In fact I have just learned it today. So forgive me on the ignorance of efficiency.

In Python 3, you can also use this function,

<pre><code>get(key[, default]) [function doc][1]
</code></pre>

It is said that it will never raise a key error.

blocks|key|597985|text|这意味着您的数组缺少您要查找的键。我使用一个函数来处理这个问题，该函数要么返回值(如果存在)，要么返回默认值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|597986|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')|code-block|syntax|javascript|597987|输出：|597988|1
2
#default|597989|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This means your array is missing the key you're looking for. I handle this with a function which either returns the value if it exists or it returns a default value instead.

<pre><code>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')
</code></pre>

Output:

<pre><code>1
2
#default
</code></pre>

blocks|key|597998|text|对于dict，只需使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|597999|if+key+in+dict|offset|length|style|CODE|598000|并且不使用关键字列表中的搜索|598001|if+key+in+dict.keys()|598002|后者将更加耗时。|598003|entityMap^0|0|0|E|0|0|0|L|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@$D|R|E|S|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|U|8|@$D|V|E|W|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|X|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|O|$]]

For dict, just use
<code>if key in dict</code>
and don't use searching in key list
<code>if key in dict.keys()</code>
The latter will be more time-consuming.

blocks|key|598034|text|我在使用嵌套for解析dict时收到此错误|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|598035|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])|code-block|syntax|javascript|598036|回溯：|598037|Traceback+(most+recent+call+last):
++++++File+"<input>",+line+3,+in+<module>
++++KeyError:+'K'|598038|因为在第二个循环中应该是cats[cat]而不是cat+(这只是一个键)|598039|所以：|598040|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])|598041|给出|598042|black
10
white
8|598043|entityMap^0|6|3|B|4|0|0|0|0|C|9|O|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1D|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|1E|8|@]|D|@]|E|$I|J]]|$1|U|3|V|5|6|7|1F|8|@]|D|@]|E|$]]|$1|W|3|X|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|Y|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|Z|$]]

I received this error when I was parsing <code>dict</code> with nested <code>for</code>:

<pre><code>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])
</code></pre>

Traceback:

<pre><code>Traceback (most recent call last):
 File "&lt;input&gt;", line 3, in &lt;module&gt;
 KeyError: 'K'
</code></pre>

Because in second loop should be <code>cats[cat]</code> instead just <code>cat</code> (what is just a key)

So:

<pre><code>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])
</code></pre>

Gives

<pre><code>black
10
white
8
</code></pre>

blocks|key|598082|text|如果您使用的是Python+3，那么让我们简单地说明一下|type|unstyled|depth|inlineStyleRanges|entityRanges|data|598083|mydict+=+{'a':'apple','b':'boy','c':'cat'}
check+=+'c'+in+mydict
if+check:
++++print('c+key+is+present')|code-block|syntax|javascript|598084|如果您需要else条件|598085|mydict+=+{'a':'apple','b':'boy','c':'cat'}
if+'c'+in+mydict:
++++print('key+present')
else:
++++print('key+not+found')|598086|对于动态键值，您还可以通过try-exception块进行处理|598087|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'}|598088|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Let us make it simple if you're using Python 3
<pre><code>mydict = {'a':'apple','b':'boy','c':'cat'}
check = 'c' in mydict
if check:
 print('c key is present')
</code></pre>
If you need else condition
<pre><code>mydict = {'a':'apple','b':'boy','c':'cat'}
if 'c' in mydict:
 print('key present')
else:
 print('key not found')
</code></pre>
For the dynamic key value, you can also handle through try-exception block
<pre><code>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'}
</code></pre>

In my python program I am getting this error:

<pre><code>KeyError: 'variablename'
</code></pre>

From this code:

<pre><code>path = meta_entry['path'].strip('/'),
</code></pre>

Can anyone please explain why this is happening?

I'm getting Key error in python

Python

在我的python程序中，我得到了这个错误：KeyError: 'variablename'从下面的代码：path = meta_entry['path'].strip('/'),有人能解释一下为什么会发生这种情况吗？

问我在python中遇到键错误
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问我在python中遇到键错误EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问我在python中遇到键错误
EN