如何在Python中将数字字符串范围转换为列表
我希望能够将字符串(如“1,2,5-7,10”)转换为python列表(如[1,2,5,6,7,10])。我环顾四周,发现了这个问题,但我想知道在Python中是否有一种干净而简单的方法可以做到这一点。这可能有些过分,但我喜欢pyparsing:如何在Python中将数字字符串范围转换为列表,python,list,string,Python,List,String,我希望能够将字符串(如“1,2,5-7,10”)转换为python列表(如[1,2,5,6,7,10])。我环顾四周,发现了这个问题,但我想知道在Python中是否有一种干净而简单的方法可以做到这一点。这可能有些过分,但我喜欢pyparsing: def f(x): result = [] for part in x.split(','): if '-' in part: a, b = part.split('-')
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]
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')
从pyparsing导入*
def回流范围(strg、loc、toks):
如果len(toks)==1:
返回整数(toks[0])
其他:
返回范围(int(toks[0]),int(toks[1])+1)
def解析字符串:
expr=Forward()
term=(Word(nums)+可选(Literal('-')。suppress()+Word(nums)))。setParseAction(返回范围)
expr我能够对这个问题做出正确的理解:
>>> 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]
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。:)
对于Python2,您甚至可以删除对列表的调用 啊,答案太冗长了!下面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:
易于修改以处理负数或返回列表。还需要从itertools导入链中获取,但如果您不使用范围对象(或sum(map(list,iter),[])
),并且您不在乎懒惰,您可以用sum(…,[])
替换它。没有理解能力打败我
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))), [])
<>最好的部分是在理解过程中使用变量<代码> i <代码>两次。
>>> convert('1,2,5-7,10')
[1, 2, 5, 6, 7, 10]
非常短,优雅(imho):
从这里开始:
这将是一个很好的生成器。通过将else替换为elif part!='',可以实现在前导或尾随逗号上不爆炸代码>
>>> convert('1,2,5-7,10')
[1, 2, 5, 6, 7, 10]
>>> 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]
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)]