查找连续字符串

Question

我试图为下面的示例找到顺序字符串，但无法获得输出。有人可以帮忙吗

数据=[“ABC”，“ABD”，“ABE”，“AB1”，“AB2”，“AB3”，“ab8”，“ab9”，“ab10”，“a123b”，“a124b”，“a125b” ”、“123-43”、“123-44”、“123-45”、“retgoven”、“ahilgdb”]

所需输出=[ABC", "ABD", "ABE", "AB1", "AB2", "AB3", "ab8", "ab9", "ab10", "a123b", "a124b", "a125b" ”、“123-43”、“123-44”、“123-45”]

我使用了下面的代码，但输出没有顺序字符串。

import re

def split_string(s):
    """Split a string into segments of numbers and letters."""
    return re.findall(r'\d+|\D+', s)

def is_sequential_string(prev_string, curr_string):
    """
    Check if curr_string is the sequential successor of prev_string.
    """
    prev_parts = split_string(prev_string)
    curr_parts = split_string(curr_string)
    
    if len(prev_parts) != len(curr_parts):
        return False

    for prev, curr in zip(prev_parts, curr_parts):
        if prev.isdigit() and curr.isdigit():
            if int(curr) != int(prev) + 1:
                return False
        elif prev.isalpha() and curr.isalpha():
            if len(prev) == 1 and len(curr) == 1:  # Only single characters
                if ord(curr) != ord(prev) + 1:
                    return False
            else:
                return False
        else:
            return False
    return True

def find_sequential_strings(data):
    """
    Find all sequential strings in the list.
    """
    string_data = [item for item in data if isinstance(item, str)]
    
    sequential_strings = []
    i = 0
    while i < len(string_data) - 1:
        temp_seq = [string_data[i]]
        while i < len(string_data) - 1 and is_sequential_string(string_data[i], string_data[i + 1]):
            temp_seq.append(string_data[i + 1])
            i += 1
        if len(temp_seq) > 1:
            sequential_strings.extend(temp_seq)
        i += 1

    return sequential_strings

# Sample data
data = [
    112300150226, 112300150226, 112300150227, 112300150228,
    1123071, 1123071, 1123072, 1123073,
    "ABC", "ABD", "ABE", "AB1", "AB2", "AB3",
    "ab8", "ab9", "ab10", "a123b", "a124b", "a125b",
    "123-43", "123-44", "123-45", 10017, 10027, 10037, 10047
]

# Find sequential strings
sequential_strings = find_sequential_strings(data)

# Display the results
print("Sequential Strings:")
if sequential_strings:
    for seq_string in sequential_strings:
        print(seq_string)
else:
    print("No sequential strings found.")

Answer 1

• 我仍然不清楚正确的输出是什么。

• 但是，我们可以定义一些方法，全部（ON），来分解每个字符串，检查数字、小写字母、大写字母，并查看它们是否按顺序排列。

• 如果您愿意或者它适用于您的序列定义，您可以利用

  ord()

  。

import string

def _sanatize(s):
    low, up, dig = '', '', ''
    for char in s:
        if char in string.ascii_uppercase:
            up += char
        if char in string.ascii_lowercase:
            low += char
        if char in string.digits:
            dig += char
    return low, up, dig


def _check(s):
    i, j = 0, 0
    invalid = False
    while i + j < len(s) - 1:
        i += 1
        if s[i + j - 1] > s[i + j]:
            invalid = True
            break
    return True if not invalid else False


def _check_parts(s):
    low, up, dig = _sanatize(s)
    if not _check(low) or not _check(up) or not _check(dig):
        return False
    return True


def get_seq(data):
    res = []
    for s in data:
        if _check_parts(s):
            res += [s]
    return res


data = ["ABC", "ABD", "ABE", "AB1", "AB2", "AB3", "ab8", "ab9", "ab10", "a123b",
        "a124b", "a125b", "123-43", "123-44", "123-45", "retgoven", "ahilgdb", "AB21"]


print(get_seq(data))

打印

['ABC'、'ABD'、'ABE'、'AB1'、'AB2'、'AB3'、'ab8'、'ab9'、'a123b'、'a124b'、'a125b'、'123-44 ', '123-45']