在由三个或更多零分隔的部分中对列表元素进行计数和求和
我有一个整数列表，我想根据某个条件将它们分开。我想获取列表元素的总和和计数，当三个或更多连续元素等于 0 时停止；然后 sum 和 count 订单从停止的地方重新开始。
例如，列表的一部分是：
[8, 2, 1, 1, 2, 0, 0, 0, 0, 0, 6, 0, 2, 0, 0, 0, 8, 0, 0, 2, 0, 0, 0, 6, 0, 0]
[8, 2, 1, 1, 2, 0, 0, 0, 0, 0, 6, 0, 2, 0, 0, 0, 8, 0, 0, 2, 0, 0, 0, 6, 0, 0]
该过程将是：
所以我想要的输出是：
[[14, 5], [8, 3], [10, 4], [6, 3]]
到目前为止我所写的计算总和没问题，但我的问题是部分内的零不计入长度。
当前（不正确）输出：
[[14, 5], [8, 2], [10, 2], [6, 2]]
代码：
arr = [8, 2, 1, 1, 2, 0, 0, 0, 0, 0, 6, 0, 2, 0, 0, 0, 8, 0, 0, 2, 0, 0, 0, 6, 0, 0]
result = []
summed, count = 0, 0
for i in range(0, len(arr) - 2):
el, el1, el2 = arr[i], arr[i + 1], arr[i + 2]
if el != 0:
summed = summed + el
count = count + 1
if el == 0 and el1 == 0 and el2 == 0:
if summed != 0:
result.append([summed, count])
summed = 0
count = 0
elif i == len(arr) - 3:
summed = el + el1 + el2
count = count + 1
result.append([summed, count])
break
print(result)
回答
很难理解你的代码是做什么的。使用字符串似乎更简单易读，您的输出只需两行即可实现（感谢@CrazyChucky 的改进）：
import re
arr = [8, 2, 1, 1, 2, 0, 0, 0, 0, 0, 6, 0, 2, 0, 0, 0, 8, 0, 0, 2, 0, 0, 0, 6, 0, 0]
# Convert to String by joining integers, and split into substrings, the separator being three zeros or more
strings = re.split(r'0{3,}', ''.join(str(i) for i in arr))
# Sums and counts using list comprehensions
output = [[sum(int(x) for x in substring), len(substring)] for substring in strings]
输出：
>>>output
>>>[[14, 5], [8, 3], [10, 4], [6, 3]]
请记住，可读性始终是任何代码中最重要的因素。一个人应该第一次阅读你的代码并理解它是如何工作的。
如果完整列表包含多于一位的数字，您可以执行以下操作：
# Convert to String by joining integers, seperating them by a commade, and split into substrings, the separator being three zeros or more
strings = re.split(r',?(?:0,){3,}', ','.join(str(i) for i in arr))
# Make a list of numbers from those strings
num_lists = [string.split(',') for string in strings]
# # Sums and counts using list comprehensions
output = [[sum(int(x) for x in num_list), len(num_list)] for num_list in num_lists]
- @MostafaGomaa The output only contains integers. I am not sure what you mean by 'data'. I am only using strings temporarely to find the output.