blocks|key|1202362|text|def+f(x):
++++result+=+[]
++++for+part+in+x.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

>>>+f('1,2,5-7,10')
[1,+2,+5,+6,+7,+10]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1202363|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 f(x):
 result = []
 for part in x.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

&gt;&gt;&gt; f('1,2,5-7,10')
[1, 2, 5, 6, 7, 10]
</code></pre>

blocks|key|4889956|text|我能够对这个问题做出一个true的理解：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|4889957|>>>+def+f(s):
++++return+sum(((list(range(*[int(j)+%2B+k+for+k,j+in+enumerate(i.split('-'))]))
+++++++++if+'-'+in+i+else+[int(i)])+for+i+in+s.split(',')),+[])

>>>+f('1,2,5-7,10')
[1,+2,+5,+6,+7,+10]

>>>+f('1,3-7,10,11-15')
[1,+3,+4,+5,+6,+7,+10,+11,+12,+13,+14,+15]|code-block|syntax|javascript|4889958|另一个假装理解的答案只是一个for循环，因为最终的列表被丢弃了。:)|4889959|对于Python2，您甚至可以删除对list的调用！|CODE|4889960|entityMap^0|C|4|0|0|0|I|4|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|U|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|V|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|W|8|@$9|X|A|Y|B|O]]|D|@]|E|$]]|$1|P|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|Q|$]]

I was able to make a true comprehension on that question:

<pre><code>&gt;&gt;&gt; def f(s):
 return sum(((list(range(*[int(j) + k for k,j in enumerate(i.split('-'))]))
 if '-' in i else [int(i)]) for i in s.split(',')), [])

&gt;&gt;&gt; f('1,2,5-7,10')
[1, 2, 5, 6, 7, 10]

&gt;&gt;&gt; f('1,3-7,10,11-15')
[1, 3, 4, 5, 6, 7, 10, 11, 12, 13, 14, 15]
</code></pre>

the other answer that pretended to have a comprehension was just a for loop because the final list was discarded. :)

For python 2 you can even remove the call to <code>list</code>!

blocks|key|4889921|text|这可能有点夸张，但我就是喜欢pyparsing：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4889922|from+pyparsing+import+*

def+return_range(strg,+loc,+toks):
++++if+len(toks)==1:
++++++++return+int(toks[0])
++++else:
++++++++return+range(int(toks[0]),+int(toks[1])%2B1)
def+parsestring(s):
++++expr+=+Forward()
++++term+=+(Word(nums)+%2B+Optional(Literal('-').suppress()+%2B+Word(nums))).setParseAction(return_range)
++++expr+<<+term+%2B+Optional(Literal(',').suppress()+%2B+expr)
++++return+expr.parseString(s,+parseAll=True)

if+__name__=='__main__':
++++print+parsestring('1,2,5-7,10')|code-block|syntax|javascript|4889923|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|$]]

This might be an overkill, but I just like pyparsing:

<pre><code>from pyparsing import *

def return_range(strg, loc, toks):
 if len(toks)==1:
 return int(toks[0])
 else:
 return range(int(toks[0]), int(toks[1])+1)
def parsestring(s):
 expr = Forward()
 term = (Word(nums) + Optional(Literal('-').suppress() + Word(nums))).setParseAction(return_range)
 expr &lt;&lt; term + Optional(Literal(',').suppress() + expr)
 return expr.parseString(s, parseAll=True)

if __name__=='__main__':
 print parsestring('1,2,5-7,10')
</code></pre>

blocks|key|2818100|text|呃，答案太冗长了！以下是一个简短而优雅的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2818101|def+rangeString(commaString):
++++def+hyphenRange(hyphenString):
++++++++x+=+[int(x)+for+x+in+hyphenString.split('-')]
++++++++return+range(x[0],+x[-1]%2B1)
++++return+chain(*[hyphenRange(r)+for+r+in+commaString.split(',')])|code-block|syntax|javascript|2818102|演示：|2818103|>>>+list(+f('1,2,5-7,10')+)
[1,+2,+5,+6,+7,+10]|2818104|易于修改以处理负数或返回列表。也需要from+itertools+import+chain，但是如果您不使用range对象(或sum(map(list,iters),[]))并且不关心懒惰，那么您可以用sum(...,[])来代替它。|offset|length|style|CODE|2818105|entityMap^0|0|0|0|0|I|R|1I|5|1R|N|2T|B|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]|$M|11|N|12|O|P]|$M|13|N|14|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

