Python列表目录，子目录和文件

Question

我正在尝试创建一个脚本来列出给定目录中的所有目录，子目录和文件 我试过这个：

import sys,os

root = "/home/patate/directory/"
path = os.path.join(root, "targetdirectory")

for r,d,f in os.walk(path):
    for file in f:
        print os.path.join(root,file)

不幸的是，它无法正常工作 我得到了所有文件，但没有完整的路径。

例如，如果dir结构为：

/home/patate/directory/targetdirectory/123/456/789/file.txt

它会打印出来：

/home/patate/directory/targetdirectory/file.txt

我需要的是第一个结果。任何帮助将不胜感激！感谢。

Answer 1

使用os.path.join连接目录和文件名称：

for path, subdirs, files in os.walk(root):
    for name in files:
        print os.path.join(path, name)

请注意在并置中使用path而不是root，因为使用root会不正确。

---

在Python 3.4中，添加了pathlib模块以便于路径操作。所以等同于os.path.join将是：

pathlib.PurePath(path, name)

pathlib的优点是您可以在路径上使用各种有用的方法。如果您使用具体的Path变体，您还可以通过它们进行实际的OS调用，例如转到目录，删除路径，打开它指向的文件等等。

Answer 2

以防万一...获取目录中的所有文件和子目录匹配某些模式（例如* .py）：

import os
from fnmatch import fnmatch

root = '/some/directory'
pattern = "*.py"

for path, subdirs, files in os.walk(root):
    for name in files:
        if fnmatch(name, pattern):
            print os.path.join(path, name)

Answer 3

这是一个单行：

import os

[val for sublist in [[os.path.join(i[0], j) for j in i[2]] for i in os.walk('./')] for val in sublist]
# Meta comment to ease selecting text

最外面的val for sublist in ...循环将列表展平为一维。 j循环收集每个文件基名的列表，并将其连接到当前路径。最后，i循环遍历所有目录和子目录。

此示例使用os.walk(...)调用中的硬编码路径./，您可以补充您喜欢的任何路径字符串。

注意：os.path.expanduser和/或os.path.expandvars可用于路径字符串，例如~/

扩展此示例：

很容易添加文件基名测试和目录名测试。

例如，测试*.jpg个文件：

... for j in i[2] if j.endswith('.jpg')] ...

此外，不包括.git目录：

... for i in os.walk('./') if '.git' not in i[0].split('/')]

Answer 4

您应该在联接中使用'r'而不是'root'

Answer 5

无法发表评论，请在此处写答案。这是我所见过的最清晰的一行：

import os
[os.path.join(path, name) for path, subdirs, files in os.walk(root) for name in files]

Answer 6

你可以看看我制作的这个样本。它使用了不推荐使用的os.path.walk函数。使用列表来存储所有文件路径

root = "Your root directory"
ex = ".txt"
where_to = "Wherever you wanna write your file to"
def fileWalker(ext,dirname,names):
    '''
    checks files in names'''
    pat = "*" + ext[0]
    for f in names:
        if fnmatch.fnmatch(f,pat):
            ext[1].append(os.path.join(dirname,f))


def writeTo(fList):

    with open(where_to,"w") as f:
        for di_r in fList:
            f.write(di_r + "\n")






if __name__ == '__main__':
    li = []
    os.path.walk(root,fileWalker,[ex,li])

    writeTo(li)

Answer 7

有点简单的单行：

import os
from itertools import product, chain

chain.from_iterable([["\\".join(w) for w in product([i[0]], i[2])] for i in os.walk(dir)])

Answer 8

由于这里的每个示例都仅使用# MyApp/main/CMakeLists.txt # build rule for executable myApp add_executable(myApp myApp.cc) # dependencies target_link_libraries(myApp gui model) （带有walk），所以我想展示一个不错的示例并与join进行比较：

listdir

如您所见，import os, time def listFiles1(root): # listdir allFiles = []; walk = [root] while walk: folder = walk.pop(0)+"/"; items = os.listdir(folder) # items = folders + files for i in items: i=folder+i; (walk if os.path.isdir(i) else allFiles).append(i) return allFiles def listFiles2(root): # listdir/join (takes ~1.4x as long) (and uses '\\' instead) allFiles = []; walk = [root] while walk: folder = walk.pop(0); items = os.listdir(folder) # items = folders + files for i in items: i=os.path.join(folder,i); (walk if os.path.isdir(i) else allFiles).append(i) return allFiles def listFiles3(root): # walk (takes ~1.5x as long) allFiles = [] for folder, folders, files in os.walk(root): for file in files: allFiles+=[folder.replace("\\","/")+"/"+file] # folder+"\\"+file still ~1.5x return allFiles def listFiles4(root): # walk/join (takes ~1.6x as long) (and uses '\\' instead) allFiles = [] for folder, folders, files in os.walk(root): for file in files: allFiles+=[os.path.join(folder,file)] return allFiles for i in range(100): files = listFiles1("src") # warm up start = time.time() for i in range(100): files = listFiles1("src") # listdir print("Time taken: %.2fs"%(time.time()-start)) # 0.28s start = time.time() for i in range(100): files = listFiles2("src") # listdir and join print("Time taken: %.2fs"%(time.time()-start)) # 0.38s start = time.time() for i in range(100): files = listFiles3("src") # walk print("Time taken: %.2fs"%(time.time()-start)) # 0.42s start = time.time() for i in range(100): files = listFiles4("src") # walk and join print("Time taken: %.2fs"%(time.time()-start)) # 0.47s 版本效率更高。 （而且listdir很慢）

