操作系统。统计和多处理。并行分析文件列表。

I'm trying to run os.stat in poolfor list of files but as result receive just nested list of files statistics:

我在试着运行操作系统。在poolfor文件列表中，但结果只接收到的文件统计信息列表:

list_of_files = ['file1.txt','file2.txt','file3.txt']
p = Pool(4) 
list_of_stats = p.map(os.stat, list_of_files)

list_of_fies contains the next value: nt.stat_result(st_mode=33206, st_ino=0L, st_dev=0L, st_nlink=0, st_uid=0, st_gid=0, st_size=8905L, st_atime=1499725858L, st_mtime=1416573654L, st_ctime=1445009618L)...

list_of_fies包含下一个值:nt.stat_result(st_mode=33206, st_ino=0L, st_dev=0L, st_nlink=0, st_uid=0, st_gid=0, st_size=8905L, st_atime=1499725858L, st_mtime=1416573654L, st_ctime=1445009618L)…

my question is: what is the simplest way to get the file name + statistics related to this files in list/dict?

我的问题是:最简单的方法是如何获取与此文件相关的文件名称和统计信息?

for instance dict: {'file1.txt': [st_mode=33206, st_ino=0L, st_dev=0L, st_nlink=0, st_uid=0, st_gid=0, st_size=8905L, st_atime=1499725858L,...]}

例如dict:{ file1。txt: [st_mode=33206, st_ino=0L, st_dev=0L, st_nlink=0, st_uid=0, st_gid=0, st_size=8905L, st_atime=1499725858L，…]}

1 个解决方案

#1

I've found the solution with using the code similar to below:

我已经找到了使用类似以下代码的解决方案:

list_of_files = [name1, name2, name3, name4]

def file_stat(file_path):
    return (file_path, os.stat(file_path))


if __name__ == "__main__":

    p = Pool(2)
    list_of_stats = p.map(file_stat, list_of_files)

    print(type(list_of_stats)) 

    print(list_of_stats)

output:

输出:

<type 'list'>

[(u'\\path\to\file\name1.xml', nt.stat_result(st_mode=33206, st_ino=0L, st_dev=0L, st_nlink=0, st_uid=0, st_gid=0, st_size=8905L, st_atime=1501540255L, st_mtime=1416573654L, st_ctime=1445009618L)), (u'\\path\to\file\name2.xml', nt.stat_result(st_mode=33206, st_ino=0L, st_dev=0L, st_nlink=0, st_uid=0, st_gid=0, st_size=47000L, st_atime=1501540327L, st_mtime=1475583701L, st_ctime=1445009618L)), ... ]