Ugh, the answers are so verbose! Here is a short and elegant answer:

<pre><code>def rangeString(commaString):
 def hyphenRange(hyphenString):
 x = [int(x) for x in hyphenString.split('-')]
 return range(x[0], x[-1]+1)
 return chain(*[hyphenRange(r) for r in commaString.split(',')])
</code></pre>

Demo:

<pre><code>&gt;&gt;&gt; list( f('1,2,5-7,10') )
[1, 2, 5, 6, 7, 10]
</code></pre>

Easily modifiable to handle negative numbers or return a list. Also will need <code>from itertools import chain</code>, but you can substitute <code>sum(...,[])</code> for it if you are not working with <code>range</code> objects (or <code>sum(map(list,iters),[])</code>) and you don't care about being lazy.

blocks|key|1202535|text|没有什么能比我的理解更好！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1202536|import+re
def+convert(x):
++++return+sum((i+if+len(i)+==+1+else+list(range(i[0],+i[1]%2B1))
+++++++++++++++for+i+in+([int(j)+for+j+in+i+if+j]+for+i+in
+++++++++++++++re.findall('(\d%2B),?(?:-(\d%2B))?',+x))),+[])|code-block|syntax|javascript|1202537|最好的部分是我在理解过程中使用了两次变量i。|offset|length|style|CODE|1202538|>>>+convert('1,2,5-7,10')
[1,+2,+5,+6,+7,+10]|1202539|entityMap^0|0|0|K|1|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

No comprehension beats mine!

<pre><code>import re
def convert(x):
 return sum((i if len(i) == 1 else list(range(i[0], i[1]+1))
 for i in ([int(j) for j in i if j] for i in
 re.findall('(\d+),?(?:-(\d+))?', x))), [])
</code></pre>

The best part is that I use variable <code>i</code> twice in the middle of the comprehension.

<pre><code>&gt;&gt;&gt; convert('1,2,5-7,10')
[1, 2, 5, 6, 7, 10]
</code></pre>

blocks|key|2818210|text|非常简短，优雅(imho)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2818211|>>>+txt+=+"1,2,5-7,10"
>>>+#+construct+list+of+xranges
>>>+xranges+=+[(lambda+l:+xrange(l[0],+l[-1]%2B1))(map(int,+r.split('-')))+for+r+in+txt.split(',')]
>>>+#+flatten+list+of+xranges
>>>+[y+for+x+in+xranges+for+y+in+x]
[1,+2,+5,+6,+7,+10]|code-block|syntax|javascript|2818212|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|$]]

Very short, and elegant (imho):

<pre><code>&gt;&gt;&gt; txt = "1,2,5-7,10"
&gt;&gt;&gt; # construct list of xranges
&gt;&gt;&gt; xranges = [(lambda l: xrange(l[0], l[-1]+1))(map(int, r.split('-'))) for r in txt.split(',')]
&gt;&gt;&gt; # flatten list of xranges
&gt;&gt;&gt; [y for x in xranges for y in x]
[1, 2, 5, 6, 7, 10]
</code></pre>

blocks|key|1202578|text|从这里：https://gist.github.com/raczben/76cd1229504d82115e6427e00cf4742c|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1202579|def+number(a,+just_try=False):
++++"""
++++Parse+any+representation+of+number+from+string.
++++"""
++++try:
++++++++#+First,+we+try+to+convert+to+integer.
++++++++#+(Note,+that+all+integer+can+be+interpreted+as+float+and+hex+number.)
++++++++return+int(a)
++++except:
++++++++#+The+order+of+the+following+convertions+doesn't+matter.
++++++++#+The+integer+convertion+has+failed+because+`a`+contains+hex+digits+[x,a-f]+or+a+decimal
++++++++#+point+['.'],+but+not+both.
++++++++try:
++++++++++++return+int(a,+16)
++++++++except:
++++++++++++try:
++++++++++++++++return+float(a)
++++++++++++except:
++++++++++++++++if+just_try:
++++++++++++++++++++return+a
++++++++++++++++else:
++++++++++++++++++++raise


