【发布时间】:2013-01-22 10:43:38
【问题描述】:
我是python新手,所以请温柔..
我正在使用 glob 收集与特定模式匹配的文件列表
for name in glob.glob('/home/myfiles/*_customer_records_2323_*.zip')
print '\t', name
输出类似于
/home/myfiles/20130110_customer_records_2323_something.zip
/home/myfiles/20130102_customer_records_2323_something.zip
/home/myfiles/20130101_customer_records_2323_something.zip
/home/myfiles/20130103_customer_records_2323_something.zip
/home/myfiles/20130104_customer_records_2323_something.zip
/home/myfiles/20130105_customer_records_2323_something.zip
/home/myfiles/20130107_customer_records_2323_something.zip
/home/myfiles/20130106_customer_records_2323_something.zip
但我希望输出仅为最新的 5 个文件(通过时间戳或操作系统报告的创建时间)
/home/myfiles/20130106_customer_records_2323_something.zip
/home/myfiles/20130107_customer_records_2323_something.zip
/home/myfiles/20130108_customer_records_2323_something.zip
/home/myfiles/20130109_customer_records_2323_something.zip
/home/myfiles/20130110_customer_records_2323_something.zip
关于如何实现这一点的想法? (是否对列表进行了排序,然后只包含最新的 5 个文件?)
更新 修改以显示默认情况下不排序 glob 的输出的方式
【问题讨论】:
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查看此答案以按时间戳列出文件顺序:stackoverflow.com/a/4500607/665869