5 Ways to Count Consecutive Elements in a Python List

1. Introduction

A follower asked how to process a Python list to count consecutive identical values. The original list is shown below.

2. Implementation

Below are five different implementations submitted by community members.

Method 1 (by 瑜亮老师)

list1=[1,1,1,0,0,0,1,1,0,1,1,1,1,1,1]
result=[0]
flag=0
for i in range(1,len(list1)):
    if list1[i]==list1[i-1]:
        flag+=1
    else:
        flag=0
    result.append(flag)
print(result)

Method 2 (by 绅)

list1 = [1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1]
list2 = []
l = 0
for i in range(len(list1)):
    l = l + 1 
    if list1[i] == list1[i-1] and i != 0 else 0
    list2.append(l)
print(list2)

Method 3 (by 逸)

list1 = [1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1]
result = [0] * len(list1)
result[0] = 0
for i in range(1, len(list1)):
    if list1[i] != list1[i-1]:
        result[i] = 0
    else:
        result[i] = result[i-1] + 1
print(result)

Method 4 (by 月神)

list1 = [1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1]
result = [0] * len(list1)  # [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
pre_num = 0
for num in range(len(list1)):
    print(num, pre_num)
    if list1[num] != list1[pre_num]:
        pre_num = num
    result[num] = num - pre_num
print(result)
print(result)

Method 5 (by 布达佩斯的永恒)

from functools import reduce
from itertools import groupby

list1 = [1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1]
ret = reduce(lambda x, y: x + y, [[i for i in range(len(list(num)))] for item, num in groupby(list1)])
print(ret)

3. Conclusion

The article presents five practical solutions for counting consecutive elements in a Python list, each with its own logic and code snippet. Readers are encouraged to try these methods, share alternative approaches, and continue discussing Python programming topics.