blocks|key|1584445|text|想一想代码到底说了些什么！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1584446|>>>+(1+or+2)
1
>>>+(2+or+1)
2|code-block|syntax|javascript|1584447|这或许可以解释这一点。:)+Python显然实现了"lazy+or"，这并不奇怪。它的执行方式类似于：|1584448|def+or(x,+y):
++++if+x:+return+x
++++if+y:+return+y
++++return+False|1584449|在第一个示例中，使用x+==+1和y+==+2。在第二个例子中，反之亦然。这就是为什么它会根据值的顺序返回不同的值。|offset|length|style|CODE|1584450|entityMap^0|0|0|0|0|A|6|H|6|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

Think about what the code actually says!

<pre><code>&gt;&gt;&gt; (1 or 2)
1
&gt;&gt;&gt; (2 or 1)
2
</code></pre>

That should probably explain it. :) Python apparently implements "lazy or", which should come as no surprise. It performs it something like this:

<pre><code>def or(x, y):
 if x: return x
 if y: return y
 return False
</code></pre>

In the first example, <code>x == 1</code> and <code>y == 2</code>. In the second example, it's vice versa. That's why it returns different values depending on the order of them.

blocks|key|1369366|text|>>>+L1+=+[2,3,4]
>>>+L2+=+[1,2]
>>>+[i+for+i+in+L1+if+i+in+L2]
[2]


>>>+S1+=+set(L1)
>>>+S2+=+set(L2)
>>>+S1.intersection(S2)
set([2])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1369367|空列表和空集都为False，因此您可以直接将该值用作真值。|unstyled|1369368|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>&gt;&gt;&gt; L1 = [2,3,4]
&gt;&gt;&gt; L2 = [1,2]
&gt;&gt;&gt; [i for i in L1 if i in L2]
[2]


&gt;&gt;&gt; S1 = set(L1)
&gt;&gt;&gt; S2 = set(L2)
&gt;&gt;&gt; S1.intersection(S2)
set([2])
</code></pre>

Both empty lists and empty sets are False, so you can use the value directly as a truth value.

blocks|key|1584555|text|我能想到的最好的办法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1584556|any([True+for+e+in+(1,+2)+if+e+in+a])|code-block|syntax|javascript|1584557|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|$]]

Best I could come up with:

<pre><code>any([True for e in (1, 2) if e in a])
</code></pre>

blocks|key|1584587|text|这将在一行代码中完成。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1584588|>>>+a=[2,3,4]
>>>+b=[1,2]
>>>+bool(sum(map(lambda+x:+x+in+b,+a)))
True|code-block|syntax|javascript|1584589|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 will do it in one line.

<pre><code>&gt;&gt;&gt; a=[2,3,4]
&gt;&gt;&gt; b=[1,2]
&gt;&gt;&gt; bool(sum(map(lambda x: x in b, a)))
True
</code></pre>

blocks|key|1369497|text|在某些情况下(例如，唯一列表元素)，可以使用集合操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369498|>>>+a=[2,3,4]
>>>+set(a)+-+set([2,3])+!=+set(a)
True
>>>+|code-block|syntax|javascript|1369499|或者，使用set.isdisjoint()，|offset|length|1369500|>>>+not+set(a).isdisjoint(set([2,3]))
True
>>>+not+set(a).isdisjoint(set([5,6]))
False
>>>+|1369501|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/stdtypes.html#set.isdisjoint^0|0|0|5|G|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

In some cases (e.g. unique list elements), set operations can be used.

