Question

我有一个用python生成的动态列表，如何过滤该列表以仅显示带有dates > 20190612的文件

"PUBLIC_DAILY_201906150000_20190616040503.zip"
"PUBLIC_DAILY_201906110000_20190612040501.zip"
"PUBLIC_DAILY_201906120000_20190613040503.zip"
"PUBLIC_DAILY_201906100000_20190611040501.zip"
"PUBLIC_DAILY_201906130000_20190614040503.zip"

结果将是这些项目的列表

"PUBLIC_DAILY_201906130000_20190614040503.zip"
"PUBLIC_DAILY_201906150000_20190616040503.zip"

谢谢

Answer 1

我们可以尝试在子字符串的帮助下进行过滤：

# list input
output = list(filter(lambda x: x[13:21] > '20190612', input))
print(output)

['PUBLIC_DAILY_201906150000_20190616040503.zip',
 'PUBLIC_DAILY_201906130000_20190614040503.zip']

这实际上是第一个日期的子字符串，然后将其与另一个字符串日期进行比较。因为这里涉及的所有日期都是固定宽度，所以我们在这里进行了字符串比较。最佳做法可能是将所有文本字符串转换为真实日期，然后进行比较。

Answer 2

尝试：

[i for i in lst if i > "PUBLIC_DAILY_20190612000"]

Answer 3

您也可以尝试类似的方法。

通常最好先将字符串转换为日期。

archives = ["PUBLIC_DAILY_201906150000_20190616040503.zip",
"PUBLIC_DAILY_201906110000_20190612040501.zip",
"PUBLIC_DAILY_201906120000_20190613040503.zip",
"PUBLIC_DAILY_201906100000_20190611040501.zip",
"PUBLIC_DAILY_201906130000_20190614040503.zip"]

date_regex = "%Y%m%d%H%M%S"
default_regex = "%Y%m%d"
default_date = "20190612"
compare_date = datetime.datetime.strptime(default_date,default_regex)
passed_list = []

for archive in archives:
    split_str = archive.split("_")
    print split_str

    date1 = datetime.datetime.strptime(split_str[2],date_regex)
    date2 = datetime.datetime.strptime(split_str[3].split(".")[0],date_regex)
    if date1 > compare_date and date2 > compare_date:
        passed_list.append(archive)

print passed_list

Answer 4

import datetime
import re

lst = ["PUBLIC_DAILY_201906150000_20190616040503.zip",
"PUBLIC_DAILY_201906110000_20190612040501.zip",
"PUBLIC_DAILY_201906120000_20190613040503.zip",
"PUBLIC_DAILY_201906100000_20190611040501.zip",
"PUBLIC_DAILY_201906130000_20190614040503.zip"]

d = datetime.datetime(year=2019, month=6, day=12)

l = [l for l in lst for g in re.findall(r'PUBLIC_DAILY_(\d{4})(\d{2})(\d{2})', l) if datetime.datetime.strptime(''.join(g), '%Y%m%d') > d]
print(l)

打印：

['PUBLIC_DAILY_201906150000_20190616040503.zip', 'PUBLIC_DAILY_201906130000_20190614040503.zip']

Answer 5

您只需将字符串切成薄片并将其转换为int，然后将其与您各自的日期进行比较数据= [“ PUBLIC_DAILY_201906150000_20190616040503.zip”， “ PUBLIC_DAILY_201906110000_20190612040501.zip”， “ PUBLIC_DAILY_201906120000_20190613040503.zip”， “ PUBLIC_DAILY_201906100000_20190611040501.zip”， “ PUBLIC_DAILY_201906130000_20190614040503.zip”]

for i in range(len(data)):
    date=int(data[i][13:21])
    if date>20190612:
    print(data[i])

如何过滤包含日期的字符串列表

5 个答案: