如何对特定模式进行字符串拆分、匹配和输出？

Question

我正在尝试解决一个我用 PHP 完成的问题，但不知道如何在 Python 中做到这一点。

在以下三行中，我们喜欢根据这两种模式进行匹配：

• 仅 vine.co 和 twitter.com URL（其他域应忽略）

• 仅逗号之前的 URL ,（每行中的最后一个 URL 应被忽略）

输入

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

---

输出将是 Python 中的一个数组（此输出基于 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 代码：

$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);

此问题基于this RegEx problem，您可以回答其中任何一个问题。

Answer 1

您可以使用此正则表达式来捕获所有具有 vine.com 或 twitter.com 域且 URL 后面有逗号的 URL，

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

如您所愿，关键是这种积极的前瞻性(?=,) 确保您的网址后紧跟一个逗号。

Regex Demo

使用re.findall提取URL的Python代码

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))

输出，

['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

因为你不需要保存重复，我建议使用集合而不是数组（但顺序会改变）：

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

代码：

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'}