英文:
How to find the first number of multiple consecutive sequence in list
问题
我使用输入
[0,2,3,4,6,7,8,16,17,18,21,23,24,26,34,35,36,37,38,40,41,46,47]
但输出
[0,2,6,16,21,23,26,34,40,46] 是错误的
应该是 [2,6,16,23,34,40,46]
如何找到列表中多个连续序列的第一个数字？
如果可能的话，是否可以修改以找到最后一个数字
例如 [4, 8, 18, 24, 38, 41, 47]
def find_consecutive_numbers(lst):result = []i = 0while i < len(lst)-2:if lst[i]+1 == lst[i+1] and lst[i+1]+1 == lst[i+2]:result.append(lst[i])i += 2else:i += 1return result
我曾尝试询问ChatGPT，但结果并不完美。
它首先使用字符串处理。
但列表中也有一些意外的数字。
英文:
I use input
[0,2,3,4,6,7,8,16,17,18,21,23,24,26,34,35,36,37,38,40,41,46,47]
But output
[0,2,6,16,21,23,26,34,40,46] which is wrong
Should be [2,6,16,23,34,40,46]
How to find the first number of multiple consecutive sequence in list?
If it is possible, can be editable to find the last one too
Such that [4, 8, 18, 24, 38, 41, 47]
def find_consecutive_numbers(lst):result = []i = 0while i < len(lst)-2:if lst[i]+1 == lst[i+1] and lst[i+1]+1 == lst[i+2]:result.append(lst[i])i += 2else:i += 1return result
I had tried to ask chatgpt, but result is not perfect.
It use string to do first.
But some unexpected numbers in list too
答案1
得分: 0
尝试按照与前一个元素的差异进行分组，并返回每个差异为1的组中的第一个元素：
from itertools import pairwise, groupbydef find_consecutive_numbers(lst, last=False):for k, g in groupby(pairwise(lst), key=lambda x: x[1]-x[0]):if k == 1:if last:*_, (_, end) = gyield endelse:yield next(g)[0]# 示例用法>>> list(find_consecutive_numbers(l))[2, 6, 16, 23, 34, 40, 46]>>> list(find_consecutive_numbers(l, last=True))[4, 8, 18, 24, 38, 41, 47]一些文档链接：* [`itertools.pairwise`][0]* [`itertools.groupby`][1]* [生成器函数 (`yield`)][2]* [`next`][3][0]: https://docs.python.org/3/library/itertools.html#itertools.pairwise[1]: https://docs.python.org/3/library/itertools.html#itertools.groupby[2]: https://stackoverflow.com/questions/231767/what-does-the-yield-keyword-do-in-python[3]: https://docs.python.org/3/library/functions.html#next
英文:
Try grouping by difference to preceding element and yield the first of each group that difference is 1:
from itertools import pairwise, groupbydef find_consecutive_numbers(lst, last=False):for k, g in groupby(pairwise(lst), key=lambda x: x[1]-x[0]):if k == 1:if last:*_, (_, end) = gyield endelse:yield next(g)[0]>>> list(find_consecutive_numbers(l))[2, 6, 16, 23, 34, 40, 46]>>> list(find_consecutive_numbers(l, last=True))[4, 8, 18, 24, 38, 41, 47]
Some docs:
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