blocks|key|5212270|text|最简单的方法是在排序时将文件名填零：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5212271|def+group_uniqmz_intensities(path):
++++pxlist+=+sorted(get_pixelist(path),+key=lambda+f:+f.rjust(17,+'0'))|code-block|syntax|javascript|5212272|在排序时，它会用0字符将每个文件名填充到17个字符；因此，将1_00000_6.54.csv填充到01_00000_6.54.csv中，而将10_00000_1.70.csv保留为原样。在词法上，01在10之前排序。|offset|length|style|CODE|5212273|我选择17作为硬编码值，以简化操作；您可以使用下面的代码自动找到所需的值：|5212274|def+group_uniqmz_intensities(path):
++++padsize+=+max(len(f)+for+f+in+pxlist)
++++pxlist+=+sorted(get_pixelist(path),+key=lambda+f:+f.rjust(padsize,+'0'))|5212275|entityMap^0|0|0|8|1|U|G|1D|H|1Y|H|2Q|2|2T|2|0|0|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]]|$1|G|3|H|5|6|7|U|8|@$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]|$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|R|$]]

The easiest would be to zero-pad the filenames when sorting:

<pre><code>def group_uniqmz_intensities(path):
 pxlist = sorted(get_pixelist(path), key=lambda f: f.rjust(17, '0'))
</code></pre>

which will pad each filename to 17 characters with <code>0</code> characters when sorting; so <code>1_00000_6.54.csv</code> is padded to <code>01_00000_6.54.csv</code> while <code>10_00000_1.70.csv</code> is left as is. Lexographically, <code>01</code> sorts before <code>10</code>.

I picked 17 as a hardcoded value to simplify things; you could find the required value automatically by using this instead:

<pre><code>def group_uniqmz_intensities(path):
 padsize = max(len(f) for f in pxlist)
 pxlist = sorted(get_pixelist(path), key=lambda f: f.rjust(padsize, '0'))
</code></pre>

blocks|key|1310299|text|由于'1‘<+'_’，你得到的是第二个排序。你可以通过给sorted一个key-function来实现你的目标：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1310300|+def+group_uniqmz_intensities(path):
+++++pxlist+=+sorted(get_pixelist(path),+key=lambda+x:+int(x.split("_")[0]))|code-block|syntax|javascript|1310301|请确保您的所有文件都遵循相同的命名方案({+ValueError+}_{rest}编号)，否则将出现编号。|1310302|编辑:+Martijn提供了一个更优雅的解决方案。|1310303|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Since '1' &lt; '_' you get the second ordering. You can achieve your goal by giving a key-function to sorted:

<pre><code> def group_uniqmz_intensities(path):
 pxlist = sorted(get_pixelist(path), key=lambda x: int(x.split("_")[0]))
</code></pre>

Please make sure ALL of your files are following the same naming scheme ({number}_{rest}.csv) otherwise there will be a ValueError.

EDIT: Martijn Pieters provides a more elegant solution.

blocks|key|1428878|text|基于字母数字排序的this+answer：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1428879|def+group_uniqmz_intensities(path):
++++pxlist+=+sorted(get_pixelist(path),+key=lambda+filename:+int(filename.partition('_')[0]))|code-block|syntax|javascript|1428880|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/2669523/50065^0|9|B|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]]]]

Based on <a href="https://stackoverflow.com/a/2669523/50065">this answer</a> for alphanumerical sorting:

<pre><code>def group_uniqmz_intensities(path):
 pxlist = sorted(get_pixelist(path), key=lambda filename: int(filename.partition('_')[0]))
</code></pre>

blocks|key|5212404|text|下面是自然排序的简单实现，假设您的字段都被_拆分|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5212405|def+int_if_possible(s):
++++try:
++++++++return+int(s)
++++except:
++++++++return+s


>>>+sorted(s,+key=lambda+s:+map(int_if_possible,+s.split('_')))
['1_00000_6.54.csv',
+'2_00000_1.70.csv',
+'3_00000_1.70.csv',
+'10_00000_1.70.csv',
+'11_00000_1.70.csv']|code-block|syntax|javascript|5212406|此实现利用了列表逐个元素进行比较的事实。如果元素可以转换为整数，我们就将它们作为整数进行比较，否则我们就会回到字符串比较。|5212407|5212408|编辑：这里介绍了一种更复杂的自然排序解决方案：Natural+string+sorting。|BOLD|5212409|它非常聪明:它使用正则表达式\d%2B\D%2B将输入字符串拆分成交替的数字和非数字。然后按数字比较数字，按字母顺序比较非数字。|5212410|entityMap|0|LINK|mutability|MUTABLE|url|http://code.activestate.com/recipes/285264-natural-string-sorting/^0|L|1|0|0|0|0|0|2|N|M|0|0|E|6|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|15|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|16|8|@$9|17|A|18|B|P]]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|Q|3|R|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

Here's a trivial implementation of natural ordering, assuming that your fields are all split by <code>_</code>:

<pre><code>def int_if_possible(s):
 try:
 return int(s)
 except:
 return s


&gt;&gt;&gt; sorted(s, key=lambda s: map(int_if_possible, s.split('_')))
['1_00000_6.54.csv',
 '2_00000_1.70.csv',
 '3_00000_1.70.csv',
 '10_00000_1.70.csv',
 '11_00000_1.70.csv']
</code></pre>

This implementation leverages the fact that lists get compared element-by-element. If the elements are convertible to ints, we compare them as ints, otherwise we fall back to string comparison.

<hr>

Edit: A more elaborate solution for natural sorting is presented here: <a href="http://code.activestate.com/recipes/285264-natural-string-sorting/" rel="nofollow">Natural string sorting</a>.

It's pretty clever: it uses a regex <code>\d+\D+</code> to split input strings into alternating numbers and non-numbers. Then numbers are compared numerically, and non-numbers alphabetically.

I want to know how I can sort the filenames as they are in the directory. For example, I have the following names:

<pre><code>1_00000_6.54.csv
2_00000_1.70.csv
3_00000_1.70.csv
...
10_00000_1.70.csv
11_00000_1.70.csv
...
</code></pre>

With the following python code I get the following order:

<pre><code> def get_pixelist(path):
 return [os.path.join(path,f) for f in os.listdir(path) if f.endswith('.csv')]

 def group_uniqmz_intensities(path):
 pxlist = sorted(get_pixelist(path))
</code></pre>

gives:

<pre><code>1_00000_6.54.csv
10_00000_1.70.csv
11_00000_1.70
...
2_00000_1.70.csv
...
3_00000_1.70.csv
...
</code></pre>

I want the order shown before.

sort files in a specific order

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

我想知道如何对目录中的文件名进行排序。例如，我有以下名称：1_00000_6.54.csv2_00000_1.70.csv3_00000_1.70.csv...10_00000_1.70.csv11_00000_1.70.csv...使用下面的python代码，我得到了以下顺序： def get_pixelist(pa...

问按特定顺序对文件进行排序
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

问按特定顺序对文件进行排序EN