Question

假设用户输入：

"daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaaaaaaaaaaafnnasm"

你如何找到最高数量的连续“a”以及如何删除“a”并且只留下其中的2个而不是之前的大量数字。

我在考虑将每个字母附加到一个新的空列表中，但我不确定这是否正确或者该怎么做。

我真的不知道从哪里开始，但这就是我的想法：

要求用户输入。
创建一个空列表
将输入中的每个字母附加到列表中

下一步我不知道。

第二次编辑（沿着这些方向）：

sentence = input("Enter your text: ")
new_sentance = " ".join(sentence.split())
length = len(new_sentance)
alist = []
while (length>0):
    alist
print ()

Answer 1

从输入字符串开始：

input = "daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaaaaaaaaaaafnnasm"

要获得最大连续出现次数，请使用：
```
max(len(s) for s in re.findall(r'a+', input))
```
用2“a”替换最长的“a”序列，你会用：
```
maxMatch = max(re.finditer(r'a+', input), key= lambda m: len(m.group()))
output = input[:maxMatch.start()] + "aa" + input[maxMatch.end():]
```
首先，我通过针对正则表达式MatchObject测试输入字符串获得a+ s的可迭代，然后使用max获取具有最大长度的MatchObject。然后，我将原始字符串的部分拼接到匹配的开头，字符串“aa”，以及匹配结束后原始字符串的部分，以便为您提供最终输出。
要用2“a”替换超过2“a”的所有次出现，你会使用：
```
output = re.sub(r'a{3,}', "aa", input)
```

Answer 2

如果您不想使用正则表达式，则采用较低级别的方法。

def count_and_reduce(s, a):
    num = 0
    maxnum = 0
    out = ''
    for c in s:
        if c == a:
            num += 1
            maxnum = max(num, maxnum)
        else:
            num = 0
        if num <= 2:
            out += c

    return maxnum, out

Answer 3

我在评论和其他问题中看到了几个正则表达式的答案，所以我将采取不同的方式。只是获得计数可以通过许多不同的方式完成。

from itertools import groupby

inp = 'daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaaaaaaaaaaafnnasm';
char_groups = groupby(inp, lambda char:char=='a')
counts = [len(list(group)) for char, group in char_groups]
# We know every other element of 'counts' is an 'a' element.
# We just need to know whether to start at zero or one.
# If inp starts with 'a', start at 0. Otherwise start at 1.
max(counts[not inp.startswith('a')::2]) # 11

我很确定我见过的两个正则表达式的答案都会用'a'替换每个字符串'aa +'。如果您只想要将'a'的最长字符串替换为'aa'而将其余部分单独替换：

char_groups = groupby(inp)
counts = [(char, len(list(group))) for char, group in char_groups]
max_idx = max(range(len(counts)), key=lambda i:counts[i][1] if counts[i][0]=='a' else 0)
result = ''.join(char*count for char, count in counts[:max_idx]) + 'aa' + ''.join(char*count for char, count in counts[max_idx+1:])
# 'daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaafnnasm'

Answer 4

我会这样做。

s = "daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaaaaaaaaaaafnnasm"
print(s)
a_len = len(s)
found_a_len = 0
keep_going = True
while a_len>0 and keep_going:
    aas = "a" * a_len
    if aas in s:
        found_a_len = a_len
        keep_going = False
    a_len=a_len -1
print ("max length of a:" , found_a_len)
keep_going = True
while keep_going:
    s=s.replace("aaa","aa")
    if "aaa" not in s:
        keep_going = False
print(s)

此输出：

daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaaaaaaaaaaafnnasm
max length of a: 11
daslakndlaajnjndibniaafijdnfijdnsijfnsdinifaafnnasm

有些人可能不喜欢我的编码风格，但对我来说，这段代码很容易理解。

Answer 5

试试这个：

import collections

def runLenEncode(s):
    start, i = 0, 0;
    answer = []
    while i<len(s):
        while i<len(s) and s[start]==s[i]:
            i += 1
        answer.append((s[start], i-start))
        start = i
    return answer

def runLenFilter(encoding, thresholds, repLens):
    answer = []
    for char, count in encoding:
        if char in thresholds and count>=thresholds[char]:
            count = repLens[char]
        answer.append(char*count)
    return ''.join(answer)

def maxFilter(encoding, repLens):
    maxCounts = collections.defaultdict(int)
    for char, count in encoding:
        if char in repLens and count > maxCounts[char]:
            maxCounts[char] = count
    maxCounts = dict(maxCounts)
    answer = []
    for char, count in encoding:
        if char in repLens and count==maxCounts[char]:
            count = repLens[char]
        answer.append(char*count)
    return ''.join(answer)

if __name__ == "__main__":
    print('starting')

    s = "daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaaaaaaaaaaafnnasm"
    encoding = runLenEncode(s)
    print("encoding:", encoding)
    thresholds = {'a':3}
    repLens = {'a':2}
    decoded = runLenFilter(encoding, thresholds, repLens)
    print('lenFilter:', decoded)
    filtered = maxFilter(encoding, repLens)
    print("maxFilter:", filtered)

    print('done')

输出：

$ python3 myFile.py
starting
encoding: [('d', 1), ('a', 1), ('s', 1), ('l', 1), ('a', 1), ('k', 1), ('n', 1), ('d', 1), ('l', 1), ('a', 5), ('j', 1), ('n', 1), ('j', 1), ('n', 1), ('d', 1), ('i', 1), ('b', 1), ('n', 1), ('i', 1), ('a', 3), ('f', 1), ('i', 1), ('j', 1), ('d', 1), ('n', 1), ('f', 1), ('i', 1), ('j', 1), ('d', 1), ('n', 1), ('s', 1), ('i', 1), ('j', 1), ('f', 1), ('n', 1), ('s', 1), ('d', 1), ('i', 1), ('n', 1), ('i', 1), ('f', 1), ('a', 11), ('f', 1), ('n', 2), ('a', 1), ('s', 1), ('m', 1)]
lenFilter: daslakndlaajnjndibniaafijdnfijdnsijfnsdinifaafnnasm
maxFilter: daslakndlaaaaajnjndibniaaafijdnfijdnsijfnsdinifaafnnasm
done

计算字符串中最大连续“a”的数量。 Python 3

5 个答案: