如何使用python查找大小超过指定限制的文件

Question

How to find a file inside a folder, where the file size should be greater than the specified limit.

for root,dirs,files in os.walk('/mnt/usbdrive/data'):
    for file in files:
        if os.stat( '/mnt/usbdrive/data/').st_size >= 1020.0:
            s3C.upload_file(os.path.join(root,file),BUCKET_NAME,file)
        else:
            print("no files found")
print ("data available")

Answer 1

Using os.path.getsize :

for root,dirs,files in os.walk('/mnt/usbdrive/data'):
    for file in files:
        if os.path.getsize(file) >= 1020.0:
            s3C.upload_file(os.path.join(root,file),BUCKET_NAME,file)
        else:
            print("no files found")
print ("data available")

EDIT:

import os
thePath = os.getcwd()
theFiles = list(os.listdir(thePath))

theDict = dict()
for something in theFiles: #Calculate size for all files here.
    theStats = os.stat(something)
    theDict[something] = theStats

for item in theDict:
    if theDict[item].st_size > 1020:
        print("File: {}, size greater than 1020, Uploading to s3 ..".format(item))
        s3C.upload_file(os.path.join(root,file),BUCKET_NAME,file)
    else:
        print("File {}, size less than 1020".format(item))

OUTPUT (from my dir and with commented s3C.upload_file() method:

File: .idea, size greater than 1020, Uploading to s3 ..
File: celebs.jpg, size greater than 1020, Uploading to s3 ..
File data.csv, size less than 1020
File: dum.jpg, size greater than 1020, Uploading to s3 ..
File: dummy.jpg, size greater than 1020, Uploading to s3 ..
File example2.csv, size less than 1020