blocks|key|608163|text|类似于Ignacio的一行代码，除了更简单/更清晰之外，应该是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|608164|max(loc+for+loc,+val+in+enumerate(li)+if+val+==+'a')|code-block|syntax|javascript|608165|对我来说，这似乎是非常明确和Pythonic式的:您正在寻找包含匹配值的最高索引。不需要nexts、lambdas、reverseds或itertools。|608166|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|$]]

A one-liner that's like Ignacio's except a little simpler/clearer would be

<pre><code>max(loc for loc, val in enumerate(li) if val == 'a')
</code></pre>

It seems very clear and Pythonic to me: you're looking for the highest index that contains a matching value. No nexts, lambdas, reverseds or itertools required.

blocks|key|608120|text|许多其他的解决方案都需要遍历整个列表。这不是问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|608121|def+find_last(lst,+elm):
++gen+=+(len(lst)+-+1+-+i+for+i,+v+in+enumerate(reversed(lst))+if+v+==+elm)
++return+next(gen,+None)|code-block|syntax|javascript|608122|编辑:事后看来，这似乎是不必要的魔法。我会做一些类似这样的事情：|608123|def+find_last(lst,+sought_elt):
++++for+r_idx,+elt+in+enumerate(reversed(lst)):
++++++++if+elt+==+sought_elt:
++++++++++++return+len(lst)+-+1+-+r_idx|608124|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|$]]

Many of the other solutions require iterating over the entire list. This does not.

<pre><code>def find_last(lst, elm):
 gen = (len(lst) - 1 - i for i, v in enumerate(reversed(lst)) if v == elm)
 return next(gen, None)
</code></pre>

Edit: In hindsight this seems like unnecessary wizardry. I'd do something like this instead:

<pre><code>def find_last(lst, sought_elt):
 for r_idx, elt in enumerate(reversed(lst)):
 if elt == sought_elt:
 return len(lst) - 1 - r_idx
</code></pre>

blocks|key|607977|text|>>>+(x+for+x+in+reversed([y+for+y+in+enumerate(li)])+if+x[1]+==+'a').next()[0]
6

>>>+len(li)+-+(x+for+x+in+(y+for+y+in+enumerate(li[::-1]))+if+x[1]+==+'a').next()[0]+-+1
6|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|607978|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><code>&gt;&gt;&gt; (x for x in reversed([y for y in enumerate(li)]) if x[1] == 'a').next()[0]
6

&gt;&gt;&gt; len(li) - (x for x in (y for y in enumerate(li[::-1])) if x[1] == 'a').next()[0] - 1
6
</code></pre>

blocks|key|608047|text|我喜欢wim's和Ignacio's+answers。然而，我认为itertools提供了一个稍微更具可读性的替代方案，尽管是lambda。(对于Python3；对于Python2，使用xrange而不是range)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|608048|>>>+from+itertools+import+dropwhile
>>>+l+=+list('apples')
>>>+l.index('p')
1
>>>+next(dropwhile(lambda+x:+l[x]+!=+'p',+reversed(range(len(l)))))
2|code-block|syntax|javascript|608049|如果未找到该项，这将引发StopIteration异常；您可以捕获该异常并引发ValueError，以使其行为为just+like+index。|608050|定义为函数，避免使用lambda快捷方式：|608051|def+rindex(lst,+item):
++++def+index_ne(x):
++++++++return+lst[x]+!=+item
++++try:
++++++++return+next(dropwhile(index_ne,+reversed(range(len(lst)))))
++++except+StopIteration:
++++++++raise+ValueError("rindex(lst,+item):+item+not+in+list")|608052|它也适用于非字符。测试结果：|608053|>>>+rindex(['apples',+'oranges',+'bananas',+'apples'],+'apples')
3|608054|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/6890170/python-how-to-find-last-occurrence-in-a-list-in-python/6890255#6890255|1|https://stackoverflow.com/questions/6890170/python-how-to-find-last-occurrence-in-a-list-in-python/6890187#6890187|2|https://docs.python.org/3/tutorial/datastructures.html#more-on-lists^0|X|9|2L|6|2U|5|3|5|0|9|9|1|0|0|C|D|13|A|1U|5|1K|F|2|0|A|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]|$9|1G|A|1H|1|1I]]|E|$]]|$1|F|3|G|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1K|8|@$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@$9|1R|A|1S|1|1T]]|E|$]]|$1|M|3|N|5|6|7|1U|8|@$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|1X|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|1Y|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|1Z|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|20|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]|12|$5|X|Y|Z|E|$10|13]]|14|$5|X|Y|Z|E|$10|15]]]]

