如何字符串分割，匹配和輸出特定模式？

Question

我正在嘗試解決我用PHP完成的問題，不確定如何在Python中完成。

在以下三行中，我們希望基於以下兩種模式進行匹配：

• 僅vine.co和twitter.com URL（其他域應忽略）

• 只有逗號之前的網址（每行一個網址就應該被忽略）

輸入

Row 1: https://vine.co/v/5W2Dg3XPX7a,https://vine.co/v/5W2Dg3XPX7a
Row 2: https://twitter.com/dog_rates/status/836677758902222849/photo/1,https://twitter.com/dog_rates/status/836677758902222849/photo/1
Row 3: https://www.gofundme.com/lolas-life-saving-surgery-funds,https://twitter.com/dog_rates/status/835264098648616962/photo/1,https://twitter.com/dog_rates/status/835264098648616962/photo/1

---

輸出將是Python中的數組（此輸出基於PHP）：

array(3) {
  [0]=>
  string(30) "https://vine.co/v/5W2Dg3XPX7a
"
  [1]=>
  string(64) "https://twitter.com/dog_rates/status/836677758902222849/photo/1
"
  [2]=>
  string(63) "https://twitter.com/dog_rates/status/835264098648616962/photo/1"
}

PHP代碼：

$input = 'Row 1: https://vine.co/v/5W2Dg3XPX7a,https://vine.co/v/5W2Dg3XPX7a
Row 2: https://twitter.com/dog_rates/status/836677758902222849/photo/1,https://twitter.com/dog_rates/status/836677758902222849/photo/1
Row 3: https://www.gofundme.com/lolas-life-saving-surgery-funds,https://twitter.com/dog_rates/status/835264098648616962/photo/1,https://twitter.com/dog_rates/status/835264098648616962/photo/1';

$array = preg_split('/Row\s\d:\s/s', $input);

$output = array();
foreach ($array as $key => $value) {
    if (strlen($value) > 1) {
        $URL_arrays = explode(',', $value);
        foreach ($URL_arrays as $key => $value) {
            if ($key = sizeof($URL_arrays) - 1) {
                unset($URL_arrays[sizeof($URL_arrays) - 1]);
            } else {
                $match = preg_match('/twitter\.com|vine\.co/s', $value);
                if ($match) {
                    array_push($output, $value);
                }
            }
        }
    }
}

var_dump($output);

此問題基於此RegEx問題 ，您可以回答其中一個。

Answer 1

您可以使用此正則表達式來捕獲所有具有vine.com或twitter.com域的URL，這些URL vine.com是逗號，

https:\/\/(?:www\.)?(?:vine\.co|twitter\.com)[^,\s]*(?=,)

如您所願，關鍵是要積極向前看(?=,) ，這可以確保URL后面緊跟一個逗號。

正則表達式演示

使用re.findall提取URL的Python代碼

import re

s = '''Row 1: https://vine.co/v/5W2Dg3XPX7a,https://vine.co/v/5W2Dg3XPX7a
Row 2: https://twitter.com/dog_rates/status/836677758902222849/photo/1,https://twitter.com/dog_rates/status/836677758902222849/photo/1
Row 3: https://www.gofundme.com/lolas-life-saving-surgery-funds,https://twitter.com/dog_rates/status/835264098648616962/photo/1,https://twitter.com/dog_rates/status/835264098648616962/photo/1'''

print(re.findall(r'https:\/\/(?:www\.)?(?:vine\.co|twitter\.com)[^,\s]*(?=,)', s))

輸出，

['https://vine.co/v/5W2Dg3XPX7a', 'https://twitter.com/dog_rates/status/836677758902222849/photo/1', 'https://twitter.com/dog_rates/status/835264098648616962/photo/1']

Answer 2

因為您不需要保留重復項，所以我建議使用集合而不是數組（但是順序會發生變化）：

{url for x in s.split('\n') for url in x.split(': ')[1].split(',')  if 'vine.co' in url or 'twitter.co' in url}

代碼 ：

s = '''Row 1: https://vine.co/v/5W2Dg3XPX7a,https://vine.co/v/5W2Dg3XPX7a
Row 2: https://twitter.com/dog_rates/status/836677758902222849/photo/1,https://twitter.com/dog_rates/status/836677758902222849/photo/1
Row 3: https://www.gofundme.com/lolas-life-saving-surgery-funds,https://twitter.com/dog_rates/status/835264098648616962/photo/1,https://twitter.com/dog_rates/status/835264098648616962/photo/1'''

print({url for x in s.split('\n') for url in x.split(': ')[1].split(',')  if 'vine.co' in url or 'twitter.co' in url})

# {'https://twitter.com/dog_rates/status/835264098648616962/photo/1', 
#  'https://twitter.com/dog_rates/status/836677758902222849/photo/1',
#  'https://vine.co/v/5W2Dg3XPX7a'}