一串数字找相邻3位数的次数

#假定给一个很长的数字，找出其中重复出现相邻的3位数
number= '1233223332321234323123'
#通过切片找出所有的可能
# split = [number[position:position+3] for position in range(len(number)-2)]
split = []
for position in range(len(number)-2):
    split.append(number[position:position+3])
print(split)
#列出一个1000个元素的列表，0-1000，每个数默认为0
seq = [0]*1000
#将出现的三位数加1
for x in split:
    seq[int(x)]+=1
for i in range(1000):
    if seq[i]>1:
        print('{} repeat {} times'.format(i,seq[i]))

#第2种方法
# inpStr = '123412345123456'
#
# # O(1) array creation.
# freq = [0] * 1000
#
# # O(n) string processing.
# for val in [int(inpStr[pos:pos+3]) for pos in range(len(inpStr) - 2)]:
#     freq[val] += 1
#
# # O(1) output of relevant array values.
# print ([(num, freq[num]) for num in range(1000) if freq[num] > 1])

#第3种
# from collections import Counter
#
# def triple_counter(s):
#     c = Counter(s[n-3: n] for n in range(3, len(s)))
#     for tri, n in c.most_common():
#         if n > 1:
#             print('%s - %i times.' % (tri, n))
#         else:
#             break
#
# if __name__ == '__main__':
#     import random
#
#     s = ''.join(random.choice('0123456789') for _ in range(1_000_000))
#     triple_counter(s)

#如果是一段字符串，找出相邻3个字母的出现次数
zimu='asdaifjweuifhqpwunsknjdbvhfbgirgubaerpubpvuibanvurbgiur'
split_zimu = []
for position in range(len(zimu )-2):
    split_zimu.append(zimu[position:position+3])
print(split_zimu)
zidian_zimu={}
for zimu_item in split_zimu:
    if zidian_zimu:
        for key in zidian_zimu.keys():
            if  key == zimu_item:
                zidian_zimu[zimu_item]+=1
            else:
                zidian_zimu[zimu_item]=1
    else:
        zidian_zimu[zimu_item] = 1
print(zidian_zimu)