检查字符串中连续出现的次数

Question

我想显示一个出现频率最高的数字或字母 在给定的字符串或数字或两者中连续。

例子：

s= 'aabskeeebadeeee'
output: e appears 4 consecutive times

我考虑过设置字符串然后为每个元素循环字符串以检查是否与元素 set element if so count =+1 并检查它旁边是否不相等将计数器值添加到具有相同索引的列表中set, if is add counter value to li list if value is greater than existing.

问题是错误索引超出或范围，虽然我想我正在看它。

s = 'aabskeeebadeeee'

c = 0
t = list(set(s)) # list of characters in s
li=[0,0,0,0,0,0]  # list for counted repeats
print(t)
for x in t:
    h = t.index(x)
    for index, i in enumerate(s):
        maximus = len(s)
        if i == x:
            c += 1
            if index < maximus:
                if s[index +1] != x:  # if next element is not x
                    if c > li[h]:   #update c if bigger than existing
                        li[h] = c
                    c = 0
            else:
                if c > li[h]:
                    li[h] = c


for i in t:
    n = t.index(i)
    print(i,li[n])

print(f'{s[li.index(max(li))]} appears {max(li)} consecutive times')

Answer 1

这是一个 O(n) 时间，O(1) 空间解决方案，它通过返回较早看到的字符来打破平局：

def get_longest_consecutive_ch(s):
    count = max_count = 0
    longest_consecutive_ch = previous_ch = None
    for ch in s:
        if ch == previous_ch:
            count += 1
        else:
            previous_ch = ch
            count = 1
        if count > max_count:
            max_count = count
            longest_consecutive_ch = ch
    return longest_consecutive_ch, max_count

s = 'aabskeeebadeeee'
longest_consecutive_ch, count = get_longest_consecutive_ch(s)
print(f'{longest_consecutive_ch} appears {count} consecutive times in {s}')

输出：

e appears 4 consecutive times in aabskeeebadeeee

Answer 2

Regex 在这里提供了一个简洁的解决方案：

inp = "aabskeeebadeeee"
matches = [m.group(0) for m in re.finditer(r'([a-z])\1*', inp)]
print(matches)
matches.sort(key=len, reverse=True)
print(matches[0])

打印出来：

['aa', 'b', 's', 'k', 'eee', 'b', 'a', 'd', 'eeee']
eeee

这里的策略是使用re.finditer 和正则表达式模式([a-z])\1* 查找所有相似字符的岛。然后，我们将结果列表按长度降序排序以找到最长的序列。

Answer 3

或者，您可以利用itertools.groupby() 的强大功能来解决此类问题（用于快速计算groups 中的类似项目。[注意，这可以应用于一些更广泛的情况，例如数字]

from itertools import groupby

>>> char_counts =  [str(len(list(g)))+k  for k, g in groupby(s)]
>>> char_counts
['2a', '1b', '1s', '1k', '3e', '1b', '1a', '1d', '4e']
>>> max(char_counts)
'4e'

# you can continue to do the rest of splitting, or printing for your needs...
>>> ans = '4e'      # example
>>> print(f' the most frequent character is {ans[-1]}, it appears {ans[:-1]} ')

输出： 出现频率最高的字符是 e，出现 4

Answer 4

此答案是由 OP Ziggy Witkowski 在 CC BY-SA 4.0 下以 edit 对问题 Check how many consecutive times appear in a string 发布的。

我不想使用任何库。

s = 'aabskaaaabadcccc'

lil = tuple(set(s)) # Set a characters in s to remove duplicates and
then make a tuple

li=[0,0,0,0,0,0]  # list for counted repeats, the index of number
repeats for character
                # will be equal to index of its character in a tuple


for i in lil:    #iter over tuple of letters
    c = 0        #counter
    h= lil.index(i)  #take an index 
    for letter in s:            #iterate ove the string characters 
        if letter == i:         # check if equal with character from tuple
            c += 1              # if equal Counter +1
            if c > li[lil.index(letter)]:   # Updated the counter if present is bigger than the one stored.
                li[lil.index(letter)] = c
        else:
            c=0
            continue

m = max(li)

for index, j in enumerate(li):              #Check if the are
characters with same max value
    if li[index] == m:
        print(f'{lil[index]} appears {m} consecutive times') 

输出：

c appears 4 consecutive times 
a appears 4 consecutive times