Question

我正在尝试在目录结构中找到最大的文件，以便我可以使用该信息来帮助创建数据库。

以下是代码：

import os
import datetime


def get_files(target):
# Get file size and modified time for all files from the target directory and down.
# Initialize files list
filelist = []
# Walk the directory structure
for root, dirs, files in os.walk(target):
    # Do not walk into directories that are mount points
    dirs[:] = filter(lambda dir: not os.path.ismount(os.path.join(root, dir)), dirs)
    for name in files:
        # Construct absolute path for files
        filename = os.path.join(root, name)
        # Test the path to account for broken symlinks
        if os.path.exists(filename):
            # File size information in bytes
            size = float(os.path.getsize(filename))
            # Get the modified time of the file
            #mtime = os.path.getmtime(filename)
            # Create a tuple of filename, size, and modified time
            construct = filename, size, #str(datetime.datetime.fromtimestamp(mtime))
            # Add the tuple to the master filelist
            filelist.append(construct)
print(sorted([filelist]))
# with open("/home/dave/sizes.txt", 'w') as size_file:
#     contents = filelist.readline()


get_files("/home/dave/TL/case")

如您所见，我是新手，不知道如何将函数的结果传递给文件。

我的最终目标是找到最大的文件及其大小。它可以转到文件或stdout。

我错过了什么？

Answer 1

只需将您的函数设为生成器函数，并使用文件大小作为max的键调用itemgetter(1)：

import os
def get_files(target):
    for root, dirs, files in os.walk(target):
        # Do not walk into directories that are mount points
        dirs[:] = filter(lambda d: not os.path.ismount(os.path.join(root, d)), dirs)
        for name in files:
            # Construct absolute path for files
            filename = os.path.join(root, name)
            # Test the path to account for broken symlinks
            if os.path.exists(filename):
                # File size information in bytes                  
                yield filename, os.path.getsize(filename)

这将允许您重复使用您喜欢的功能：

In [5]: from operator import itemgetter

In [6]: max(get_files("."),key=itemgetter(1))
Out[6]: 
('./node_modules/browser-sync/node_modules/socket.io/node_modules/socket.io-parser/bg.gif',
 1277113)

如果您想按字母顺序按名称对文件进行排序：

sorted(get_files("path"))

按大小排序：

 sorted(get_files("path"), key=itemgetter(1))

Answer 2

这是一种冗长的方法。首先，我创建了一个文件名和文件大小tuples的列表。然后我遍历列表并保存最大的文件名称和大小。

import os


fileSizeTupleList = []
largestSize = 0

for i in os.listdir(os.curdir):
    if os.path.isfile(i):
        fileSizeTupleList.append((i, os.path.getsize(i)))

for fileName, fileSize in fileSizeTupleList:
    if fileSize > largestSize:
        largestSize = fileSize
        largestFile = fileName

print(largestFile, largestSize)

这是一种递归方法：

import os


fileSizeTupleList = []
largestSize = 0

for root, dirs, files in os.walk(os.curdir):
    for file in files:
        fileSizeTupleList.append((file, os.path.getsize(os.path.join(root, file))))

for fileName, fileSize in fileSizeTupleList:
    if fileSize > largestSize:
        largestSize = fileSize
        largestFile = fileName

print(largestFile, largestSize)

如何从这个python 3代码中获取最大的文件？

2 个答案: