问题：如何打开文件夹中的每个文件？
我有一个python脚本parse.py，该脚本在脚本中打开一个文件，例如file1，然后执行一些操作，可能会打印出字符总数。
filename = 'file1'
f = open(filename, 'r')
content = f.read()
print filename, len(content)
现在，我正在使用stdout将结果定向到我的输出文件-输出
python parse.py >> output但是，我不想按文件手动处理此文件，有没有办法自动处理每个文件？喜欢
ls | awk '{print}' | python parse.py >> output 然后问题是如何从standardin中读取文件名？还是已经有一些内置函数可以轻松执行ls和此类工作？
谢谢！
回答 0
操作系统
您可以使用以下命令列出当前目录中的所有文件os.listdir：
import os
for filename in os.listdir(os.getcwd()):
with open(os.path.join(os.cwd(), filename), 'r') as f: # open in readonly mode
# do your stuff
球状
或者，您可以根据glob模块的文件模式仅列出一些文件：
import glob
for filename in glob.glob('*.txt'):
with open(os.path.join(os.cwd(), filename), 'r') as f: # open in readonly mode
# do your stuff
不必是当前目录，您可以在所需的任何路径中列出它们：
path = '/some/path/to/file'
for filename in glob.glob(os.path.join(path, '*.txt')):
with open(os.path.join(os.cwd(), filename), 'r') as f: # open in readonly mode
# do your stuff
管道
或者您甚至可以使用指定的管道来使用fileinput
import fileinput
for line in fileinput.input():
# do your stuff
然后将其与管道一起使用：
ls -1 | python parse.py回答 1
你应该尝试使用os.walk
yourpath = 'path'
import os
for root, dirs, files in os.walk(yourpath, topdown=False):
for name in files:
print(os.path.join(root, name))
stuff
for name in dirs:
print(os.path.join(root, name))
stuff回答 2
我一直在寻找这个答案：
import os,glob
folder_path = '/some/path/to/file'
for filename in glob.glob(os.path.join(folder_path, '*.htm')):
with open(filename, 'r') as f:
text = f.read()
print (filename)
print (len(text))您也可以选择“ * .txt”或文件名的另一端
回答 3
您实际上可以只使用os模块来完成这两个操作：
- 列出文件夹中的所有文件
- 按文件类型，文件名等对文件进行排序
这是一个简单的例子：
import os #os module imported here
location = os.getcwd() # get present working directory location here
counter = 0 #keep a count of all files found
csvfiles = [] #list to store all csv files found at location
filebeginwithhello = [] # list to keep all files that begin with 'hello'
otherfiles = [] #list to keep any other file that do not match the criteria
for file in os.listdir(location):
try:
if file.endswith(".csv"):
print "csv file found:\t", file
csvfiles.append(str(file))
counter = counter+1
elif file.startswith("hello") and file.endswith(".csv"): #because some files may start with hello and also be a csv file
print "csv file found:\t", file
csvfiles.append(str(file))
counter = counter+1
elif file.startswith("hello"):
print "hello files found: \t", file
filebeginwithhello.append(file)
counter = counter+1
else:
otherfiles.append(file)
counter = counter+1
except Exception as e:
raise e
print "No files found here!"
print "Total files found:\t", counter现在，您不仅列出了文件夹中的所有文件，而且（可选）按起始名称，文件类型等排序。刚才遍历每个列表并做您的工作。
回答 4
import pyautogui
import keyboard
import time
import os
import pyperclip
os.chdir("target directory")
# get the current directory
cwd=os.getcwd()
files=[]
for i in os.walk(cwd):
for j in i[2]:
files.append(os.path.abspath(j))
os.startfile("C:\Program Files (x86)\Adobe\Acrobat 11.0\Acrobat\Acrobat.exe")
time.sleep(1)
for i in files:
print(i)
pyperclip.copy(i)
keyboard.press('ctrl')
keyboard.press_and_release('o')
keyboard.release('ctrl')
time.sleep(1)
keyboard.press('ctrl')
keyboard.press_and_release('v')
keyboard.release('ctrl')
time.sleep(1)
keyboard.press_and_release('enter')
keyboard.press('ctrl')
keyboard.press_and_release('p')
keyboard.release('ctrl')
keyboard.press_and_release('enter')
time.sleep(3)
keyboard.press('ctrl')
keyboard.press_and_release('w')
keyboard.release('ctrl')
pyperclip.copy('')回答 5
下面的代码读取包含我们正在运行的脚本的目录中所有可用的文本文件。然后，它将打开每个文本文件，并将文本行中的单词存储到列表中。存储单词后，我们逐行打印每个单词
import os, fnmatch
listOfFiles = os.listdir('.')
pattern = "*.txt"
store = []
for entry in listOfFiles:
if fnmatch.fnmatch(entry, pattern):
_fileName = open(entry,"r")
if _fileName.mode == "r":
content = _fileName.read()
contentList = content.split(" ")
for i in contentList:
if i != '\n' and i != "\r\n":
store.append(i)
for i in store:
print(i)