def+str2numlist(s):
++++"""
++++Convert+a+string+parameter+to+iterable+object.
++++"""
++++return+[y+for+x+in+s.split(',')+for+y+in+str_ranges_to_list(x)+]


def+str_ranges_to_list(s):
++++"""
++++Convert+a+string+parameter+to+iterable+object.
++++"""
++++s+=+s.strip()
++++try:
++++++++begin,end=s.split(':')
++++++++return+range(number(begin),+number(end))
++++except+ValueError:+#+not+enough+values+to+unpack
++++++++return+[number(s)]|code-block|syntax|javascript|1202580|entityMap|0|LINK|mutability|MUTABLE|url|https://gist.github.com/raczben/76cd1229504d82115e6427e00cf4742c^0|4|1S|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 here: <a href="https://gist.github.com/raczben/76cd1229504d82115e6427e00cf4742c" rel="nofollow noreferrer">https://gist.github.com/raczben/76cd1229504d82115e6427e00cf4742c</a>
<pre><code>def number(a, just_try=False):
 &quot;&quot;&quot;
 Parse any representation of number from string.
 &quot;&quot;&quot;
 try:
 # First, we try to convert to integer.
 # (Note, that all integer can be interpreted as float and hex number.)
 return int(a)
 except:
 # The order of the following convertions doesn't matter.
 # The integer convertion has failed because `a` contains hex digits [x,a-f] or a decimal
 # point ['.'], but not both.
 try:
 return int(a, 16)
 except:
 try:
 return float(a)
 except:
 if just_try:
 return a
 else:
 raise


def str2numlist(s):
 &quot;&quot;&quot;
 Convert a string parameter to iterable object.
 &quot;&quot;&quot;
 return [y for x in s.split(',') for y in str_ranges_to_list(x) ]


def str_ranges_to_list(s):
 &quot;&quot;&quot;
 Convert a string parameter to iterable object.
 &quot;&quot;&quot;
 s = s.strip()
 try:
 begin,end=s.split(':')
 return range(number(begin), number(end))
 except ValueError: # not enough values to unpack
 return [number(s)]
</code></pre>

blocks|key|2818231|text|正在初始化字符串|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2818232|test_str+=+"1,+4-6,+8-10,+11"|code-block|syntax|javascript|2818233|打印原始字符串|2818234|print("The+original+string+is+:+"+%2B+test_str)|2818235|将字符串范围转换为列表|2818236|res+=+sum(((list(range(*[int(b)+%2B+c+
+++++++++++for+c,+b+in+enumerate(a.split('-'))]))
+++++++++++if+'-'+in+a+else+[int(a)])+for+a+in+test_str.split(',+')),+[])|2818237|打印结果|2818238|print("List+after+conversion+from+string+:+"+%2B+str(res))|2818239|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|$]]

Initializing string
<pre><code>test_str = &quot;1, 4-6, 8-10, 11&quot;
</code></pre>
printing original string
<pre><code>print(&quot;The original string is : &quot; + test_str)
</code></pre>
Convert String ranges to list
<pre><code>res = sum(((list(range(*[int(b) + c 
 for c, b in enumerate(a.split('-'))]))
 if '-' in a else [int(a)]) for a in test_str.split(', ')), [])
</code></pre>
printing result
<pre><code>print(&quot;List after conversion from string : &quot; + str(res))
</code></pre>

I would like to be able to convert a string such as "1,2,5-7,10" to a python list such as [1,2,5,6,7,10]. I looked around and found <a href="https://stackoverflow.com/questions/2668107">this</a>, but I was wondering if there is a clean and simple way to do this in Python.

How to convert numeric string ranges to a list in Python

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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我希望能够将像"1,2,5-7,10“这样的字符串转换成像1,2,5,6,7,10这样的Python列表。我环顾四周，找到了，但我想知道在Python语言中是否有一种干净而简单的方法来做到这一点。

问如何在Python中将数字字符串范围转换为列表
EN

回答 8

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问如何在Python中将数字字符串范围转换为列表
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