blocks|key|1177319|text|使用列表理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177320|[item+for+item+in+x+if+item+not+in+y]|code-block|syntax|javascript|1177321|如果你想使用-中缀语法，你可以这样做：|offset|length|style|CODE|1177322|class+MyList(list):
++++def+__init__(self,+*args):
++++++++super(MyList,+self).__init__(args)

++++def+__sub__(self,+other):
++++++++return+self.__class__(*[item+for+item+in+self+if+item+not+in+other])|1177323|然后，您可以像这样使用它：|1177324|x+=+MyList(1,+2,+3,+4)
y+=+MyList(2,+5,+2)
z+=+x+-+y+++|1177325|但是，如果您不是绝对需要列表属性(例如，排序)，则只需按照其他答案建议的方式使用集合。|1177326|entityMap^0|0|0|6|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Use a list comprehension:

<pre><code>[item for item in x if item not in y]
</code></pre>

If you want to use the <code>-</code> infix syntax, you can just do:

<pre><code>class MyList(list):
 def __init__(self, *args):
 super(MyList, self).__init__(args)

 def __sub__(self, other):
 return self.__class__(*[item for item in self if item not in other])
</code></pre>

you can then use it like: 

<pre><code>x = MyList(1, 2, 3, 4)
y = MyList(2, 5, 2)
z = x - y 
</code></pre>

But if you don't absolutely need list properties (for example, ordering), just use sets as the other answers recommend.

blocks|key|962988|text|使用set+difference|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|962989|>>>+z+=+list(set(x)+-+set(y))
>>>+z
[0,+8,+2,+4,+6]|code-block|syntax|javascript|962990|或者你可以只设置x和y，这样你就不需要做任何转换了。|962991|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/stdtypes.html#set.difference^0|2|E|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

Use <a href="http://docs.python.org/library/stdtypes.html#set.difference" rel="noreferrer">set difference</a>

<pre><code>&gt;&gt;&gt; z = list(set(x) - set(y))
&gt;&gt;&gt; z
[0, 8, 2, 4, 6]
</code></pre>

Or you might just have x and y be sets so you don't have to do any conversions.

blocks|key|1177376|text|如果重复和排序项目有问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177377|[i+for+i+in+a+if+not+i+in+b+or+b.remove(i)]|offset|length|style|CODE|1177378|a+=+[1,2,3,3,3,3,4]
b+=+[1,3]
result:+[2,+3,+3,+3,+4]|code-block|syntax|javascript|1177379|entityMap^0|0|0|17|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

if duplicate and ordering items are problem :

<code>[i for i in a if not i in b or b.remove(i)]</code>

<pre><code>a = [1,2,3,3,3,3,4]
b = [1,3]
result: [2, 3, 3, 3, 4]
</code></pre>

blocks|key|963035|text|这是一个“集合减法”操作。为此，请使用set数据结构。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|963036|在Python+2.7中：|963037|x+=+{1,2,3,4,5,6,7,8,9,0}
y+=+{1,3,5,7,9}
print+x+-+y|code-block|syntax|javascript|963038|输出：|963039|>>>+print+x+-+y
set([0,+8,+2,+4,+6])|963040|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

That is a "set subtraction" operation. Use the set data structure for that.

In Python 2.7:

<pre><code>x = {1,2,3,4,5,6,7,8,9,0}
y = {1,3,5,7,9}
print x - y
</code></pre>

Output:

<pre><code>&gt;&gt;&gt; print x - y
set([0, 8, 2, 4, 6])
</code></pre>

blocks|key|963341|text|如果列表允许重复的元素，您可以使用集合中的计数器：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|963342|from+collections+import+Counter
result+=+list((Counter(x)-Counter(y)).elements())|code-block|syntax|javascript|963343|如果需要保留x中元素的顺序：|963344|result+=+[+v+for+c+in+[Counter(y)]+for+v+in+x+if+not+c[v]+or+c.subtract([v])+]|963345|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|$]]

If the lists allow duplicate elements, you can use Counter from collections:

<pre><code>from collections import Counter
result = list((Counter(x)-Counter(y)).elements())
</code></pre>

If you need to preserve the order of elements from x:

<pre><code>result = [ v for c in [Counter(y)] for v in x if not c[v] or c.subtract([v]) ]
</code></pre>

blocks|key|1177618|text|我认为实现这一点的最简单方法是使用set()。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177619|>>>+x+=+[1,2,3,4,5,6,7,8,9,0]++
>>>+y+=+[1,3,5,7,9]++
>>>+list(set(x)-+set(y))
[0,+2,+4,+6,+8]|code-block|syntax|javascript|1177620|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|$]]

I think the easiest way to achieve this is by using set().

