blocks|key|676498|text|我认为这会输出你所需要的，它有点长，但是演示了这个过程。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|676499|string+=+'1,2,3-6,10'
splt_a+=+string.split(",")
o+=+[]

for+s+in+splt_a:
++++if+"-"+in+s:
++++++++splt_b+=+s.split("-")
++++++++for+i+in+range(int(splt_b[0]),+int(splt_b[1])+%2B+1):
++++++++++++o.append(i)
++++else:
++++++++o.append(int(s))

print(o)+#+[1,+2,+3,+4,+5,+6,+10]|code-block|syntax|javascript|676500|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|$]]

I think this would output what you need, it's a bit long winded but demonstrates the process.

<pre><code>string = '1,2,3-6,10'
splt_a = string.split(",")
o = []

for s in splt_a:
 if "-" in s:
 splt_b = s.split("-")
 for i in range(int(splt_b[0]), int(splt_b[1]) + 1):
 o.append(i)
 else:
 o.append(int(s))

print(o) # [1, 2, 3, 4, 5, 6, 10]
</code></pre>

blocks|key|676474|text|您可以使用regex模式查找每个范围，并使用一个函数将范围替换为数字列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|676475|import+re
s+=+'1,2,3-6,10'

def+replace_range_by_ints(m):
++++a+=+m.group(1)
++++b+=+m.group(2)
++++return+','.join(str(i)+for+i+in+range(int(a),+int(b)+%2B+1))

pattern+=+re.compile('(\d%2B)\-(\d%2B)')
print(re.sub(pattern,+replace_range_by_ints,+s))
#+1,2,3,4,5,6,10|code-block|syntax|javascript|676476|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/re.html^0|5|5|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]]]]

You could use a <a href="https://docs.python.org/3/library/re.html" rel="noreferrer">regex</a> pattern to find every range and a function to replace a range by a list of numbers:

<pre><code>import re
s = '1,2,3-6,10'

def replace_range_by_ints(m):
 a = m.group(1)
 b = m.group(2)
 return ','.join(str(i) for i in range(int(a), int(b) + 1))

pattern = re.compile('(\d+)\-(\d+)')
print(re.sub(pattern, replace_range_by_ints, s))
# 1,2,3,4,5,6,10
</code></pre>

blocks|key|343822|text|如果检测到-+(有边界固定)，一行将创建一个范围，否则只会将其转换为整数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|343823|import+itertools

result+=+[x+for+y+in+'-1,1,2,3-6,10'.split(",")+for+x+in+(range(int(y.split("-")[0]),int(y.split("-")[1])%2B1)+if+y.find("-")>0+else+[int(y)])+]|code-block|syntax|javascript|343824|结果(作为整数列表)：|343825|[-1,+1,+2,+3,+4,+5,+6,+10]|343826|请注意，它也处理负数(y.find("-")>0确保-在检测范围时不处于起始位置)，但不处理负范围(必须为此选择另一个分隔符，如:)。|343827|entityMap^0|5|1|0|0|0|0|B|D|Q|1|1S|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

one-liner which creates a range if <code>-</code> is detected (with boundary fixing) else just converts to integer:

<pre><code>import itertools

result = [x for y in '-1,1,2,3-6,10'.split(",") for x in (range(int(y.split("-")[0]),int(y.split("-")[1])+1) if y.find("-")&gt;0 else [int(y)]) ]
</code></pre>

result (as a list of integers):

<pre><code>[-1, 1, 2, 3, 4, 5, 6, 10]
</code></pre>

note that it handles negative numbers as well (<code>y.find("-")&gt;0</code> ensures that <code>-</code> is not at start when detecting range), but not negative ranges (would have to choose another separator for this like <code>:</code>)

blocks|key|676526|text|您可以使用以下函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|676527|def+range_to_list(l):
+++++result+=+[]
+++++for+part+in+l.split(','):
+++++++++if+'-'+in+part:
+++++++++++++a,+b+=+part.split('-')
+++++++++++++a,+b+=+int(a),+int(b)
+++++++++++++result.extend(range(a,+b+%2B+1))
+++++++++else:
+++++++++++++a+=+int(part)
+++++++++++++result.append(a)
+++++return+result|code-block|syntax|javascript|676528|然后你会得到：|676529|>>>+myStr+=++'1,2,3-6,10'
>>>+range_to_list(myStr)
[1,+2,+3,+4,+5,+6,+10]|676530|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|$]]

You can use the following function:

