blocks|key|1613677|text|最好的方法可能是使用list+method+.index。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1613678|对于列表中的对象，您可以执行以下操作：|1613679|def+__eq__(self,+other):
++++return+self.Value+==+other.Value|code-block|syntax|javascript|1613680|任何你需要的特殊处理。|1613681|还可以将for/in语句与enumerate(arr)一起使用|1613682|查找值大于100的项的索引的示例。|1613683|for+index,+item+in+enumerate(arr):
++++if+item+>+100:
++++++++return+index,+item|1613684|Source|1613685|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/tutorial/datastructures.html#more-on-lists|1|http://www.daniweb.com/forums/thread174502.html^0|A|I|0|0|0|0|0|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|6|7|18|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|19|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|1B|8|@]|9|@]|C|$]]|$1|O|3|P|5|6|7|1C|8|@]|9|@]|C|$]]|$1|Q|3|R|5|H|7|1D|8|@]|9|@]|C|$I|J]]|$1|S|3|T|5|6|7|1E|8|@]|9|@$A|1F|B|1G|1|1H]]|C|$]]|$1|U|3|-4|5|6|7|1I|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]|12|$5|X|Y|Z|C|$10|13]]]]

The best way is probably to use the <a href="https://docs.python.org/tutorial/datastructures.html#more-on-lists" rel="noreferrer">list method .index</a>. 

For the objects in the list, you can do something like:

<pre><code>def __eq__(self, other):
 return self.Value == other.Value
</code></pre>

with any special processing you need.

You can also use a for/in statement with enumerate(arr)

Example of finding the index of an item that has value > 100.

<pre><code>for index, item in enumerate(arr):
 if item &gt; 100:
 return index, item
</code></pre>

<a href="http://www.daniweb.com/forums/thread174502.html" rel="noreferrer">Source</a>

blocks|key|1398942|text|来自Dive+Into+Python|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1398943|>>>+li
['a',+'b',+'new',+'mpilgrim',+'z',+'example',+'new',+'two',+'elements']
>>>+li.index("example")
5|code-block|syntax|javascript|1398944|entityMap|0|LINK|mutability|MUTABLE|url|http://www.diveintopython.net/native_data_types/lists.html^0|2|G|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]]]]

From <a href="http://www.diveintopython.net/native_data_types/lists.html" rel="noreferrer">Dive Into Python</a>:

<pre><code>&gt;&gt;&gt; li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
&gt;&gt;&gt; li.index("example")
5
</code></pre>

blocks|key|1399287|text|如果你只是想知道一个元素是否包含在列表中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1399288|>>>+li
['a',+'b',+'new',+'mpilgrim',+'z',+'example',+'new',+'two',+'elements']
>>>+'example'+in+li
True
>>>+'damn'+in+li
False|code-block|syntax|javascript|1399289|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|$]]

If you just want to find out if an element is contained in the list or not:

<pre><code>&gt;&gt;&gt; li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
&gt;&gt;&gt; 'example' in li
True
&gt;&gt;&gt; 'damn' in li
False
</code></pre>

blocks|key|1399145|text|这是另一种使用列表理解的方法(有些人可能会觉得有争议)。对于简单的测试，例如对对象属性的比较(我需要很多)，它是非常容易接近的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1399146|el+=+[x+for+x+in+mylist+if+x.attr+==+"foo"][0]|code-block|syntax|javascript|1399147|当然，这假设列表中存在(实际上是唯一的)合适的元素。|1399148|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|$]]

Here is another way using list comprehension (some people might find it debatable). It is very approachable for simple tests, e.g. comparisons on object attributes (which I need a lot):

<pre><code>el = [x for x in mylist if x.attr == "foo"][0]
</code></pre>

Of course this assumes the existence (and, actually, uniqueness) of a suitable element in the list.

blocks|key|1399205|text|假设您想要在numpy数组中查找一个值，我猜这样的方法可能会有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1399206|Numpy.where(arr=="value")[0]|code-block|syntax|javascript|1399207|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 you want to find a value in a numpy array,
I guess something like this might work:

<pre><code>Numpy.where(arr=="value")[0]
</code></pre>

blocks|key|1613645|text|有一个名为i+=+array.index(value)的index方法，但我认为您不能指定自定义比较运算符。不过，编写自己的函数来实现这一点并不难：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1613646|def+custom_index(array,+compare_function):
++++for+i,+v+in+enumerate(array):
++++++++if+compare_function(v):
++++++++++++return+i|code-block|syntax|javascript|1613647|entityMap^0|5|M|S|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

There is the <code>index</code> method, <code>i = array.index(value)</code>, but I don't think you can specify a custom comparison operator. It wouldn't be hard to write your own function to do so, though:

<pre><code>def custom_index(array, compare_function):
 for i, v in enumerate(array):
 if compare_function(v):
 return i
</code></pre>

blocks|key|1614121|text|我使用函数返回匹配元素的索引(Python+2.6)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1614122|def+index(l,+f):
+++++return+next((i+for+i+in+xrange(len(l))+if+f(l[i])),+None)|code-block|syntax|javascript|1614123|然后通过lambda函数使用它来通过任何所需的等式检索所需的元素，例如使用元素名称。|1614124|element+=+mylist[index(mylist,+lambda+item:+item["name"]+==+"my+name")]|1614125|如果我需要在代码中的几个地方使用它，我只需要定义特定的find函数，例如通过名称查找元素：|1614126|def+find_name(l,+name):
+++++return+l[index(l,+lambda+item:+item["name"]+==+name)]|1614127|然后，它非常容易和可读性：|1614128|element+=+find_name(mylist,"my+name")|1614129|entityMap^0|0|0|0|0|0|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]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

I use function for returning index for the matching element (Python 2.6):

<pre><code>def index(l, f):
 return next((i for i in xrange(len(l)) if f(l[i])), None)
</code></pre>

Then use it via lambda function for retrieving needed element by any required equation e.g. by using element name.