<pre><code>&gt;&gt;&gt; x = [1,2,3,4,5,6,7,8,9,0] 
&gt;&gt;&gt; y = [1,3,5,7,9] 
&gt;&gt;&gt; list(set(x)- set(y))
[0, 2, 4, 6, 8]
</code></pre>

blocks|key|1177420|text|在集合中查找值比在列表中查找它们更快：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177421|[item+for+item+in+x+if+item+not+in+set(y)]|code-block|syntax|javascript|1177422|我相信这将比以下扩展稍微好一点：|1177423|[item+for+item+in+x+if+item+not+in+y]|1177424|两者都保留了列表的顺序。|1177425|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Looking up values in sets are faster than looking them up in lists:

<pre><code>[item for item in x if item not in set(y)]
</code></pre>

I believe this will scale slightly better than:

<pre><code>[item for item in x if item not in y]
</code></pre>

Both preserve the order of the lists.

blocks|key|963108|text|尝尝这个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|963109|def+subtract_lists(a,+b):
++++"""+Subtracts+two+lists.+Throws+ValueError+if+b+contains+items+not+in+a+"""
++++#+Terminate+if+b+is+empty,+otherwise+remove+b[0]+from+a+and+recurse
++++return+a+if+len(b)+==+0+else+[a[:i]+%2B+subtract_lists(a[i%2B1:],+b[1:])+
++++++++++++++++++++++++++++++++++for+i+in+[a.index(b[0])]][0]

>>>+x+=+[1,2,3,4,5,6,7,8,9,0]
>>>+y+=+[1,3,5,7,9]
>>>+subtract_lists(x,y)
[2,+4,+6,+8,+0]
>>>+x+=+[1,2,3,4,5,6,7,8,9,0,9]
>>>+subtract_lists(x,y)
[2,+4,+6,+8,+0,+9]+++++#9+is+only+deleted+once
>>>|code-block|syntax|javascript|963110|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|$]]

Try this.

<pre><code>def subtract_lists(a, b):
 """ Subtracts two lists. Throws ValueError if b contains items not in a """
 # Terminate if b is empty, otherwise remove b[0] from a and recurse
 return a if len(b) == 0 else [a[:i] + subtract_lists(a[i+1:], b[1:]) 
 for i in [a.index(b[0])]][0]

&gt;&gt;&gt; x = [1,2,3,4,5,6,7,8,9,0]
&gt;&gt;&gt; y = [1,3,5,7,9]
&gt;&gt;&gt; subtract_lists(x,y)
[2, 4, 6, 8, 0]
&gt;&gt;&gt; x = [1,2,3,4,5,6,7,8,9,0,9]
&gt;&gt;&gt; subtract_lists(x,y)
[2, 4, 6, 8, 0, 9] #9 is only deleted once
&gt;&gt;&gt;
</code></pre>

blocks|key|1177631|text|我们也可以使用set方法来找出两个列表之间的差异|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177632|x+=+[1,+2,+3,+4,+5,+6,+7,+8,+9,+0]
y+=+[1,+3,+5,+7,+9]
list(set(x).difference(y))
[0,+2,+4,+6,+8]|code-block|syntax|javascript|1177633|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|$]]

We can use set methods as well to find the difference between two list
<pre><code>x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
y = [1, 3, 5, 7, 9]
list(set(x).difference(y))
[0, 2, 4, 6, 8]
</code></pre>

blocks|key|1177523|text|@aaronasterling提供的答案看起来不错，但它与list：x+=+MyList(1,+2,+3,+4)+vs+x+=+MyList([1,+2,+3,+4])的默认接口不兼容。因此，下面的代码可以用作对python列表更友好的代码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1177524|class+MyList(list):
++++def+__init__(self,+*args):
++++++++super(MyList,+self).__init__(*args)

