我有包含30个数字的列表
list = [-21,-22,-33,-55,-454,65,48,-516,614,6,2,-64,-64,-87,6,45,87,15,11,03,-34,-6,-68,-959,-653,24,658,68,9,-2181]
现在首先我要计算连续3个正数或负数的数量。为此，我正在使用此程序：
list = [-21,-22,-33,-55,-454,65,48,-516,614,6,2,-64,-64,-87,6,45,87,15,11,03,-34,-6,-68,-959,-653,24,658,68,9,-2181]
counts = []
count = 0
daysCounter = 1
plus_counter = 0
minus_counter = 0
row_counter = 0
answer_counter = 1
for each in list: # for the "dev column"
if each > 0:
minus_counter = 0
plus_counter += 1
if plus_counter == 3:
count = answer_counter
row_counter = answer_counter
counts.append(count)
plus_counter = 0
answer_counter += 1
else:
counts.append(0)
elif each < 0:
plus_counter = 0
minus_counter += 1
if minus_counter == 3:
count = answer_counter
row_counter = answer_counter
counts.append(count)
minus_counter = 0
answer_counter += 1
else:
counts.append(0)
row_counter += 1
print counts
输出：
[0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 3, 0, 0, 4, 0, 0, 5, 0, 0, 6, 0, 0, 0, 0, 7, 0, 0]
这是正确的，但我想将计数器重置在该%10 == 0位置。基本上，如果列表包含30个元素，那么我想在0到10个元素之间，然后在11到20个之间，再到21到30个元素之间进行计数。所需的输出：
[0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0]
您的问题对我来说没有任何意义，您说要在每个模块10索引上重设，但是然后继续说出要包含的30个元素的列表
这总共是31个元素，但是您说列表包含30个元素。从0到29进行索引。所以我在这里假设您的确意味着每10个元素IE 0到9、10到19、20到29。这使我的输出与您的输出不一致，但是我只能在此假设您误算了索引。
nums = [
-21, -22, -33, -55, -454, 65, 48, -516, 614, 6,
2, -64, -64, -87, 6, 45, 87, 15, 11, 3,
-34, -6, -68, -959, -653, 24, 658, 68, 9, -2181
]
nths = 10
sequential_limit = 3
sequential_count = sequential_finds = 0
indexer = sequential_limit - 1
sequential_list = [0 for _ in range(indexer)]
skip = 0
for index, num in enumerate(nums[indexer:], indexer):
result = 0
if index % nths == 0:
sequential_count = sequential_finds = 0
skip = indexer
if skip:
skip -= 1
else:
negative = sum(1 for next_num in nums[index - indexer:index + 1] if next_num < 0)
positive = sum(1 for next_num in nums[index - indexer:index + 1] if next_num >= 0)
if sequential_limit in (positive, negative):
sequential_finds += 1
sequential_count = 0
skip = indexer
result = sequential_finds
sequential_list.append(result)
print(sequential_list)
输出值
[0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 3, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0]
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