blocks|key|2446929|text|如果您已经有了字典，那么您将列出所有的键：key_list+=+yourDic.keys()。使用该列表，您可以循环遍历您的字典。如果您知道其中一个值，这就更容易了，但是下面我假设您不知道。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2446930|yourDic+=+{'a':1,+'b':4,+'c':1,+'d':1,+}
key_list+=+yourDic.keys()
previous_value+=+yourDic[key_list[0]]+#+Making+it+so+loop+gets+past+first+test
count+=+0
for+key+in+key_list:
++++test_value+=+yourDic[key]
++++if+(test_value+!=+previous_value)+and+count+==+1:+#+Checks+first+key
++++++++print+key_list[count+-+1]
++++++++break
++++elif+(test_value+!=+previous_value):
++++++++print+key
++++++++break
++++else:
++++++++previous_value+=+test_value
++++++++count+%2B=+1|code-block|syntax|javascript|2446931|因此，一旦您找到不同的值，它就会打印密钥。如果希望它也打印值，则只需要一个print+test_value语句。|2446932|entityMap^0|L|P|0|0|11|G|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

If you already have your dictionary, then you make a list of all of the keys: <code>key_list = yourDic.keys()</code>. Using that list, you can then loop through your dictionary. This is easier if you know one of the values, but below I assume that you do not.

<pre><code>yourDic = {'a':1, 'b':4, 'c':1, 'd':1, }
key_list = yourDic.keys()
previous_value = yourDic[key_list[0]] # Making it so loop gets past first test
count = 0
for key in key_list:
 test_value = yourDic[key]
 if (test_value != previous_value) and count == 1: # Checks first key
 print key_list[count - 1]
 break
 elif (test_value != previous_value):
 print key
 break
 else:
 previous_value = test_value
 count += 1
</code></pre>

So, once you find the value that is different, it will print the key. If you want it to print the value, too, you just need a <code>print test_value</code> statement

blocks|key|2446947|text|如果您有一个字典，您可以快速地检查并找到第一个值，这与接下来的两个值循环在字典的键之间是不同的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2446948|下面是一个例子：|2446949|def+find_different(d):
++++k+=+d.keys()
++++for+i+in+xrange(0,+len(k)):
++++++++if+d[k[i]]+!=+d[k[(i%2B1)%25len(k)]]+and+d[k[i]]+!=+d[k[(i%2B2)%25len(k)]]:
++++++++++++return+k[i]

>>>+mydict+=+{'a':1,+'b':1,+'c':2,+'d':1}
>>>+find_different(mydict)
'c'|code-block|syntax|javascript|2446950|否则，如果您拥有的是一个单键字典列表，那么您可以很好地用一个函数来映射您的列表，该函数从元素中“提取”值，然后使用相同的逻辑检查每个元素。|2446951|下面是另一个工作示例：|2446952|def+find_different(l):
++++mask+=+map(lambda+x:+x[x.keys()[0]],+l)
++++for+i+in+xrange(0,+len(l)):
++++++++if+mask[i]+!=+mask[(i%2B1)%25len(l)]+and+mask[i]+!=+mask[(i%2B2)%25len(l)]:
++++++++++++return+l[i].keys()[0]

>>>+mylist+=+[{'a':1},{'b':1},{'c':2},{'d':1}]
>>>+find_different(mylist)
'c'|2446953|注意到：这些解决方案在Python3中不起作用，因为map函数不返回list，字典的.keys()方法也不返回。|offset|length|style|BOLD|CODE|2446954|entityMap^0|0|0|0|0|0|0|0|3|Q|3|Y|4|16|7|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|10|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|11|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|O|3|P|5|6|7|13|8|@$Q|14|R|15|S|T]|$Q|16|R|17|S|U]|$Q|18|R|19|S|U]|$Q|1A|R|1B|S|U]]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|W|$]]

If you have a dictionary you can quickly check and find the first value which is different from the next two values cycling around the keys of your dictionary.

Here's an example:

<pre><code>def find_different(d):
 k = d.keys()
 for i in xrange(0, len(k)):
 if d[k[i]] != d[k[(i+1)%len(k)]] and d[k[i]] != d[k[(i+2)%len(k)]]:
 return k[i]

&gt;&gt;&gt; mydict = {'a':1, 'b':1, 'c':2, 'd':1}
&gt;&gt;&gt; find_different(mydict)
'c'
</code></pre>

Otherwise, if what you have is a list of single-key dictionaries, then you can do it quite nicely mapping your list with a function which "extracts" the values from your elements, then check each one using the same logic.

Here's another working example:

<pre><code>def find_different(l):
 mask = map(lambda x: x[x.keys()[0]], l)
 for i in xrange(0, len(l)):
 if mask[i] != mask[(i+1)%len(l)] and mask[i] != mask[(i+2)%len(l)]:
 return l[i].keys()[0]

&gt;&gt;&gt; mylist = [{'a':1},{'b':1},{'c':2},{'d':1}]
&gt;&gt;&gt; find_different(mylist)
'c'
</code></pre>

NOTE: these solutions do not work in Python 3 as the <code>map</code> function doesn't return a <code>list</code> and neither does the <code>.keys()</code> method of dictionaries.

blocks|key|665911|text|假设您的“配对列表”(实际上是字典列表，叹息)无法更改：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|665912|from+collections+import+defaultdict

def+get_pair(d):
++++return+(d.keys()[0],+d.values()[0])

def+extract_unique(l):
++++d+=+defaultdict(list)
++++for+key,+value+in+map(get_pair,+l):
++++++++d[value].append(key)
++++return+filter(lambda+(v,l):+len(l)+==+1,+d.items())[0][1]|code-block|syntax|javascript|665913|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Assuming that your "list of pairs" (actually list of dictionaries, sigh) cannot be changed:

<pre><code>from collections import defaultdict

def get_pair(d):
 return (d.keys()[0], d.values()[0])

def extract_unique(l):
 d = defaultdict(list)
 for key, value in map(get_pair, l):
 d[value].append(key)
 return filter(lambda (v,l): len(l) == 1, d.items())[0][1]
</code></pre>

I have a python dictionary containing n key-value pairs, out of which n-1 values are identical and 1 is not. I need to find the key of the distinct element.

For example: consider a python list <code>[{a:1},{b:1},{c:2},{d:1}]</code>. I need the to get <code>'c'</code> as the output.

I can use a for loop to compare consecutive elements and then use two more for loops to compare those elements with the other elements. But is there a more efficient way to go about it or perhaps a built-in function which I am unaware of?

Finding if there are distinct elements in a python dictionary

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我有一个包含n个键值对的python字典，其中n-1值是相同的，1不是。我需要找到不同元素的关键。例如:考虑一个python [{a:1},{b:1},{c:2},{d:1}]。我需要'c'作为输出。我可以使用for循环来比较连续的元素，然后再使用两个for循环来比较这些元素和其他元素。但是，是否有更有效的方法来解决这...

问查找python字典中是否有不同的元素
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问查找python字典中是否有不同的元素EN