Python 根据名称移动文件

Question

值得称赞，我目前使用的代码来自 cji、here.

的回复

我正在尝试从源文件夹中递归提取所有文件，并将它们移动到文件名前五个字符的文件夹中 0:5

我的代码如下：

import os
import shutil

srcpath = "SOURCE"
srcfiles = os.listdir(srcpath)

destpath = "DESTINATION"

# extract the three letters from filenames and filter out duplicates
destdirs = list(set([filename[0:5] for filename in srcfiles]))


def create(dirname, destpath):
    full_path = os.path.join(destpath, dirname)
    os.mkdir(full_path)
    return full_path

def move(filename, dirpath):
    shutil.move(os.path.join(srcpath, filename)
                ,dirpath)

# create destination directories and store their names along with full paths
targets = [(folder, create(folder, destpath)) for folder in destdirs]

for dirname, full_path in targets:
    for filename in srcfiles:
        if dirname == filename[0:5]:
            move(filename, full_path)

现在，用下面的代码更改 srcfiles = os.listdir(srcpath) 和 destdirs = list(set([filename[0:5] for filename in srcfiles])) 得到一个变量中的路径和文件名的前五个字符另一个。

srcfiles = []
destdirs = []

for root, subFolders, files in os.walk(srcpath):
    for file in files:
       srcfiles.append(os.path.join(root,file))
    for name in files:
       destdirs.append(list(set([name[0:5] for file in srcfiles])))

我将如何着手修改原始代码以使用它...或者如果有人对我将如何着手这样做有更好的想法。谢谢。

Answer 1

我真的不能很容易地测试它，但我认为这段代码应该可以工作：

import os
import shutil

srcpath = "SOURCE"
destpath = "DESTINATION"

for root, subFolders, files in os.walk(srcpath):
    for file in files:
        subFolder = os.path.join(destpath, file[:5])
        if not os.path.isdir(subFolder):
            os.makedirs(subFolder)
        shutil.move(os.path.join(root, file), subFolder)

Python 根据名称移动文件

Python Move Files Based On Name

python

file

move

python-2.7