blocks|key|1446865|text|对于你的问题，你可以做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1446866|[(i,+j)+for+i,j+in+enumerate(l)+if+j+>+2]|code-block|syntax|javascript|1446867|输出：|1446868|[(2,+3),
+(3,+3),
+(4,+3),
+(5,+3),
+(6,+3),
+(7,+3),
+(8,+3),
+(9,+3),
+(10,+3),
+(11,+3),
+(12,+3),
+(13,+3),
+(14,+3),
+(15,+3)]|1446869|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|$]]

for your problem you can do 

<pre><code>[(i, j) for i,j in enumerate(l) if j &gt; 2]
</code></pre>

output:

<pre><code>[(2, 3),
 (3, 3),
 (4, 3),
 (5, 3),
 (6, 3),
 (7, 3),
 (8, 3),
 (9, 3),
 (10, 3),
 (11, 3),
 (12, 3),
 (13, 3),
 (14, 3),
 (15, 3)]
</code></pre>

blocks|key|1319323|text|lst+=+[1,+1,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+1,+1,+1,+1,+1,+1,+1,+1]
print+[(i,+e)+for+(i,+e)+in+enumerate(lst)+if+e+>+2]

[(2,+3),
+(3,+3),
+(4,+3),
+(5,+3),
+(6,+3),
+(7,+3),
+(8,+3),
+(9,+3),
+(10,+3),
+(11,+3),
+(12,+3),
+(13,+3),
+(14,+3),
+(15,+3)]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1319324|unstyled|1319325|如果您的列表很大，更好的方法是通过()使用generator动态生成每个元素，而不是在内存中一次性生成所有元素：|offset|length|style|CODE|1319326|result+=+((i,+e)+for+(i,+e)+in+enumerate(lst)+if+e+>+2)

for+i+in+result:
++++print+i|1319327|然后输出将是：|1319328|(2,+3)
(3,+3)
(4,+3)
(5,+3)
(6,+3)
(7,+3)
(8,+3)
(9,+3)
(10,+3)
(11,+3)
(12,+3)
(13,+3)
(14,+3)
(15,+3)|1319329|请注意，生成器只能读取一次，这意味着如果您再次遍历result，则不会打印任何内容。|1319330|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/tutorial/classes.html#generators^0|0|0|H|2|L|9|0|0|0|0|0|P|6|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|12|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|13|8|@$H|14|I|15|J|K]]|9|@$H|16|I|17|1|18]]|A|$]]|$1|L|3|M|5|6|7|19|8|@]|9|@]|A|$B|C]]|$1|N|3|O|5|E|7|1A|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1B|8|@]|9|@]|A|$B|C]]|$1|R|3|S|5|E|7|1C|8|@$H|1D|I|1E|J|K]]|9|@]|A|$]]|$1|T|3|-4|5|E|7|1F|8|@]|9|@]|A|$]]]|U|$V|$5|W|X|Y|A|$Z|10]]]]

<pre><code>lst = [1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1]
print [(i, e) for (i, e) in enumerate(lst) if e &gt; 2]

[(2, 3),
 (3, 3),
 (4, 3),
 (5, 3),
 (6, 3),
 (7, 3),
 (8, 3),
 (9, 3),
 (10, 3),
 (11, 3),
 (12, 3),
 (13, 3),
 (14, 3),
 (15, 3)]
</code></pre>

<hr>

If your list is huge, a better way would be to use a <a href="http://docs.python.org/2/tutorial/classes.html#generators" rel="nofollow">generator</a> via <code>()</code> to generate each element on the fly instead of all at once in memory:

<pre><code>result = ((i, e) for (i, e) in enumerate(lst) if e &gt; 2)

for i in result:
 print i
</code></pre>

then the output will be:

<pre><code>(2, 3)
(3, 3)
(4, 3)
(5, 3)
(6, 3)
(7, 3)
(8, 3)
(9, 3)
(10, 3)
(11, 3)
(12, 3)
(13, 3)
(14, 3)
(15, 3)
</code></pre>

Note that a generator can only be read once, meaning if you loop over <code>result</code> again nothing will get printed.

blocks|key|3762358|text|lst+=+[1,+1,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+3,+1,+1,+1,+1,+1,+1,+1,+1]
print+[(i,+e)+for+(i,+e)+in+enumerate(lst)+if+e+>+2]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3762359|很显然，我是用c语言写的。|unstyled|3762360|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>lst = [1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1]
print [(i, e) for (i, e) in enumerate(lst) if e &gt; 2]
</code></pre>

sory prevoiusly i wrote it in c-Language.

blocks|key|1446931|text|在列表上做一个理解|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1446932|interesting_bonds+=+[index,+bonds+for+index,+bonds+in+enumerate(number_of_bonds)+if+bonds+>+2]
for+bonds+in+interesting_bonds:
++++print+bonds|code-block|syntax|javascript|1446933|列表理解是获取您感兴趣的元素的更为"pythonic“的方法。您也可以使用内置的filter。|offset|length|style|CODE|1446934|编辑：正如Jon在下面的评论中指出的那样，过滤后的枚举不会从原始列表中给出正确的索引。所以只需对枚举列表进行理解即可。|BOLD|1446935|entityMap^0|0|0|14|6|0|0|2|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@$I|U|J|V|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|W|8|@$I|X|J|Y|K|O]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|Q|$]]

Do a comprehension on the list

<pre><code>interesting_bonds = [index, bonds for index, bonds in enumerate(number_of_bonds) if bonds &gt; 2]
for bonds in interesting_bonds:
 print bonds
</code></pre>

List comprehensions are the more "pythonic" way to get elements that interest you. You could also use the builtin <code>filter</code>.