I like both <a href="https://stackoverflow.com/questions/6890170/python-how-to-find-last-occurrence-in-a-list-in-python/6890255#6890255">wim's</a> and <a href="https://stackoverflow.com/questions/6890170/python-how-to-find-last-occurrence-in-a-list-in-python/6890187#6890187">Ignacio's</a> answers. However, I think <code>itertools</code> provides a slightly more readable alternative, lambda notwithstanding. (For Python 3; for Python 2, use <code>xrange</code> instead of <code>range</code>).

<pre><code>&gt;&gt;&gt; from itertools import dropwhile
&gt;&gt;&gt; l = list('apples')
&gt;&gt;&gt; l.index('p')
1
&gt;&gt;&gt; next(dropwhile(lambda x: l[x] != 'p', reversed(range(len(l)))))
2
</code></pre>

This will raise a <code>StopIteration</code> exception if the item isn't found; you could catch that and raise a <code>ValueError</code> instead, to make this behave <a href="https://docs.python.org/3/tutorial/datastructures.html#more-on-lists" rel="nofollow noreferrer">just like <code>index</code></a>.

Defined as a function, avoiding the <code>lambda</code> shortcut:

<pre><code>def rindex(lst, item):
 def index_ne(x):
 return lst[x] != item
 try:
 return next(dropwhile(index_ne, reversed(range(len(lst)))))
 except StopIteration:
 raise ValueError("rindex(lst, item): item not in list")
</code></pre>

It works for non-chars too. Tested:

<pre><code>&gt;&gt;&gt; rindex(['apples', 'oranges', 'bananas', 'apples'], 'apples')
3
</code></pre>

blocks|key|823520|text|使用dict|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|823521|您可以利用字典键是唯一的这一事实，并且当使用元组构建字典键时，将只使用特定键的最后一个赋值。正如在其他答案中所述，这对于小列表来说很好，但它为所有唯一值创建了一个字典，对于大列表可能效率不高。|823522|dict(map(reversed,+enumerate(li)))["a"]

6|code-block|syntax|javascript|823523|entityMap^0|2|4|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

<h3>With <code>dict</code></h3>

You can use the fact that dictionary keys are unique and when building one with tuples only the last assignment of a value for a particular key will be used. As stated in other answers, this is fine for small lists but it creates a dictionary for all unique values and might not be efficient for large lists.

<pre><code>dict(map(reversed, enumerate(li)))["a"]

6
</code></pre>

blocks|key|823290|text|我来到这里，希望有人已经完成了编写最有效版本的list.rindex的工作，它提供了list.index的完整接口(包括可选的start和stop参数)。我没有在这个问题的答案中找到，或者here，或者here，或者here。所以我自己拼凑了这些...利用其他答案中的建议来回答这个问题和其他问题。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|823291|def+rindex(seq,+value,+start=None,+stop=None):
++"""L.rindex(value,+[start,+[stop]])+->+integer+--+return+last+index+of+value.
++Raises+ValueError+if+the+value+is+not+present."""
++start,+stop,+_+=+slice(start,+stop).indices(len(seq))
++if+stop+==+0:
++++#+start+=+0
++++raise+ValueError('{!r}+is+not+in+list'.format(value))
++else:
++++stop+-=+1
++++start+=+None+if+start+==+0+else+start+-+1
++return+stop+-+seq[stop:start:-1].index(value)|code-block|syntax|javascript|823292|在其他几个答案中建议的使用len(seq)+-+1+-+next(i+for+i,v+in+enumerate(reversed(seq))+if+v+==+value)的技术可以更有效地利用空间:它不需要创建完整列表的反向副本。但在我的(随意的)测试中，它慢了大约50%25。|823293|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/522372|1|https://stackoverflow.com/questions/9836425|2|https://stackoverflow.com/questions/12985096^0|N|B|16|A|1R|5|1X|4|2M|4|0|2T|4|1|30|4|2|0|0|D|20|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]|$9|1A|A|1B|1|1C]|$9|1D|A|1E|1|1F]]|E|$]]|$1|F|3|G|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1H|8|@$9|1I|A|1J|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]|W|$5|P|Q|R|E|$S|X]]]]

