Python实现遇1重置的循环内元素出现次数计数方法

实现方案 #

你可以用单次遍历的生成器写法实现，逻辑清晰同时符合Pythonic风格，性能也足够好：

list_sequences = [1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 2, 3, 2, 3, 1]

def count_occurrences(seq):
    count_1 = 0
    count_2 = count_3 = 0
    for num in seq:
        if num == 1:
            count_1 += 1
            count_2 = count_3 = 0
            yield count_1
        elif num == 2:
            count_2 += 1
            yield count_2
        elif num == 3:
            count_3 += 1
            yield count_3

occurrence_list = list(count_occurrences(list_sequences))
print(occurrence_list)
# 输出：[1, 1, 1, 2, 2, 2, 1, 1, 3, 1, 1, 2, 2, 3, 3, 4, 4, 4]

逻辑说明 #

• 维护三个计数器：count_1用来累计1的总出现次数，count_2和count_3用来统计上一次遇到1之后，2和3各自的出现次数
• 遍历原列表每一个元素：
  • 遇到1时，count_1自增1，同时重置2和3的计数器，返回当前count_1的值
  • 遇到2时，count_2自增1，返回当前count_2的值
  • 遇到3时，count_3自增1，返回当前count_3的值

内容的提问来源于stack exchange，提问作者user3194861