獲取 [ 括號和 python 中的特殊字符之間的字符串

Question

我有一個與此非常相似的問題。

我真的很想知道為什么我的結果是： NaN 。

我有一個 dataframe 這個專欄：

Action
Player[J♡, K♧] won the $5.40 main pot with a Straight
Player [5, 2] won the $21.00 main pot with a flush

我想用牌建立一個新列，誰被玩過：

[J♡, K♧]
[5, 2]

甚至：

[J, K]
[5, 2]

但是，當我玩正則表達式並使用： dfpot['cards'] = dfpot['Action'].str.extract(r'\[([A-Za-z0-9_]+)\]', expand=False)

我只有NaN 。

Answer 1

您可以將字符添加到捕獲組中的字符 class 中，如模式\[([A-Za-z0-9_♤♡♢♧, ]+)\]或使模式更具體：

\[([A-Za-z0-9_][♤♡♢♧]?,\s*[A-Za-z0-9_][♤♡♢♧]?)]

模式匹配：

• \[匹配[
• (捕獲組 1
  • [A-Za-z0-9_]匹配列出的字符之一
  • [♤♡♢♧]? 可選擇匹配列出的字符之一
  • ,\s*[A-Za-z0-9_][♤♡♢♧]? 匹配一個逗號和與逗號之前相同的邏輯
• )關閉第 1 組
• ]匹配]

正則表達式演示

例如

import pandas as pd

dfpot = pd.DataFrame({'Action':['Player[J♡, K♧] won the $5.40 main pot with a Straight', 'Player [5, 2] won the $21.00 main pot with a flush']})
dfpot['cards'] = dfpot['Action'].str.extract(r'\[([A-Za-z0-9_][♤♡♢♧]?,\s*[A-Za-z0-9_][♤♡♢♧]?)]', expand=False)
print(dfpot)

Output

                                              Action   cards
0  Player[J♡, K♧] won the $5.40 main pot with a S...  J♡, K♧
1  Player [5, 2] won the $21.00 main pot with a f...    5, 2

Answer 2

嘗試模式（我假設您在文本中使用()而不是[] ，正如在正則表達式演示中發布的那樣）：

\([^,]+,[^\)]+\)

解釋：

\( - 匹配(字面意思

[^,]+ - 匹配除,之外的一個或多個字符

, - 匹配,字面意思

[^\)]+ - 匹配除)以外的一個或多個字符

\) - 匹配)字面意思

正則表達式演示

Answer 3

利用

>>> import pandas as pd
>>> df = pd.DataFrame({'Action':['Player[J♡, K♧] won the $5.40 main pot with a Straight', 'Player [5, 2] won the $21.00 main pot with a flush']})
>>> df['cards'] = df['Action'].str.findall(r'(\w+)(?=[^][]*])')
>>> df
                                              Action   cards
0  Player[J♡, K♧] won the $5.40 main pot with a S...  [J, K]
1  Player [5, 2] won the $21.00 main pot with a f...  [5, 2]
>>> 

正則表達式： (\w+)(?=[^][]*])

解釋

--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    \w+                      word characters (a-z, A-Z, 0-9, _) (1 or
                             more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
  )                        end of \1
--------------------------------------------------------------------------------
  (?=                      look ahead to see if there is:
--------------------------------------------------------------------------------
    [^][]*                   any character except: ']', '[' (0 or
                             more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
    ]                        ']'
--------------------------------------------------------------------------------
  )                        end of look-ahead