<pre><code>&gt;&gt;&gt; a=[2,3,4]
&gt;&gt;&gt; set(a) - set([2,3]) != set(a)
True
&gt;&gt;&gt; 
</code></pre>

Or, using <a href="http://docs.python.org/library/stdtypes.html#set.isdisjoint" rel="nofollow noreferrer">set.isdisjoint()</a>,

<pre><code>&gt;&gt;&gt; not set(a).isdisjoint(set([2,3]))
True
&gt;&gt;&gt; not set(a).isdisjoint(set([5,6]))
False
&gt;&gt;&gt; 
</code></pre>

blocks|key|1369516|text|也许更懒一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369517|a+=+[1,2,3,4]
b+=+[2,7]

print+any((True+for+x+in+a+if+x+in+b))|code-block|syntax|javascript|1369518|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|$]]

Maybe a bit more lazy:

<pre><code>a = [1,2,3,4]
b = [2,7]

print any((True for x in a if x in b))
</code></pre>

blocks|key|1584689|text|托比亚斯你先我一步。我在想your+solution的这个细微的变化|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1584690|>>>+a+=+[1,2,3,4]
>>>+b+=+[2,7]
>>>+any(x+in+a+for+x+in+b)
True|code-block|syntax|javascript|1584691|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/740359/4518341^0|D|D|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Ah, Tobias you beat me to it. I was thinking of this slight variation on <a href="https://stackoverflow.com/a/740359/4518341">your solution</a>:
<pre><code>&gt;&gt;&gt; a = [1,2,3,4]
&gt;&gt;&gt; b = [2,7]
&gt;&gt;&gt; any(x in a for x in b)
True
</code></pre>

blocks|key|1369654|text|当你想“检查a是否在b中”时，想想哈希(在本例中是set)。最快的方法是对要检查的列表进行散列，然后检查其中的每一项。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369655|这就是为什么Joe+Koberg的答案是快速的:检查集合相交非常快。|1369656|然而，当你没有很多数据时，创建集合可能是浪费时间。所以，你可以制作一组列表，然后只检查每一项：|1369657|tocheck+=+[1,2]+#+items+to+check
a+=+[2,3,4]+#+the+list

a+=+set(a)+#+convert+to+set+(O(len(a)))
print+[i+for+i+in+tocheck+if+i+in+a]+#+check+items+(O(len(tocheck)))|code-block|syntax|javascript|1369658|当您要检查的项目数量很少时，差异可以忽略不计。但是在一个大列表中检查大量的数字...|1369659|测试：|1369660|from+timeit+import+timeit

methods+=+['''tocheck+=+[1,2]+#+items+to+check
a+=+[2,3,4]+#+the+list
a+=+set(a)+#+convert+to+set+(O(n))
[i+for+i+in+tocheck+if+i+in+a]+#+check+items+(O(m))''',

'''L1+=+[2,3,4]
L2+=+[1,2]
[i+for+i+in+L1+if+i+in+L2]''',

'''S1+=+set([2,3,4])
S2+=+set([1,2])
S1.intersection(S2)''',

'''a+=+[1,2]
b+=+[2,3,4]
any(x+in+a+for+x+in+b)''']

for+method+in+methods:
++++print+timeit(method,+number=10000)

print

methods+=+['''tocheck+=+range(200,300)+#+items+to+check
a+=+range(2,+10000)+#+the+list
a+=+set(a)+#+convert+to+set+(O(n))
[i+for+i+in+tocheck+if+i+in+a]+#+check+items+(O(m))''',

'''L1+=+range(2,+10000)
L2+=+range(200,300)
[i+for+i+in+L1+if+i+in+L2]''',

'''S1+=+set(range(2,+10000))
S2+=+set(range(200,300))
S1.intersection(S2)''',

'''a+=+range(200,300)
b+=+range(2,+10000)
any(x+in+a+for+x+in+b)''']

for+method+in+methods:
++++print+timeit(method,+number=1000)|1369661|速度：|1369662|M1:+0.0170331001282+#+make+one+set
M2:+0.0164539813995+#+list+comprehension
M3:+0.0286040306091+#+set+intersection
M4:+0.0305438041687+#+any

M1:+0.49850320816+#+make+one+set
M2:+25.2735087872+#+list+comprehension
M3:+0.466138124466+#+set+intersection
M4:+0.668627977371+#+any|1369663|最快的方法是创建一组(列表)，但交集在大型数据集上的效果最好！|1369664|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|H|7|14|8|@]|9|@]|A|$I|J]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|H|7|16|8|@]|9|@]|A|$I|J]]|$1|U|3|V|5|6|7|17|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

When you think "check to see if a in b", think hashes (in this case, sets). The fastest way is to hash the list you want to check, and then check each item in there.

