Question

我希望能够将诸如“1,2,5-7,10”之类的字符串转换为诸如[1,2,5,6,7,10]之类的python列表。我环顾四周找到this，但我想知道在Python中是否有一种简洁明了的方法。

Answer 1

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

Answer 2

我能够对这个问题做出真正的理解：

>>> 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(',')), [])

>>> 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]

假装理解的另一个答案只是for循环，因为最终的列表被丢弃了。：）

对于python 2，您甚至可以删除对list！

的调用

Answer 3

这可能是一种矫枉过正，但我只是喜欢pyparsing：

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 << term + Optional(Literal(',').suppress() + expr)
    return expr.parseString(s, parseAll=True)

if __name__=='__main__':
    print parsestring('1,2,5-7,10')

Answer 4

唉，答案很冗长！这是一个简短而优雅的答案：

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(',')])

演示：

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

可轻松修改以处理负数或返回列表。还需要from itertools import chain，但如果您不使用sum(...,[])个对象（或range）并且您不关心懒惰，则可以将sum(map(list,iters),[])替换为{{1}}。

Answer 5

没有理解胜过我的！

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))), [])

最好的部分是我在理解的过程中两次使用变量i。

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

Answer 6

很短，很优雅（imho）：

>>> txt = "1,2,5-7,10"
>>> # construct list of xranges
>>> xranges = [(lambda l: xrange(l[0], l[-1]+1))(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]

Answer 7

从这里：https://gist.github.com/raczben/76cd1229504d82115e6427e00cf4742c

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)]

如何将数字字符串范围转换为Python中的列表

7 个答案: