python regex - 如何拆分日期模式上的字符串 - 仅保留日期模式第二次出现后的值

Question

我有一个值列表，我需要用日期模式 dd mmm yyyy 将它们分开

input_list = ['01 Feb 2023 CNY  0.00  559,929.32 **01 Feb 2023 CNY  0.00  0.00**',
 '01 Feb 2023 HKD  0.00  22,441,926.65 **01 Feb 2023 HKD  0.00  0.00**',
 '01 Feb 2023 USD  0.00  585,686.14 **01 Feb 2023 USD  0.00  0.00**']

希望得到结果: df 结果:

currency     1stvalue   2ndvalue
CNY           0.00        0.00
HKD           0.00        0.00
USD           0.00        0.00

任何指导或建议都会有帮助，谢谢

我找到了类似的方法，但仍然无法弄清楚: Regex to splitstring on date and keep it

类似示例:

import re
rx = r"\b\d+/\d+/\d+.*?(?=\b\d+/\d+/\d+|$)"
s = "28/11/2016 Mushroom 05/12/2016 Carrot 12/12/2016 Broccoli 19/12/2016 Potato"
results = re.findall(rx, s)
print(results)

Answer 1

你可以尝试:

import re

pattern = r"\d{2} [A-Z,a-z]{3} [0-9]{4} ([A-Z]{3})  (\d\.\d\d)"                                                                                             
string = "01 Feb 2023 CNY  0.00  559,929.32 **01 Feb 2023 CNY  0.00  0.00**"
responses = re.findall(pattern, string)

# printer
for i, response in enumerate(responses):
    if i == 0:
        print(response[0], end="")
    print(" ", response[1], end="")