This is why Joe Koberg's answer is fast: checking set intersection is very fast.

When you don't have a lot of data though, making sets can be a waste of time. So, you can make a set of the list and just check each item:

<pre><code>tocheck = [1,2] # items to check
a = [2,3,4] # the list

a = set(a) # convert to set (O(len(a)))
print [i for i in tocheck if i in a] # check items (O(len(tocheck)))
</code></pre>

When the number of items you want to check is small, the difference can be negligible. But check lots of numbers against a large list...

tests:

<pre><code>from timeit import timeit

methods = ['''tocheck = [1,2] # items to check
a = [2,3,4] # the list
a = set(a) # convert to set (O(n))
[i for i in tocheck if i in a] # check items (O(m))''',

'''L1 = [2,3,4]
L2 = [1,2]
[i for i in L1 if i in L2]''',

'''S1 = set([2,3,4])
S2 = set([1,2])
S1.intersection(S2)''',

'''a = [1,2]
b = [2,3,4]
any(x in a for x in b)''']

for method in methods:
 print timeit(method, number=10000)

print

methods = ['''tocheck = range(200,300) # items to check
a = range(2, 10000) # the list
a = set(a) # convert to set (O(n))
[i for i in tocheck if i in a] # check items (O(m))''',

'''L1 = range(2, 10000)
L2 = range(200,300)
[i for i in L1 if i in L2]''',

'''S1 = set(range(2, 10000))
S2 = set(range(200,300))
S1.intersection(S2)''',

'''a = range(200,300)
b = range(2, 10000)
any(x in a for x in b)''']

for method in methods:
 print timeit(method, number=1000)
</code></pre>

speeds:

<pre><code>M1: 0.0170331001282 # make one set
M2: 0.0164539813995 # list comprehension
M3: 0.0286040306091 # set intersection
M4: 0.0305438041687 # any

M1: 0.49850320816 # make one set
M2: 25.2735087872 # list comprehension
M3: 0.466138124466 # set intersection
M4: 0.668627977371 # any
</code></pre>

The method that is consistently fast is to make one set (of the list), but the intersection works on large data sets the best!

blocks|key|1369734|text|a+=+{2,3,4}
if+{1,2}+&+a:
++++pass|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1369735|代码高尔夫版本。如果这样做有意义，可以考虑使用set。我发现这比列表理解更具可读性。|unstyled|1369736|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>a = {2,3,4}
if {1,2} &amp; a:
 pass
</code></pre>

Code golf version. Consider using a set if it makes sense to do so.
I find this more readable than a list comprehension.

blocks|key|1369773|text|没有列表理解的1行。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369774|>>>+any(map(lambda+each:+each+in+[2,3,4],+[1,2]))
True
>>>+any(map(lambda+each:+each+in+[2,3,4],+[1,5]))
False
>>>+any(map(lambda+each:+each+in+[2,3,4],+[2,4]))
True|code-block|syntax|javascript|1369775|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|$]]

1 line without list comprehensions.

<pre><code>&gt;&gt;&gt; any(map(lambda each: each in [2,3,4], [1,2]))
True
&gt;&gt;&gt; any(map(lambda each: each in [2,3,4], [1,5]))
False
&gt;&gt;&gt; any(map(lambda each: each in [2,3,4], [2,4]))
True
</code></pre>

blocks|key|1369790|text|很简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369791|_new_list+=+[]
for+item+in+a:
++++if+item+in+b:
++++++++_new_list.append(item)
++++else:
++++++++pass|code-block|syntax|javascript|1369792|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|$]]

Simple.

