这就是我如何在一个文件夹中找到一个文件列表(在这些图像中)并在一个文件夹中搜索它(与子文件夹):

# This script concatenates JavaScript files into a unified JavaScript file to reduce server round-trips

import os
import string
import math
import ntpath
import sys

#import pyodbc

import gzip
import shutil

import hashlib

# BUF_SIZE is totally arbitrary, change for your app!
BUF_SIZE = 65536  # Let’s read stuff in 64 kilobyte chunks

# Iterate over all JavaScript files in the folder and combine them
filenames = []
shortfilenames = []

imgfilenames = []
imgshortfilenames = []

# Get a unified path so we can stop dancing with user paths.
# Determine where files are on this machine (%TEMP% directory and application installation directory)
if '.exe' in sys.argv[0]: # if getattr(sys, 'frozen', False):
    RootPath = os.path.abspath(os.path.join(__file__, "..\\"))

elif __file__:
    RootPath = os.path.abspath(os.path.join(__file__, "..\\"))

print ("\n storage of image files RootPath: %s\n" %RootPath)

FolderPath = "D:\\TFS-FARM1\\StoneSoup_STS\\SDLC\\Build\\Code\\StoneSoup_Refactor\\StoneSoupUI\\Images"
print ("\n storage of image files in folder to search: %s\n" %FolderPath)

for root, directories, filenames2 in os.walk(FolderPath):
    for filename in filenames2:
        fullname = os.path.join(root, filename)
        filenames.append(fullname)
        shortfilenames.append(filename)

for i, fname in enumerate(shortfilenames):
        print("%s - %s" % (i+1, fname))

for root, directories, filenames2 in os.walk(RootPath):
    for filename in filenames2:
        fullname = os.path.join(root, filename)
        imgfilenames.append(fullname)
        imgshortfilenames.append(filename)

for i, fname in enumerate(imgshortfilenames):
        print("%s - %s" % (i+1, fname))

for i, fname in enumerate(imgshortfilenames):
        if fname in shortfilenames:
            print("%s - %s exists" % (i+1, fname))
        else:
            print("%s - %s ABSENT" % (i+1, fname))

如果你正在检查的原因是,所以你可以做一些类似的文件_存在: open_it(),它更安全地使用一个尝试周围试图打开它。

如果您不打算立即打开文件,您可以使用 os.path.isfile。

如果路径是现有常规文件,则返回是真实的,这跟随了象征性链接,因此 islink() 和 isfile() 都可以对同一条路径是真实的。

import os.path
os.path.isfile(fname) 

如果你需要确保它是一个文件。

从 Python 3.4 开始,Pathlib 模块提供一个以对象为导向的方法(在 Python 2.7 中返回Pathlib2):

from pathlib import Path

my_file = Path("/path/to/file")
if my_file.is_file():
    # file exists

要查看一个目录,请:

if my_file.is_dir():
    # directory exists

要检查路径对象是否存在,无论它是否是一个文件或目录,使用存在():

if my_file.exists():
    # path exists

您也可以在尝试区块中使用 resolve(strict=True):

try:
    my_abs_path = my_file.resolve(strict=True)
except FileNotFoundError:
    # doesn't exist
else:
    # exists

在2016年,最好的方式仍然是使用os.path.isfile:

>>> os.path.isfile('/path/to/some/file.txt')

在Python 3中,您可以使用Pathlib:

import pathlib
path = pathlib.Path('/path/to/some/file.txt')
if path.is_file():
    ...

exist() and is_file() 对“路”对象的方法可以用来检查是否有一个特定的路径存在并是一个文件。

Python 3 程序检查是否存在文件:

# File name:  check-if-file-exists.py

from pathlib import Path

filePath = Path(input("Enter path of the file to be found: "))

if filePath.exists() and filePath.is_file():
    print("Success: File exists")
else:
    print("Error: File does not exist")

出口:

$ python3 check-if-file-exists.py 输入文件的路径可以找到: /Users/macuser1/stack-overflow/index.html 成功: 文件存在 $ python3 check-if-file-exists.py 输入文件的路径可以找到: hghjg jghj 错误: 文件不存在

这就是我如何在一个文件夹中找到一个文件列表(在这些图像中)并在一个文件夹中搜索它(与子文件夹):

# This script concatenates JavaScript files into a unified JavaScript file to reduce server round-trips

import os
import string
import math
import ntpath
import sys

#import pyodbc

import gzip
import shutil

import hashlib

# BUF_SIZE is totally arbitrary, change for your app!
BUF_SIZE = 65536  # Let’s read stuff in 64 kilobyte chunks

# Iterate over all JavaScript files in the folder and combine them
filenames = []
shortfilenames = []

imgfilenames = []
imgshortfilenames = []

# Get a unified path so we can stop dancing with user paths.
# Determine where files are on this machine (%TEMP% directory and application installation directory)
if '.exe' in sys.argv[0]: # if getattr(sys, 'frozen', False):
    RootPath = os.path.abspath(os.path.join(__file__, "..\\"))

elif __file__:
    RootPath = os.path.abspath(os.path.join(__file__, "..\\"))

print ("\n storage of image files RootPath: %s\n" %RootPath)

FolderPath = "D:\\TFS-FARM1\\StoneSoup_STS\\SDLC\\Build\\Code\\StoneSoup_Refactor\\StoneSoupUI\\Images"
print ("\n storage of image files in folder to search: %s\n" %FolderPath)

for root, directories, filenames2 in os.walk(FolderPath):
    for filename in filenames2:
        fullname = os.path.join(root, filename)
        filenames.append(fullname)
        shortfilenames.append(filename)

for i, fname in enumerate(shortfilenames):
        print("%s - %s" % (i+1, fname))

for root, directories, filenames2 in os.walk(RootPath):
    for filename in filenames2:
        fullname = os.path.join(root, filename)
        imgfilenames.append(fullname)
        imgshortfilenames.append(filename)

for i, fname in enumerate(imgshortfilenames):
        print("%s - %s" % (i+1, fname))

for i, fname in enumerate(imgshortfilenames):
        if fname in shortfilenames:
            print("%s - %s exists" % (i+1, fname))
        else:
            print("%s - %s ABSENT" % (i+1, fname))

它似乎没有一个有意义的功能区别在尝试/排除和isfile(),所以你应该使用哪一个有意义。

如果你想阅读一个文件,如果它存在,

try:
    f = open(filepath)
except IOError:
    print 'Oh dear.'

但是,如果你只是想重新命名一个文件,如果它存在,因此不需要打开它,

if os.path.isfile(filepath):
    os.rename(filepath, filepath + '.old')

如果你想写到文件,如果它不存在,

# Python 2
if not os.path.isfile(filepath):
    f = open(filepath, 'w')

# Python 3: x opens for exclusive creation, failing if the file already exists
try:
    f = open(filepath, 'wx')
except IOError:
    print 'file already exists'

如果你需要文件锁,那是另一个问题。