++++def+__sub__(self,+other):
++++++++return+self.__class__([item+for+item+in+self+if+item+not+in+other])|code-block|syntax|javascript|1177525|示例：|1177526|x+=+MyList([1,+2,+3,+4])
y+=+MyList([2,+5,+2])
z+=+x+-+y|1177527|entityMap^0|Y|M|1O|O|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

The answer provided by @aaronasterling looks good, however, it is not compatible with the default interface of list: <code>x = MyList(1, 2, 3, 4)</code> vs <code>x = MyList([1, 2, 3, 4])</code>. Thus, the below code can be used as a more python-list friendly:

<pre><code>class MyList(list):
 def __init__(self, *args):
 super(MyList, self).__init__(*args)

 def __sub__(self, other):
 return self.__class__([item for item in self if item not in other])
</code></pre>

Example: 

<pre><code>x = MyList([1, 2, 3, 4])
y = MyList([2, 5, 2])
z = x - y
</code></pre>

blocks|key|1177656|text|如果值是唯一的，您也可以尝试这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177657|list(set(x)+-+set(y))|code-block|syntax|javascript|1177658|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|$]]

You could also try this if the values are unique anyway:
<pre><code>list(set(x) - set(y))
</code></pre>

blocks|key|963225|text|此示例减去两个列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|963226|#+List+of+pairs+of+points
list+=+[]
list.append([(602,+336),+(624,+365)])
list.append([(635,+336),+(654,+365)])
list.append([(642,+342),+(648,+358)])
list.append([(644,+344),+(646,+356)])
list.append([(653,+337),+(671,+365)])
list.append([(728,+13),+(739,+32)])
list.append([(756,+59),+(767,+79)])

itens_to_remove+=+[]
itens_to_remove.append([(642,+342),+(648,+358)])
itens_to_remove.append([(644,+344),+(646,+356)])

print("Initial+List+Size:+",+len(list))

for+a+in+itens_to_remove:
++++for+b+in+list:
++++++++if+a+==+b+:
++++++++++++list.remove(b)

print("Final+List+Size:+",+len(list))|code-block|syntax|javascript|963227|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|$]]

This example subtracts two lists:

<pre><code># List of pairs of points
list = []
list.append([(602, 336), (624, 365)])
list.append([(635, 336), (654, 365)])
list.append([(642, 342), (648, 358)])
list.append([(644, 344), (646, 356)])
list.append([(653, 337), (671, 365)])
list.append([(728, 13), (739, 32)])
list.append([(756, 59), (767, 79)])

itens_to_remove = []
itens_to_remove.append([(642, 342), (648, 358)])
itens_to_remove.append([(644, 344), (646, 356)])

print("Initial List Size: ", len(list))

for a in itens_to_remove:
 for b in list:
 if a == b :
 list.remove(b)

print("Final List Size: ", len(list))
</code></pre>

blocks|key|963433|text|from+collections+import+Counter

y+=+Counter(y)
x+=+Counter(x)

print(list(x-y))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|963434|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre class="lang-py prettyprint-override"><code>from collections import Counter

y = Counter(y)
x = Counter(x)

print(list(x-y))
</code></pre>

I want to do something similar to this:

<pre><code>&gt;&gt;&gt; x = [1,2,3,4,5,6,7,8,9,0] 
&gt;&gt;&gt; x 
[1, 2, 3, 4, 5, 6, 7, 8, 9, 0] 
&gt;&gt;&gt; y = [1,3,5,7,9] 
&gt;&gt;&gt; y 
[1, 3, 5, 7, 9] 
&gt;&gt;&gt; y - x # (should return [2,4,6,8,0])
</code></pre>

But this is not supported by python lists
What is the best way of doing it?

Python list subtraction operation

数据结构

Python

我想做一些类似的事情：>>> x = [1,2,3,4,5,6,7,8,9,0]  >>> x  [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]  >>> y = [1,3,5,7,9]  >>> y  [1, 3, 5, 7, 9]  >>> y - x   # (should return [2,4,6,8,0])但是python列表并不支持这一点，最好的方法是什么？

问Python列表减法运算
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回答 13

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