用逗號分割字符串，但忽略括號內的逗號

Question

我正在嘗試使用 python 以逗號分隔字符串：

s = "year:2020,concepts:[ab553,cd779],publisher:elsevier"

但我想忽略方括號 [] 中的任何逗號。 所以上面的結果是：

["year:2020", "concepts:[ab553,cd779]", "publisher:elsevier"]

有人對如何執行此操作有建議嗎？ 我試着像這樣使用 re.split：

params = re.split(",(?![\w\d\s])", param)

但它不能正常工作。

Answer 1

result = re.split(r",(?!(?:[^,\[\]]+,)*[^,\[\]]+])", subject, 0)

,                 # Match the character “,” literally
(?!               # Assert that it is impossible to match the regex below starting at this position (negative lookahead)
   (?:               # Match the regular expression below
      [^,\[\]]          # Match any single character NOT present in the list below
                           # The literal character “,”
                           # The literal character “[”
                           # The literal character “]”
         +                 # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
      ,                 # Match the character “,” literally
   )
      *                 # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
   [^,\[\]]          # Match any single character NOT present in the list below
                        # The literal character “,”
                        # The literal character “[”
                        # The literal character “]”
      +                 # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
   ]                 # Match the character “]” literally
)

更新以支持括號中的 2 個以上的項目。 例如

year:2020,concepts:[ab553,cd779],publisher:elsevier,year:2020,concepts:[ab553,cd779,xx345],publisher:elsevier

Answer 2

您可以使用用戶定義的 function 而不是拆分來解決此問題：

s = "year:2020,concepts:[ab553,cd779],publisher:elsevier"


def split_by_commas(s):
    lst = list()
    last_bracket = ''
    word = ""
    for c in s:
        if c == '[' or c == ']':
            last_bracket = c
        if c == ',' and last_bracket == ']':
            lst.append(word)
            word = ""
            continue
        elif c == ',' and last_bracket == '[':
            word += c
            continue
        elif c == ',':
            lst.append(word)
            word = ""
            continue
        word += c
    lst.append(word)
    return lst
main_lst = split_by_commas(s)

print(main_lst)

上述代碼運行結果：

['year:2020', 'concepts:[ab553,cd779]', 'publisher:elsevier']

Answer 3

此正則表達式適用於您的示例：

,(?=[^,]+?:)

在這里，我們使用正向前瞻來查找逗號，后跟非逗號和冒號字符，然后是冒號。 這會正確找到您正在搜索的<comma><key>模式。 當然，如果允許鍵有逗號，則必須進一步調整。

你可以在這里查看正則表達式

Answer 4

使用僅具有前瞻的模式來斷言右側的字符，如果左側有伴隨字符，則不會斷言。

除了使用拆分之外，您還可以匹配方括號之間的 1 個或多個重復值，或者匹配除逗號之外的任何字符。

(?:[^,]*\[[^][]*])+[^,]*|[^,]+

正則表達式演示

s = "year:2020,concepts:[ab553,cd779],publisher:elsevier"
params = re.findall(r"(?:[^,]*\[[^][]*])+[^,]*|[^,]+", s)
print(params)

Output

['year:2020', 'concepts:[ab553,cd779]', 'publisher:elsevier']

Answer 5

我采用了@Bemwa 的解決方案（不適用於我的用例）

def split_by_commas(s):
    lst = list()
    brackets = 0
    word = ""
    for c in s:
        if c == "[":
            brackets += 1
        elif c == "]":
            if brackets > 0:
                brackets -= 1
        elif c == "," and not brackets:
            lst.append(word)
            word = ""
            continue
        word += c
    lst.append(word)
    return lst