Answer 9

这只是一个补充，这样您就可以将数据转换为 CSV 格式

import sys,os
try:
    import pandas as pd
except:
    os.system("pip3 install pandas")
    
root = "/home/kiran/Downloads/MainFolder" # it may have many subfolders and files inside
lst = []
from fnmatch import fnmatch
pattern = "*.csv"      #I want to get only csv files 
pattern = "*.*"        # Note: Use this pattern to get all types of files and folders 
for path, subdirs, files in os.walk(root):
    for name in files:
        if fnmatch(name, pattern):
            lst.append((os.path.join(path, name)))
df = pd.DataFrame({"filePaths":lst})
df.to_csv("filepaths.csv")

Answer 10

非常简单的解决方案是运行几个子进程调用以将文件导出为 CSV 格式：

import subprocess

# Global variables for directory being mapped

location = '.' # Enter the path here.
pattern = '*.py' # Use this if you want to only return certain filetypes
rootDir = location.rpartition('/')[-1]
outputFile = rootDir + '_directory_contents.csv'

# Find the requested data and export to CSV, specifying a pattern if needed.
find_cmd = 'find ' + location + ' -name ' + pattern +  ' -fprintf ' + outputFile + '  "%Y%M,%n,%u,%g,%s,%A+,%P\n"'
subprocess.call(find_cmd, shell=True)

该命令生成逗号分隔的值，可以在 Excel 中轻松分析。

f-rwxrwxrwx,1,cathy,cathy,2642,2021-06-01+00:22:00.2970880000,content-audit.py

生成的 CSV 文件没有标题行，但您可以使用第二个命令添加它们。

# Add headers to the CSV
headers_cmd = 'sed -i.bak 1i"Permissions,Links,Owner,Group,Size,ModifiedTime,FilePath" ' + outputFile
subprocess.call(headers_cmd, shell=True)

根据您返回的数据量，您可以使用 Pandas 进一步对其进行按摩。以下是我发现有用的一些内容，尤其是当您要处理多个级别的目录时。

将这些添加到您的导入中：

import numpy as np
import pandas as pd

然后将其添加到您的代码中：

# Create DataFrame from the csv file created above.
df = pd.read_csv(outputFile)
    
# Format columns
# Get the filename and file extension from the filepath 
df['FileName'] = df['FilePath'].str.rsplit("/",1).str[-1]
df['FileExt'] = df['FileName'].str.rsplit('.',1).str[1]

# Get the full path to the files. If the path doesn't include a "/" it's the root directory
df['FullPath'] = df["FilePath"].str.rsplit("/",1).str[0]
df['FullPath'] = np.where(df['FullPath'].str.contains("/"), df['FullPath'], rootDir)

# Split the path into columns for the parent directory and its children
df['ParentDir'] = df['FullPath'].str.split("/",1).str[0]
df['SubDirs'] = df['FullPath'].str.split("/",1).str[1]
# Account for NaN returns, indicates the path is the root directory
df['SubDirs'] = np.where(df.SubDirs.str.contains('NaN'), '', df.SubDirs)

# Determine if the item is a directory or file.
df['Type'] = np.where(df['Permissions'].str.startswith('d'), 'Dir', 'File')

# Split the time stamp into date and time columns
df[['ModifiedDate', 'Time']] = df.ModifiedTime.str.rsplit('+', 1, expand=True)
df['Time'] = df['Time'].str.split('.').str[0]

# Show only files, output includes paths so you don't necessarily need to display the individual directories.
df = df[df['Type'].str.contains('File')]

# Set columns to show and their order.
df=df[['FileName','ParentDir','SubDirs','FullPath','DocType','ModifiedDate','Time', 'Size']]

filesize=[] # Create an empty list to store file sizes to convert them to something more readable.

# Go through the items and convert the filesize from bytes to something more readable.
for items in df['Size'].items():
    filesize.append(convert_bytes(items[1]))
    df['Size'] = filesize 

# Send the data to an Excel workbook with sheets by parent directory
with pd.ExcelWriter("scripts_directory_contents.xlsx") as writer:
    for directory, data in df.groupby('ParentDir'):
    data.to_excel(writer, sheet_name = directory, index=False) 
        

# To convert sizes to be more human readable
def convert_bytes(size):
    for x in ['b', 'K', 'M', 'G', 'T']:
        if size < 1024:
            return "%3.1f %s" % (size, x)
        size /= 1024

    return size