美元金额，千位以逗号分隔

Question

我是 python 新手。我正在尝试使用正则表达式从子字符串中提取以美元计价的金额。它在大多数情况下都有效，但是我面临着一些我无法解决的问题。

结果金额是一个字符串，由于逗号而无法识别为金额。它也不适用于小于$1（例如0.89）的小额金额。没有前导$。非常感谢任何帮助。

这是我所拥有的：

df['Amount']=df['description'].str.extract('(\d{1,3}?(\,\d{3})*\.\d{2})')

这是一个应该被解析的字符串：

000000000463 NYC DOF OPA CONCENTRATION ACCT. *00029265 07/01/2013 AP5378 1,107,844.38 Ven000000000463 Vch:00029265

我正在尝试在数据框对象的单独列中提取金额 1,107,844.38。我没有任何应该被拒绝的字符串。

Answer 1

给定您的示例字符串：

"000000000463 NYC DOF OPA CONCENTRATION ACCT. *00029265 07/01/2013 AP5378 1,107,844.38 Ven000000000463 Vch:00029265"

这是我想出的：

match = re.search(r"(?P<amount>\$?(?:\d+,)*\d+\.\d+)", subject)
if match:
    result = match.group("amount")  # result will be "1,107,844.38"
else:
    result = ""

提取金额。它还处理像0.38 这样的小额金额，像123456789.38 这样没有千位分隔符逗号的金额，并且金额前面可能有也可能没有美元符号$。

正则表达式详细信息

(?<amount>\$?(?:\d+,)*\d+\.\d+)  Match the regular expression below and capture its match into backreference with name “amount” 
\$?                              Match the character “$” literally
?                                Between zero and one times, as many times as possible, giving back as needed (greedy) 
(?:\d+,)*                        Match the regular expression below 
*                                Between zero and unlimited times, as many times as possible, giving back as needed (greedy) 
\d+                              Match a single digit 0..9 
+                                Between one and unlimited times, as many times as possible, giving back as needed (greedy) 
,                                Match the character “,” literally 
\d+                              Match a single digit 0..9 
+                                Between one and unlimited times, as many times as possible, giving back as needed (greedy)
\.                               Match the character “.” literally 
\d+                              Match a single digit 0..9 
+                                Between one and unlimited times, as many times as possible, giving back as needed (greedy) 

Answer 2

你可以试试像这样的正则表达式

rx = r"\b(?<!/)(\d{1,3}(?:,\d{3})*(?:\.\d{2})?)\b(?!/)"
df['Amount']=df['description'].str.extract(rx)

见regex demo

详情

• \b - 单词边界
• (?<!/) - 没有 / 紧邻当前位置的左侧（以避免匹配日期时间值）
• \d{1,3} - 1 到 3 位数字
• (?:,\d{3})* - , 的 0+ 次重复和 3 位数字
• (?:\.\d{2})? - 可选的 . 和 2 位数字
• \b - 单词边界
• (?!/) - 没有 / 紧邻当前位置（以避免匹配日期时间值）