I came here hoping to find someone had already done the work of writing the most efficient version of <code>list.rindex</code>, which provided the full interface of <code>list.index</code> (including optional <code>start</code> and <code>stop</code> parameters). I didn't find that in the answers to this question, or <a href="https://stackoverflow.com/questions/522372">here</a>, or <a href="https://stackoverflow.com/questions/9836425">here</a>, or <a href="https://stackoverflow.com/questions/12985096">here</a>. So I put this together myself... making use of suggestions from other answers to this and the other questions.

<pre><code>def rindex(seq, value, start=None, stop=None):
 """L.rindex(value, [start, [stop]]) -&gt; integer -- return last index of value.
 Raises ValueError if the value is not present."""
 start, stop, _ = slice(start, stop).indices(len(seq))
 if stop == 0:
 # start = 0
 raise ValueError('{!r} is not in list'.format(value))
 else:
 stop -= 1
 start = None if start == 0 else start - 1
 return stop - seq[stop:start:-1].index(value)
</code></pre>

The technique using <code>len(seq) - 1 - next(i for i,v in enumerate(reversed(seq)) if v == value)</code>, suggested in several other answers, can be more space-efficient: it needn't create a reversed copy of the full list. But in my (offhand, casual) testing, it's about 50% slower.

blocks|key|608189|text|使用一个简单的循环：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|608190|def+reversed_index(items,+value):
++++for+pos,+curr+in+enumerate(reversed(items)):
++++++++if+curr+==+value:
++++++++++++return+len(items)+-+pos+-+1
++++raise+ValueError("{0!r}+is+not+in+list".format(value))|code-block|syntax|javascript|608191|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|$]]

Use a simple loop:

<pre><code>def reversed_index(items, value):
 for pos, curr in enumerate(reversed(items)):
 if curr == value:
 return len(items) - pos - 1
 raise ValueError("{0!r} is not in list".format(value))
</code></pre>

blocks|key|608456|text|last_occurence=len(yourlist)-yourlist[::-1].index(element)-1|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|608457|就像that.no需要导入或创建函数一样简单。|unstyled|608458|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>last_occurence=len(yourlist)-yourlist[::-1].index(element)-1
</code></pre>

just easy as that.no need to import or create a function.

blocks|key|608519|text|lastIndexOf+=+lambda+array,+item:+len(array)+-+(array[::-1].index(item))+-+1|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|608520|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><code>lastIndexOf = lambda array, item: len(array) - (array[::-1].index(item)) - 1
</code></pre>

blocks|key|608568|text|如果列表很小，您可以计算所有索引并返回最大的索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|608569|index+=+max(i+for+i,+x+in+enumerate(elements)+if+x+==+'foo')|code-block|syntax|javascript|608570|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 the list is small, you can compute all indices and return the largest:
<pre><code>index = max(i for i, x in enumerate(elements) if x == 'foo')
</code></pre>

blocks|key|823607|text|下面是一个函数，用于查找列表中元素的最后一个匹配项。一个列表和一个元素被传递给函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|823608|li+=+["a",+"b",+"a",+"c",+"x",+"d",+"a",+"6"]
element+=+"a"

def+last_occurrence(li,element):
++++for+i+in+range(len(li)-1,0,-1):
++++++++if+li[i]+==+element:
++++++++++++return+i

