我想迭代非连续范围。换句话说,根据用户输入跳过某些索引/元素,如此。
>>> 0
>>> 3
>>> 4
>>> 6
>>> 7
>>> 8
>>> 9
我无法弄清楚我是如何尝试这种方法的。
import re
>>> crazyrange = '0, 3-4, 6, 7-9' # User input string
>>> crazyrange = re.split (', ', crazyrange)
>>> crazyrange
['0', '3-4', '6', '7-9']
>>>
>>> masterlist = []
>>> def get_range (values):
rangelist = []
for i in range (int (values [0]), int (values [1]) + 1):
rangelist.append (i)
return rangelist
>>>
>>> for i in crazyrange:
if '-' in i:
values = i.split ('-')
masterlist.extend (get_range (values))
else:
masterlist.append (int (i))
>>>
>>> masterlist
[0, 3, 4, 6, 7, 8, 9]
有更好的方法吗?
答案 0 :(得分:2)
你的代码看起来很不错,但是你不需要re
,你可以通过使用python的range
函数和yield
来自生成器函数的项来整理它:
crazyrange = '0, 3-4, 6, 7-9' # User input string
def get_range(range_string):
items = range_string.replace(" ", "").split(",")
for i in items:
if '-' in i:
start, end = i.split('-')
for j in range(int(start), int(end)+1): # range() is not inclusive
yield j
else:
yield int(i)
for x in get_range(crazyrange):
print x
print list(get_range(crazyrange))
# [0, 3, 4, 6, 7, 8, 9]
还有一个没有显式循环的选项:
crazyrange = '0, 3-4, 6, 7-9' # User input string
import itertools
items = [i.split("-") for i in crazyrange.split(", ")]
range_list = [range(int(i[0]), int(i[-1])+1) for i in items]
print list(itertools.chain(*range_list))
或者这个选项可能是最简单的:
crazyrange = '0, 3-4, 6, 7-9' # User input string
def get_range(item):
pair = item.split('-') # may only be one item, but that's OK
return range(int(pair[0]), int(pair[-1])+1)
items = crazyrange.split(', ')
print [i for item in items for i in get_range(item)]
# [0, 3, 4, 6, 7, 8, 9]
答案 1 :(得分:1)
使用list comprehension
:
crazyrange = '0, 3-4, 6, 7-9'
ranges = [range(int(k.split("-")[0]), int(k.split("-")[-1]) +1) for k in crazyrange.split(',')]
print(ranges)
print([j for k in ranges for j in k])
输出:
[range(0, 1), range(3, 5), range(6, 7), range(7, 10)]
[0, 3, 4, 6, 7, 8, 9]