英文:
Hi, am having trouble on how to code [False, False, True, True, True, False, True, True, False] into [[2,3],[6,2]] in python
问题
将[False, False, True, True, True, False, True, True, False] 编码成 [[2, 3], [6, 2]] 的Python代码应该如何写，其中第一个子列表中的2是True的索引起始位置，3是从其索引开始的True元素的总和。被卡在这个问题上。
英文:
How should I code [False, False, True, True, True, False, True, True, False] into [[2, 3], [6, 2]] in python, where for the first sublist, 2 is where the True index starts and 3 is the sum of True elements from where its index starts.
Stuck with this question
答案1
得分: 1
你可以使用 itertools.groupby 和 enumerate：
from itertools import groupby[(next(g)[0], 1+len(list(g))) for k, g in groupby(enumerate(lst), key=lambda x: x[1]) if k]# [(2, 3), (6, 2)]
英文:
You can use itertools.groupbyand enumerate:
>>> from itertools import groupby[(next(g)[0], 1+len(list(g))) for k, g in groupby(enumerate(lst), key=lambda x: x[1]) if k]# [(2, 3), (6, 2)]
答案2
得分: 0
使用简单的循环和标志来跟踪上一个状态：
l = [False, False, True, True, True, False, True, True, False]out = []previous = Falsefor i, x in enumerate(l): # 遍历值和位置if x: # 如果为Trueif not previous: # 并且之前为False，添加新的列表out.append([i, 1])else: # 否则增加计数器out[-1][1] += 1previous = x # 更新下一次迭代的标志
输出：[[2, 3], [6, 2]]
为了好玩，可以使用 itertools.groupby 来实现一行的替代方法：
from itertools import groupbyout = [[(l:=list(g))[0][0], len(l)]for k, g in groupby(enumerate(l), key=lambda x: x[1]) if k]
英文:
Use a simple loop and a flag to keep track of the last state:
l = [False, False, True, True, True, False, True, True, False]out = []previous = Falsefor i, x in enumerate(l): # iterate over values and positionsif x: # if Trueif not previous: # and we had a False before, add new listout.append([i, 1])else: # else increment the counterout[-1][1] += 1previous = x # update flag for next iteration
Output: [[2, 3], [6, 2]]
For fun, a one-liner alternative using itertools.groupby:
from itertools import groupbyout = [[(l:=list(g))[0][0], len(l)]for k, g in groupby(enumerate(l), key=lambda x: x[1]) if k]
答案3
得分: -1
检查这段代码
lst = [False, False, True, True, True, False, True, True, False]result = []start_index = Nonetrue_count = 0for i, value in enumerate(lst):if value:if start_index is None:start_index = itrue_count += 1elif start_index is not None:result.append([start_index, true_count])start_index = Nonetrue_count = 0if start_index is not None:result.append([start_index, true_count])print(result)
英文:
check this code
lst = [False, False, True, True, True, False, True, True, False]result = []start_index = Nonetrue_count = 0for i, value in enumerate(lst):if value:if start_index is None:start_index = itrue_count += 1elif start_index is not None:result.append([start_index, true_count])start_index = Nonetrue_count = 0if start_index is not None:result.append([start_index, true_count])print(result)
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论