<pre><code>element = mylist[index(mylist, lambda item: item["name"] == "my name")]
</code></pre>

If i need to use it in several places in my code i just define specific find function e.g. for finding element by name:

<pre><code>def find_name(l, name):
 return l[index(l, lambda item: item["name"] == name)]
</code></pre>

And then it is quite easy and readable:

<pre><code>element = find_name(mylist,"my name")
</code></pre>

blocks|key|1399000|text|list的index方法将为您完成此操作。如果您想保证顺序，请首先使用sorted()对列表进行排序。Sorted接受cmp或key参数来指定如何进行排序：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1399001|a+=+[5,+4,+3]
print+sorted(a).index(5)|code-block|syntax|javascript|1399002|或者：|1399003|a+=+['one',+'aardvark',+'a']
print+sorted(a,+key=len).index('a')|1399004|entityMap^0|Z|8|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|$]]

The index method of a list will do this for you. If you want to guarantee order, sort the list first using <code>sorted()</code>. Sorted accepts a cmp or key parameter to dictate how the sorting will happen:

<pre><code>a = [5, 4, 3]
print sorted(a).index(5)
</code></pre>

Or:

<pre><code>a = ['one', 'aardvark', 'a']
print sorted(a, key=len).index('a')
</code></pre>

blocks|key|1613783|text|这个怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1613784|def+global_index(lst,+test):
++++return+(+pair[0]+for+pair+in+zip(range(len(lst)),+lst)+if+test(pair[1])+)|code-block|syntax|javascript|1613785|用法：|1613786|>>>+global_index([1,+2,+3,+4,+5,+6],+lambda+x:+x>3)
<generator+object+<genexpr>+at+...>
>>>+list(_)
[3,+4,+5]|1613787|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|$]]

how's this one?

<pre><code>def global_index(lst, test):
 return ( pair[0] for pair in zip(range(len(lst)), lst) if test(pair[1]) )
</code></pre>

Usage:

<pre><code>&gt;&gt;&gt; global_index([1, 2, 3, 4, 5, 6], lambda x: x&gt;3)
&lt;generator object &lt;genexpr&gt; at ...&gt;
&gt;&gt;&gt; list(_)
[3, 4, 5]
</code></pre>

blocks|key|1399334|text|我通过改编一些tutos找到了这个。感谢google，感谢你们所有人;)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1399335|def+findall(L,+test):
++++i=0
++++indices+=+[]
++++while(True):
++++++++try:
++++++++++++#+next+value+in+list+passing+the+test
++++++++++++nextvalue+=+filter(test,+L[i:])[0]

++++++++++++#+add+index+of+this+value+in+the+index+list,
++++++++++++#+by+searching+the+value+in+L[i:]+
++++++++++++indices.append(L.index(nextvalue,+i))

++++++++++++#+iterate+i,+that+is+the+next+index+from+where+to+search
++++++++++++i=indices[-1]%2B1
++++++++#when+there+is+no+further+"good+value",+filter+returns+[],
++++++++#+hence+there+is+an+out+of+range+exeption
++++++++except+IndexError:
++++++++++++return+indices|code-block|syntax|javascript|1399336|一个非常简单的用法：|1399337|a+=+[0,0,2,1]
ind+=+findall(a,+lambda+x:x>0))

[2,+3]|1399338|附言:我的英语|1399339|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|$]]

I found this by adapting some tutos. Thanks to google, and to all of you ;)

<pre><code>def findall(L, test):
 i=0
 indices = []
 while(True):
 try:
 # next value in list passing the test
 nextvalue = filter(test, L[i:])[0]

 # add index of this value in the index list,
 # by searching the value in L[i:] 
 indices.append(L.index(nextvalue, i))

 # iterate i, that is the next index from where to search
 i=indices[-1]+1
 #when there is no further "good value", filter returns [],
 # hence there is an out of range exeption
 except IndexError:
 return indices
</code></pre>

A very simple use:

<pre><code>a = [0,0,2,1]
ind = findall(a, lambda x:x&gt;0))

[2, 3]
</code></pre>

P.S. scuse my english

What is a good way to find the index of an element in a list in Python? 
Note that the list may not be sorted.

Is there a way to specify what comparison operator to use?

Python: finding an element in a list

在Python中查找列表中元素的索引的好方法是什么？请注意，该列表可能不会排序。有没有办法指定要使用的比较运算符？

问Python:在列表中查找元素
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回答 10

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