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))

这里是一个单线的Python命令为Linux命令线环境,我觉得这是非常有用的,因为我不是那么热的Bash男孩。

python -c "import os.path; print os.path.isfile('/path_to/file.xxx')"

使用:

import os

# For testing purposes the arguments defaulted to the current folder and file.
# returns True if file found
def file_exists(FOLDER_PATH='../', FILE_NAME=__file__):
    return os.path.isdir(FOLDER_PATH) \
        and os.path.isfile(os.path.join(FOLDER_PATH, FILE_NAME))

它基本上是一个文件夹检查,然后使用 os.path.join 使用适当的目录分离器进行文件检查。

import os
os.path.exists(path) # Returns whether the path (directory or file) exists or not
os.path.isfile(path) # Returns whether the file exists or not

您可以使用Python的“OS”图书馆:

>>> import os
>>> os.path.exists("C:\\Users\\####\\Desktop\\test.txt") 
True
>>> os.path.exists("C:\\Users\\####\\Desktop\\test.tx")
False

你可以写布莱恩的建议,没有尝试:

from contextlib import suppress

with suppress(IOError), open('filename'):
    process()

suppress 是 Python 3.4 的组成部分。在旧版本中,您可以快速写下自己的 suppress:

from contextlib import contextmanager

@contextmanager
def suppress(*exceptions):
    try:
        yield
    except exceptions:
        pass