正則表達式的findall檢索基礎上開始和結束字符子串

Question

我有以下字符串：

6[Sup. 1e+02] 

我試圖找回的只是1e+02一子。該變量首先引用上面指定的字符串。以下是我所嘗試過的。

re.findall(' \d*]', first) 

Answer 1

您需要使用下面的正則表達式：

\b\d+e\+\d+\b 

說明：

• \b - 單詞邊界
• \d+ - 數字，1個或多個
• e - 字面e
• \+ - 字面+
• \d+ - 位數，1個或多個
• \b - 字邊界

參見demo

樣品的編號：

import re 
p = re.compile(ur'\b\d+e\+\d+\b') 
test_str = u"6[Sup. 1e+02]" 
re.findall(p, test_str) 

參見IDEONE demo

Answer 2

import re 
first = "6[Sup. 1e+02]" 
result = re.findall(r"\s+(.*?)\]", first) 
print result 

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輸出：

['1e+02'] 

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演示 http://ideone.com/Kevtje

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正則表達式說明：

\s+(.*?)\] 

Match a single character that is a 「whitespace character」 (ASCII space, tab, line feed, carriage return, vertical tab, form feed) «\s+» 
    Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+» 
Match the regex below and capture its match into backreference number 1 «(.*?)» 
    Match any single character that is NOT a line break character (line feed) «.*?» 
     Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?» 
Match the character 「]」 literally «\]»