<pre><code>_new_list = []
for item in a:
 if item in b:
 _new_list.append(item)
 else:
 pass
</code></pre>

blocks|key|1369815|text|在python+3中，我们可以开始使用解包星号。给出两个列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369816|bool(len({*a}+&+{*b}))|code-block|syntax|javascript|1369817|编辑:纳入alkanen的建议|1369818|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

In python 3 we can start make use of the unpack asterisk. Given two lists:

<pre><code>bool(len({*a} &amp; {*b}))
</code></pre>

Edit: incorporate alkanen's suggestion

blocks|key|1584991|text|我不得不说，我的情况可能不是你想要的，但它可能为你的想法提供了另一种选择。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1584992|我已经尝试了set()和any()方法，但仍然存在速度问题。所以我记得Raymond+Hettinger说过python中的一切都是一本字典，只要有可能就使用dict。这就是我尝试过的。|1584993|我使用带有int的defaultdict来表示否定结果，并使用第一个列表中的项作为第二个列表的键(转换为defaultdict)。因为您可以使用dict即时查找，所以您可以立即知道该项目是否存在于defaultdict中。我知道你不能总是改变你的第二个列表的数据结构，但是如果你能够从一开始就改变，那么它就会快得多。您可能需要将list2+(较大的列表)转换为defaultdict，其中key是要从较小的列表中检查的潜在值，值可以是1(命中)或0(无命中，默认)。|1584994|from+collections+import+defaultdict
already_indexed+=+defaultdict(int)

def+check_exist(small_list,+default_list):
++++for+item+in+small_list:
++++++++if+default_list[item]+==+1:
++++++++++++return+True
++++return+False

if+check_exist(small_list,+already_indexed):
++++continue
else:
++++for+x+in+small_list:
++++++++already_indexed[x]+=+1|code-block|syntax|javascript|1584995|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I have to say that my situation might not be what you are looking for, but it may provide an alternative to your thinking.

I have tried both the set() and any() method but still have problems with speed. So I remembered Raymond Hettinger said everything in python is a dictionary and use dict whenever you can. So that's what I tried. 

I used a defaultdict with int to indicate negative results and used the item in the first list as the key for the second list (converted to defaultdict). Because you have instant lookup with dict, you know immediately whether that item exist in the defaultdict. I know you don't always get to change data structure for your second list, but if you are able to from the start, then it's much faster. You may have to convert list2 (larger list) to a defaultdict, where key is the potential value you want to check from small list, and value is either 1 (hit) or 0 (no hit, default). 

<pre><code>from collections import defaultdict
already_indexed = defaultdict(int)

def check_exist(small_list, default_list):
 for item in small_list:
 if default_list[item] == 1:
 return True
 return False

if check_exist(small_list, already_indexed):
 continue
else:
 for x in small_list:
 already_indexed[x] = 1
</code></pre>

I'm trying to find a short way to see if any of the following items is in a list, but my first attempt does not work. Besides writing a function to accomplish this, is the any short way to check if one of multiple items is in a list.

<pre><code>&gt;&gt;&gt; a = [2,3,4]
&gt;&gt;&gt; print (1 or 2) in a
False
&gt;&gt;&gt; print (2 or 1) in a
True
</code></pre>

How to check if one of the following items is in a list?

数据结构

Python

我试图找到一种简单的方法来查看列表中是否有以下项目，但我的第一次尝试不起作用。除了编写一个函数来实现这一点之外，还可以使用任何简单的方法来检查多个项目中是否有一个项目在列表中。>>> a = [2,3,4]>>> print (1 or 2) in aFalse>>> print (2 or 1) in aTrue

问如何检查以下项目是否在列表中？
EN

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问如何检查以下项目是否在列表中？
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