blocks|key|4381270|text|我会将这些列表zip在一起，逐个元素地比较它们。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4381271|main_list+=+[b+for+a,+b+in+zip(list1,+list2)+if+a!=+b]|code-block|syntax|javascript|4381272|entityMap^0|7|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I would <code>zip</code> the lists together to compare them element by element.

<pre><code>main_list = [b for a, b in zip(list1, list2) if a!= b]
</code></pre>

blocks|key|4381772|text|您可以使用集合：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4381773|main_list+=+list(set(list_2)+-+set(list_1))|code-block|syntax|javascript|4381774|输出：|4381775|>>>+list_1=["a",+"b",+"c",+"d",+"e"]
>>>+list_2=["a",+"f",+"c",+"m"]
>>>+set(list_2)+-+set(list_1)
set(['m',+'f'])
>>>+list(set(list_2)+-+set(list_1))
['m',+'f']|4381776|根据@JonClements的评论，这里是一个更整洁的版本：|4381777|>>>+list_1=["a",+"b",+"c",+"d",+"e"]
>>>+list_2=["a",+"f",+"c",+"m"]
>>>+list(set(list_2).difference(list_1))
['m',+'f']|4381778|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|$]]

You can use sets:

<pre><code>main_list = list(set(list_2) - set(list_1))
</code></pre>

Output:

<pre><code>&gt;&gt;&gt; list_1=["a", "b", "c", "d", "e"]
&gt;&gt;&gt; list_2=["a", "f", "c", "m"]
&gt;&gt;&gt; set(list_2) - set(list_1)
set(['m', 'f'])
&gt;&gt;&gt; list(set(list_2) - set(list_1))
['m', 'f']
</code></pre>

Per @JonClements' comment, here is a tidier version:

<pre><code>&gt;&gt;&gt; list_1=["a", "b", "c", "d", "e"]
&gt;&gt;&gt; list_2=["a", "f", "c", "m"]
&gt;&gt;&gt; list(set(list_2).difference(list_1))
['m', 'f']
</code></pre>

blocks|key|4381785|text|main_list=[]
list_1=["a",+"b",+"c",+"d",+"e"]
list_2=["a",+"f",+"c",+"m"]

for+i+in+list_2:
++++if+i+not+in+list_1:
++++++++main_list.append(i)

print(main_list)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4381786|输出：|unstyled|4381787|['f',+'m']|4381788|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>main_list=[]
list_1=["a", "b", "c", "d", "e"]
list_2=["a", "f", "c", "m"]

for i in list_2:
 if i not in list_1:
 main_list.append(i)

print(main_list)
</code></pre>

output:

<pre><code>['f', 'm']
</code></pre>

blocks|key|4381407|text|如果应该考虑出现的次数，则可能需要使用类似于collections.Counter的内容|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4381408|list_1=["a",+"b",+"c",+"d",+"e"]
list_2=["a",+"f",+"c",+"m"]+
from+collections+import+Counter
cnt1+=+Counter(list_1)
cnt2+=+Counter(list_2)
final+=+[key+for+key,+counts+in+cnt2.items()+if+cnt1[key]+!=+counts]

>>>+final
['f',+'m']|code-block|syntax|javascript|4381409|正如承诺的那样，这也可以将不同的出现次数处理为“差异”：|4381410|list_1=["a",+"b",+"c",+"d",+"e",+'a']
cnt1+=+Counter(list_1)
cnt2+=+Counter(list_2)
final+=+[key+for+key,+counts+in+cnt2.items()+if+cnt1[key]+!=+counts]

>>>+final
['a',+'f',+'m']|4381411|entityMap^0|M|J|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

If the number of occurences should be taken into account you probably need to use something like <code>collections.Counter</code>:
<pre><code>list_1=[&quot;a&quot;, &quot;b&quot;, &quot;c&quot;, &quot;d&quot;, &quot;e&quot;]
list_2=[&quot;a&quot;, &quot;f&quot;, &quot;c&quot;, &quot;m&quot;] 
from collections import Counter
cnt1 = Counter(list_1)
cnt2 = Counter(list_2)
final = [key for key, counts in cnt2.items() if cnt1[key] != counts]

