从字符串中删除连续的子字符串而不导入任何包

Question

我想删除 连续的“a”子字符串 并将它们替换为字符串 中的 一个“a”不导入任何包。例如，我想从 aaabbcccaaaa 得到 abbccca。有什么建议么？谢谢

Answer 1

使用循环有什么问题？

oldstring = 'aaabbcccaaaa'
# Initialise the first character as the same as the initial string 
# as this will always be the same.
newstring = oldstring[0]

# Loop through each character starting at the second character
# check if the preceding character is an a, if it isn't add it to 
# the new string. If it is an a then check if the current character
# is an a too. If the current character isn't an a then add it to 
# the new string.
for i in range(1, len(oldstring)):
    if oldstring[i-1] != 'a':
        newstring += oldstring[i]
    else:
       if oldstring[i] != 'a':
           newstring += oldstring[i]

print(newstring)

Answer 2

此方法将从您的字符串中删除确定的重复字符：

def remove_dulicated_char(string, char):
    new_s = ""
    prev = ""
    for c in string:
        if len(new_s) == 0:
            new_s += c
            prev = c
        if c == prev and c == char:
            continue
        else:
            new_s += c
            prev = c
    return new_s

print(remove_dulicated_char("aaabbcccaaaa", "a"))

Answer 3

使用 python 正则表达式就可以了。如果你不知道正则表达式。他们是非常强大的这种搭配

重新导入

str = 'aaabbcccaaaa'

print(re.sub('a+', 'a', str))

Answer 4

您可以使用函数以递归方式删除出现的字符串的双精度值，直到只剩下重复字符串的一次出现：

val = 'aaabbcccaaaaaaaaaaa'

def remove_doubles(v):
    v = v.replace('aa', 'a')
    if 'aa' in v:
        v = remove_doubles(v)
        if 'aa' in v:
            v = remove_doubles(v)
        else: return v
    else: return v

print(remove_doubles(val))

Answer 5

有很多方法可以做到这一点。这是另一个：

def remove_duplicates(s, x):
    t = [s[0]]
    for c in s[1:]:
        if c != x or t[-1] != x:
            t.append(c)
    return ''.join(t)

print(remove_duplicates('aaabbcccaaaa', 'a'))

从字符串中删除连续的子字符串而不导入任何包

remove consecutive substrings from a string without importing any packages

python

string

replace

substring