Python如何从含Approved关键词的数组字符串中提取目标整数

需求说明 #

现有存储4个字符串元素的数组，需要仅从包含Approved关键词的字符串中提取末尾带3个0的整数。

测试数据 #

my_array = ['STK72184 4/28/2022 50 from Exchange Balance, 50 from Earning Balance & 10 from Bonus 25000 Regular 10/20/2023 Approved 4/28/2022',
            'STK725721 4/27/2022 50 from Exchange Balance, 40 from Earning Balance & 10 from Bonus Balance 5000 Regular 10/19/2023 Closed 4/27/2022',
            'STK725721 4/27/2022 50 from Exchange Balance, 40 from Earning Balance & 10 from Bonus Balance 15000 Regular 10/19/2023 Closed 4/27/2022',
            'STK722222 4/26/2022 50 from Exchange Balance, 40 from Earning Balance & 10 from Bonus Balance 10000 Regular 10/18/2023 Approved 4/26/2022']

原有实现问题 #

原有代码未做关键词过滤，会遍历所有数组元素提取匹配值，输出结果不符合要求：

import re

nums = [int(re.search(r'\d+000', s)[0]) for s in my_array]
print(nums)
# 输出: [25000, 5000, 15000, 10000]

预期输出 #

[25000,10000]

解决方案 #

在列表推导中增加筛选条件，只处理包含Approved的字符串即可，修正后代码：

import re

nums = [int(re.search(r'\d+000', s)[0]) for s in my_array if 'Approved' in s]
print(nums)

运行后即可得到预期输出。

鲁棒性优化：如果后续数据中存在带Approved但无匹配整数的字符串，直接取[0]会抛出异常，可以改用如下写法规避报错：

import re
nums = []
for s in my_array:
    if 'Approved' not in s:
        continue
    match = re.search(r'\d+000', s)
    if match:
        nums.append(int(match.group()))

内容的提问来源于stack exchange，提问作者Rishi Pandey