<pre><code>def range_to_list(l):
 result = []
 for part in l.split(','):
 if '-' in part:
 a, b = part.split('-')
 a, b = int(a), int(b)
 result.extend(range(a, b + 1))
 else:
 a = int(part)
 result.append(a)
 return result
</code></pre>

And then you'll get:

<pre><code>&gt;&gt;&gt; myStr = '1,2,3-6,10'
&gt;&gt;&gt; range_to_list(myStr)
[1, 2, 3, 4, 5, 6, 10]
</code></pre>

blocks|key|342528|text|为什么不为这个构造构建一个简单的解析器呢？我想您计划无论如何处理来自外部数据源的某种输入，否则您的问题就没有什么意义了:手动解决一些规范要容易一些，而不是编写特殊代码来处理。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|342529|这一做法将遵循以下步骤：|342530|为输出准备一个空列表|unordered-list-item|342531|在"，“处拆分输入字符串|342532|对于每一项，请检查"-“是否存在|342533|如果没有=>，则将项转换为int并将其放入输出列表中。|342534|如果是，=>+|342535|在"-“处拆分字符串|342536|向输出列表中添加一系列int值|342537|342538|342539|​|342540|342541|返回输出列表|342542|我想这是最简单的方法。使用字符串中提供的用于拆分字符串的split方法和将字符串转换为int的int(...)函数来编写代码应该非常简单。|offset|length|style|CODE|342543|构建这样一个简单的解析器将有一个很大的优势:您可以同时验证您收到的输入是完全有效的。您甚至可以检查所提供的值是否按递增顺序排列。如果我看一下您提供的示例，这似乎是一个基本条件。|342544|entityMap^0|0|0|0|0|1|1|2|2|2|1|0|0|0|0|S|5|1B|8|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1A|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1B|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|1C|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|1D|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1E|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|1F|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|1G|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|1H|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1K|8|@]|9|@]|A|$]]|$1|W|3|V|5|6|7|1L|8|@]|9|@]|A|$]]|$1|X|3|Y|5|F|7|1M|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1N|8|@$11|1O|12|1P|13|14]|$11|1Q|12|1R|13|14]]|9|@]|A|$]]|$1|15|3|16|5|6|7|1S|8|@]|9|@]|A|$]]|$1|17|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|18|$]]

Why don't you build a simple parser for that construct? I guess you're planning to process some kind of input from external data sources anyway as otherwise your question wouldn't make much sense: It's more easy to resolve a few specifications like that manually instead of writing special code to handle it.

This approach would follow these steps:

<ul>
<li>prepare an empty list for output</li>
<li>split the input string at ","</li>
<li>for each item check if a "-" is present

<ul>
<li>if no => convert the item to an int and put it into the output list</li>
<li>if yes =>

<ul>
<li>split the string at "-"</li>
<li>add a range of int values to the output list</li>
</ul></li>
</ul></li>
<li>return the output list</li>
</ul>

I guess this is the easiest way. Should be very simple to code that yourself using the <code>split</code> method provided at strings to split strings and the <code>int(...)</code> function to convert strings to int.

Building such a simple parser will have a big advantage anyway: You can at the same time verify that the input you receive is perfectly valid. You even can check if the values provided are in increasing order. That seems to be a basic condition if I look at the example(s) you provided.

I'm working a lot with ranges and have hit a wall on this one. I'm trying to turn a range in to the full list of numbers.

<pre><code>string = '1,2,3-6,10'
print string 
</code></pre>

what i want print string to result in is '1,2,3,4,5,6,10'. So bascially showing the full range.

so the bulk of the question is how to turn 3-6 in to 3,4,5,6 using python

any help would be much appreciated!

Python: turning 1,2,3-7,8,9,10 in to string looking like this 1,2,3,4,5,6,7,8,9,10

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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我在靶场上做了很多工作，在这件事上我遇到了麻烦。我在试着把范围扩大到完整的数字列表。string = '1,2,3-6,10'print string 我希望打印字符串的结果是'1,2,3,4,5,6,10‘。所以基本地展示了整个范围。因此，主要问题是如何使用python将3-6转化为3,4,5,6。任何帮助都将不胜感激！

问Python:将1,2,3-7,8,9,10变成字符串，如1,2,3,4,5,6,7,8,9,10所示
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回答 5

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问Python:将1,2,3-7,8,9,10变成字符串，如1,2,3,4,5,6,7,8,9,10所示EN

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问Python:将1,2,3-7,8,9,10变成字符串，如1,2,3,4,5,6,7,8,9,10所示
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