Edit: As Jon pointed out in a comment below, enumerating after the filtering won't give the right indices from the original list. So just do the comprehension on the enumerated list.

blocks|key|1325616|text|您可以只添加一个变量i，它在每次循环再次开始时计数为%2B1。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1325617|i+=+0
for+bonds+in+number_of_bonds:+++++++
+++++if+bonds+>=+2:
+++++print+bonds
+++++print+i
+++++i+%2B=+1|code-block|syntax|javascript|1325618|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 just add a variable i, which counts +1 everytime the loop starts again.

<pre><code>i = 0
for bonds in number_of_bonds: 
 if bonds &gt;= 2:
 print bonds
 print i
 i += 1
</code></pre>

blocks|key|3762431|text|for+i+in+range(len(number_of_bonds)):
++++if+number_of_bonds[i]+>=+2:
++++++++print+i,+number_of_bonds[i]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3762432|为什么不在for循环中使用计数器呢？|unstyled|3762433|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>for i in range(len(number_of_bonds)):
 if number_of_bonds[i] &gt;= 2:
 print i, number_of_bonds[i]
</code></pre>

why not just use a counter in the for loop

blocks|key|1447018|text|在函数风格中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1447019|filter(lambda+(a,b):+b+>=+2,+enumerate(L))|code-block|syntax|javascript|1447020|结果：|1447021|[(2,+3),+(3,+3),+(4,+3),+(5,+3),+(6,+3),+(7,+3),+(8,+3),+(9,+3),+(10,+3),+(11,+3),+(12,+3),+(13,+3),+(14,+3),+(15,+3)]|1447022|它使用enumerate+function和filter+function。|offset|length|style|CODE|1447023|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/functions.html#enumerate|1|http://docs.python.org/2/library/functions.html#filter^0|0|0|0|0|3|9|M|6|3|I|0|M|F|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@$M|15|N|16|O|P]|$M|17|N|18|O|P]]|9|@$M|19|N|1A|1|1B]|$M|1C|N|1D|1|1E]]|A|$]]|$1|Q|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

In functional style:

<pre><code>filter(lambda (a,b): b &gt;= 2, enumerate(L))
</code></pre>

Result:

<pre><code>[(2, 3), (3, 3), (4, 3), (5, 3), (6, 3), (7, 3), (8, 3), (9, 3), (10, 3), (11, 3), (12, 3), (13, 3), (14, 3), (15, 3)]
</code></pre>

This uses <a href="http://docs.python.org/2/library/functions.html#enumerate" rel="nofollow"><code>enumerate</code> function</a> and <a href="http://docs.python.org/2/library/functions.html#filter" rel="nofollow"><code>filter</code> function</a>.

blocks|key|3762519|text|所以，如果你想打印original列表中的位置(在之前做任何过滤)，这是我认为你想要做的，那么首先枚举项目，然后对每个项目过滤掉不感兴趣的项目。否则，如果你想打印元素在新的过滤列表中的位置，那么kunev的答案是正确的。下面是我想说的代码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3762520|for+i,bond+in+enumerate(number_of_bonds):
++++if+bond+>=2:
++++++++print+(i,bond)|code-block|syntax|javascript|3762521|entityMap^0|9|8|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|$]]

So, if you want to print the position in the original list (previous to do any filtering), which is what I think you want to do, then the idea to enumerate the items first and then, for each of them, filter out the ones which are not of the interest. Otherwise, if you want to print the position of the element in the new filtered list, then kunev answer is the right one.
Here's the code for what I mean:

<pre><code>for i,bond in enumerate(number_of_bonds):
 if bond &gt;=2:
 print (i,bond)
</code></pre>

I have a list that goes:

[1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1]

I'd like to print all the elements of the list with a value of greater than 2 but also print the position in the list, I can print the elements by themselves with:

<pre><code>for bonds in number_of_bonds: 
 if bonds &gt;= 2:
 print bonds
</code></pre>

But if I use enumerate it will instead print all the items in the list with their position like this:

<pre><code>(0, 1)
(1, 1)
(2, 3)
</code></pre>

etc

So what is the correct way to do this?

Numbering elements in a 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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我有一个清单是这样写的：1、1、3、3、3、3、3、3、3、3、3、3、3、3、3、3、1、1、1、1、1、1、1我想打印列表中值大于2的所有元素，但也打印列表中的位置，我可以使用以下命令打印元素：for bonds in number_of_bonds:           if bonds >= 2:       ...

问给列表中的元素编号
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问给列表中的元素编号EN