++++return+-1

last_occ+=+last_occurrence(li,+element)
if+(last_occ+!=+-1):
++++print("The+last+occurrence+at+index+:+",last_occ)
else:
++++print("Element+not+found")|code-block|syntax|javascript|823609|在last_occurrence函数中，range使用了一个for循环。它将以相反的顺序迭代列表。如果当前索引的元素与searched元素匹配，则函数将返回index。如果在比较列表中的所有元素后，找不到searched元素，function将返回-1。|offset|length|style|CODE|823610|entityMap^0|0|0|1|F|K|5|U|3|1N|8|26|5|2U|8|3G|2|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|N|$]]

Here is a function for finding the last occurrence of an element in a list. A list and an element are passed to the function.
<pre><code>li = [&quot;a&quot;, &quot;b&quot;, &quot;a&quot;, &quot;c&quot;, &quot;x&quot;, &quot;d&quot;, &quot;a&quot;, &quot;6&quot;]
element = &quot;a&quot;

def last_occurrence(li,element):
 for i in range(len(li)-1,0,-1):
 if li[i] == element:
 return i

 return -1

last_occ = last_occurrence(li, element)
if (last_occ != -1):
 print(&quot;The last occurrence at index : &quot;,last_occ)
else:
 print(&quot;Element not found&quot;)
</code></pre>
Inside the <code>last_occurrence</code> function a <code>for</code> loop is used with <code>range</code>. which will iterate the list in reverse order. if the element of the current index match the <code>searched</code> element, the function will return the <code>index</code>.
In case after comparing all elements in the list the <code>searched</code> element is not found function will return <code>-1</code>.

blocks|key|608203|text|def+rindex(lst,+val):
++++try:
++++++++return+next(len(lst)-i+for+i,+e+in+enumerate(reversed(lst),+start=1)+if+e+==+val)
++++except+StopIteration:
++++++++raise+ValueError('{}+is+not+in+list'.format(val))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|608204|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><code>def rindex(lst, val):
 try:
 return next(len(lst)-i for i, e in enumerate(reversed(lst), start=1) if e == val)
 except StopIteration:
 raise ValueError('{} is not in list'.format(val))
</code></pre>

blocks|key|823342|text|val+=1，2，2，2，2，4，5。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|823343|如果需要查找2的最后一次出现|823344|last_occurence+=+(len(val)+-1)+-+list(reversed(val)).index(2)|offset|length|style|CODE|823345|entityMap^0|0|0|0|1P|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@$F|O|G|P|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

val = [1,2,2,2,2,2,4,5].

If you need to find last occurence of 2

<code>last_occurence = (len(val) -1) - list(reversed(val)).index(2)</code>

blocks|key|608409|text|下面是使用enumerate和列表理解获取最后一个索引的一小段代码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|608410|li+=+["a",+"b",+"a",+"c",+"x",+"d",+"a",+"6"]
[l[0]+for+l+in+enumerate(li)+if+l[1]+==+"a"][-1]|code-block|syntax|javascript|608411|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/functions.html#enumerate^0|5|9|5|9|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Here's a little one-liner for obtaining the last index, using <a href="https://docs.python.org/3/library/functions.html#enumerate" rel="nofollow noreferrer"><code>enumerate</code></a> and a list comprehension:

<pre><code>li = ["a", "b", "a", "c", "x", "d", "a", "6"]
[l[0] for l in enumerate(li) if l[1] == "a"][-1]
</code></pre>

Say I have this list:

<pre><code>li = ["a", "b", "a", "c", "x", "d", "a", "6"]
</code></pre>

As far as help showed me, there is not a builtin function that returns the last occurrence of a string (like the reverse of <code>index</code>). So basically, how can I find the last occurrence of <code>"a"</code> in the given list?

How to find the last occurrence of an item in a Python list

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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假设我有这个列表：li = ["a", "b", "a", "c", "x", "d", "a", "6"]据help介绍，没有一个内置函数可以返回字符串的最后一个匹配项(就像index的反面一样)。那么基本上，我如何在给定的列表中找到"a"的最后一个匹配项？

问如何查找Python列表中项目的最后一个匹配项
EN

回答 14

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回答 14

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问如何查找Python列表中项目的最后一个匹配项
EN