好吧,如果您知道所有文件的大小(至少有点)为 100MB,并且假设没有其他任何因素大幅改变计算机上的磁盘使用情况,则无需在每次迭代时检查可用空间。
另外,如果所有文件都具有相同的名称,除了末尾的计数器之外,您可以跳过 os.stat 调用(这对于快速连续创建的文件也可能无用)并根据计数器对文件名进行排序:
import os
def free_space_up_to(free_bytes_required=161061273600, rootfolder="/data/", filesize=104857600, basename="filename-"):
'''Deletes rootfolder/basename*, oldest first, until there are free_bytes_required available on the partition.
Assumes that all files have file_size, and are all named basename{0,1,2,3,...}
Returns number of deleted files.
'''
statv = os.statvfs(rootfolder)
required_space = free_bytes_required - statv.f_bfree*statv.f_bsize
basepath = os.path.join(rootfolder, basename)
baselen = len(basepath)
if required_space <= 0:
return 0
# "1 +" here for quickly rounding
files_to_delete = 1 + required_space/filesize
# List all matching files. If needed, replace with os.walk for recursively
# searching into subdirectories of rootfolder
file_list = [os.path.join(rootfolder, f) for f in os.listdir(rootfolder)
if f.startswith(basename)]
file_list.sort(key=lambda i: int(i[baselen:]), reverse=True)
# Alternatively, if the filenames can't be trusted, sort based on modification time
#file_list.sort(key=lambda i: os.stat(i).st_mtime)
for f in file_list[:files_to_delete]:
os.remove(f)
return files_to_delete
(未经彻底测试,我建议进行测试运行,用“print”替换“os.remove”;))