如何修复Python中处理日期的错误？

Question

我刚开始学习Python，正在尝试一些我在网上找到的挑战。我想从一个字符串中提取日期，这样我就可以把它放到一列里。给我的输入是：

"131594", "", "BIDGROUP", 1, 0, 0, 2, "", 0:00, 0:00, 01JAN2009, 01JAN2009, 01JAN2009, 01JAN2009, false, 0,

"131594", "AWARD", "UNTOUCHABLE", 1, 1, 0, 1, "", 0:00, 0:00, 10JUN2014, 13JUN2014 23:59, 01JAN2009, 01JAN2009, false, 100,

"131594", "AWARD", "ADVANCED_TRIP", 1, 2, 0, 0, "740025Jun2014,705406Jun2014,737722Jun2014,696130Jun2014", 0:00, 0:00, 01JAN2009, 01JAN2009, 01JAN2009, 01JAN2009, false, 15,

首先，我查找元素“ADVANCE_TRIP”，然后对于每个标识符，我在字符串中找到的地方，需要创建一个新的行，命名为“TRIP_ID”，并保留之前提到的日期。我尝试后得到的结果是：

"131594", "", "BIDGROUP", 1, 0, 0, 2, "", 0:00, 0:00, 01JAN2009, 01JAN2009, 01JAN2009, 01JAN2009, false, 0,

"131594", "AWARD", "UNTOUCHABLE", 1, 1, 0, 1, "", 0:00, 0:00, 10JUN2014, 13JUN2014 23:59, 01JAN2009, 01JAN2009, false, 100,

"131594", "AWARD", "TRIP_ID", 1, 2, 0, 0, "7400", 0:00, 0:00, 01JAN2009, 01JAN2009, 01JAN2009, 01JAN2009, false, 15,

"131594", "AWARD", "TRIP_ID", 1, 3, 0, 0, "7054", 0:00, 0:00, 01JAN2009, 01JAN2009, 01JAN2009, 01JAN2009, false, 15,

"131594", "AWARD", "TRIP_ID", 1, 4, 0, 0, "7377", 0:00, 0:00, 01JAN2009, 01JAN2009, 01JAN2009, 01JAN2009, false, 15,

"131594", "AWARD", "TRIP_ID", 1, 5, 0, 0, "6961", 0:00, 0:00, 01JAN2009, 01JAN2009, 01JAN2009, 01JAN2009, false, 15,

现在，正确的输出应该是这样的：

"131594", "", "BIDGROUP", 1, 0, 0, 5, "", 0:00, 0:00, 01JAN2009, 01JAN2009, 01JAN2009, 01JAN2009, false, 0,

"131594", "AWARD", "UNTOUCHABLE", 1, 1, 0, 1, "", 0:00, 0:00, 10JUN2014, 13JUN2014 23:59, 01JAN2009, 01JAN2009, false, 100,

"131594", "AWARD", "TRIP_ID", 1, 2, 0, 0, "7400", 0:00, 0:00, 25Jun2014, 01JAN2009, 01JAN2009, 01JAN2009, false, 15,

"131594", "AWARD", "TRIP_ID", 1, 3, 0, 0, "7054", 0:00, 0:00, 06Jun2014, 01JAN2009, 01JAN2009, 01JAN2009, false, 15,

"131594", "AWARD", "TRIP_ID", 1, 4, 0, 0, "7377", 0:00, 0:00, 22Jun2014, 01JAN2009, 01JAN2009, 01JAN2009, false, 15,

"131594", "AWARD", "TRIP_ID", 1, 5, 0, 0, "6961", 0:00, 0:00, 30Jun2014, 01JAN2009, 01JAN2009, 01JAN2009, false, 15,

我唯一不明白的是，如何提取每个“TRIP_ID”标识符旁边的日期，并把它放到相应的列中，也就是第十一列。例如，在我的输出中，我有：“7400”，0:00，0:00，01JAN2009，01JAN2009，但应该是：“7400”，0:00，0:00，25Jun2014，01JAN2009。

这是我写的代码：

import sys

lines = []

for line in sys.stdin:
    lines.append(line.strip())

output_lines = []

for line in lines:
    elements = line.split(", ")
    if elements[2] == '"ADVANCED_TRIP"':
        elements[2] = '"TRIP_ID"'
        trip_ids = elements[7].split(",")
        for i, trip_id in enumerate(trip_ids):
            trip_id = trip_id.strip('"')
            output_line = elements[:7] + [f'"{trip_id[:4]}"'] + elements[8:]
            output_line[4] = str(int(output_line[4]) + i)
            output_lines.append(output_line)
    else:
        output_lines.append(elements)

for output_line in output_lines:
    print(", ".join(output_line))

有没有人知道我该如何继续？

Answer 1

你走在正确的道路上。要正确提取每个“TRIP_ID”相关的日期，你可以使用正则表达式来识别字符串中的日期格式……你可以试试这个

import sys
import re
from datetime import datetime

lines = []

for line in sys.stdin:
    lines.append(line.strip())

output_lines = []

for line in lines:
    elements = line.split(", ")
    if elements[2] == '"ADVANCED_TRIP"':
        elements[2] = '"TRIP_ID"'
        trip_ids = elements[7].split(",")
        dates = re.findall(r'\d{2}[A-Za-z]{3}\d{4}', line)  # Extract dates (e.g., 01JAN2009)
        for i, (trip_id, date) in enumerate(zip(trip_ids, dates)):
            trip_id = trip_id.strip('"')
            output_line = elements[:7] + [f'"{trip_id[:4]}"'] + elements[8:]
            output_line[4] = str(int(output_line[4]) + i)
            output_line[10] = date  # Replace the placeholder date with the extracted date
            output_lines.append(output_line)
    else:
        output_lines.append(elements)

for output_line in output_lines:
    print(", ".join(output_line))

希望这对你有帮助！祝你在学习Python的过程中好运！