&gt;&gt;&gt; final
['f', 'm']
</code></pre>
As promised this can also handle differing number of occurences as &quot;difference&quot;:
<pre><code>list_1=[&quot;a&quot;, &quot;b&quot;, &quot;c&quot;, &quot;d&quot;, &quot;e&quot;, 'a']
cnt1 = Counter(list_1)
cnt2 = Counter(list_2)
final = [key for key, counts in cnt2.items() if cnt1[key] != counts]

&gt;&gt;&gt; final
['a', 'f', 'm']
</code></pre>

blocks|key|4381883|text|不确定为什么当你有原生方法时，上面的解释会如此复杂：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4381884|main_list+=+list(set(list_2)-set(list_1))|code-block|syntax|javascript|4381885|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|$]]

Not sure why the above explanations are so complicated when you have native methods available:

<pre><code>main_list = list(set(list_2)-set(list_1))
</code></pre>

blocks|key|4381448|text|从ser1中删除ser2中存在的项目。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4381449|输入|4381450|ser1+=+pd.Series(1，2，3，4，5)+ser2+=+pd.Series(4，5，6，7，8)|4381451|解决方案|4381452|ser1~ser1.isin(ser2)|4381453|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|O|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

From ser1 remove items present in ser2.

<h1>Input</h1>

ser1 = pd.Series([1, 2, 3, 4, 5])
ser2 = pd.Series([4, 5, 6, 7, 8])

<h1>Solution</h1>

ser1[~ser1.isin(ser2)]

blocks|key|4381492|text|我使用了两种方法，我发现其中一种方法比其他方法更有用。这是我的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4381493|我的输入数据：|4381494|crkmod_mpp+=+['M13','M18','M19','M24']
testmod_mpp+=+['M13','M14','M15','M16','M17','M18','M19','M20','M21','M22','M23','M24']|code-block|syntax|javascript|4381495|Method1：np.setdiff1d我喜欢这种方法，因为它保留了位置|offset|length|style|CODE|4381496|test=+list(np.setdiff1d(testmod_mpp,crkmod_mpp))
print(test)
['M15',+'M16',+'M22',+'M23',+'M20',+'M14',+'M17',+'M21']|4381497|Method2:虽然它给出了与Method1相同的答案，但扰乱了秩序|4381498|test+=+list(set(testmod_mpp).difference(set(crkmod_mpp)))
print(test)
['POA23',+'POA15',+'POA17',+'POA16',+'POA22',+'POA18',+'POA24',+'POA21']|4381499|Method1+np.setdiff1d完美地满足了我的需求。此答案用于获取信息。|4381500|entityMap^0|0|0|0|8|C|0|0|0|0|8|C|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@$K|12|L|13|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|17|8|@$K|18|L|19|M|N]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|X|$]]

I used two methods and I found one method useful over other. Here is my answer: 

My input data: 

<pre><code>crkmod_mpp = ['M13','M18','M19','M24']
testmod_mpp = ['M13','M14','M15','M16','M17','M18','M19','M20','M21','M22','M23','M24']
</code></pre>

Method1: <code>np.setdiff1d</code> I like this approach over other because it preserves the position

<pre><code>test= list(np.setdiff1d(testmod_mpp,crkmod_mpp))
print(test)
['M15', 'M16', 'M22', 'M23', 'M20', 'M14', 'M17', 'M21']
</code></pre>

Method2: Though it gives same answer as in Method1 but disturbs the order 

<pre><code>test = list(set(testmod_mpp).difference(set(crkmod_mpp)))
print(test)
['POA23', 'POA15', 'POA17', 'POA16', 'POA22', 'POA18', 'POA24', 'POA21']
</code></pre>

Method1 <code>np.setdiff1d</code> meets my requirements perfectly. 
This answer for information.

I need to compare two lists in order to create a new list of specific elements found in one list but not in the other. For example:

<pre><code>main_list=[]
list_1=["a", "b", "c", "d", "e"]
list_2=["a", "f", "c", "m"] 
</code></pre>

I want to loop through list_1 and append to main_list all the elements from list_2 that are not found in list_1.

The result should be:

<pre><code>main_list=["f", "m"]
</code></pre>

How can I do it with python?

Python find elements in one list that are not in the other

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。

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我需要比较两个列表，以便创建在一个列表中找到但在另一个列表中找不到的特定元素的新列表。例如：main_list=[]list_1=["a", "b", "c", "d", "e"]list_2=["a", "f", "c", "m"] 我想遍历list_1，并将list_2中在list_1中找不到的所有元素附加到mai...

问Python在一个列表中查找不在另一个列表中的元素
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