Python 如何打开文件夹中的每个文件？

How to open every file in a folder?

提问by B.Mr.W.

I have a python script parse.py, which in the script open a file, say file1, and then do something maybe print out the total number of characters.

我有一个 python 脚本 parse.py，它在脚本中打开一个文件，比如 file1，然后做一些事情可能会打印出字符的总数。

filename = 'file1'
f = open(filename, 'r')
content = f.read()
print filename, len(content)

Right now, I am using stdout to direct the result to my output file - output

现在，我正在使用 stdout 将结果定向到我的输出文件 - output

python parse.py >> output

However, I don't want to do this file by file manually, is there a way to take care of every single file automatically? Like

但是，我不想手动逐个文件地执行此文件，有没有办法自动处理每个文件？喜欢

ls | awk '{print}' | python parse.py >> output 

Then the problem is how could I read the file name from standardin? or there are already some built-in functions to do the ls and those kind of work easily?

那么问题是如何从standardin读取文件名？或者已经有一些内置函数可以轻松完成 ls 和那些工作？

Thanks!

谢谢！

采纳答案by Viktor Kerkez

Os

操作系统

You can list all files in the current directory using os.listdir:

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

Glob

球体

Or you can list only some files, depending on the file pattern using the globmodule:

或者您可以仅列出一些文件，具体取决于使用glob模块的文件模式：

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

It doesn't have to be the current directory you can list them in any path you want:

它不必是当前目录，您可以将它们列在您想要的任何路径中：

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

PipeOr you can even use the pipe as you specified using fileinput

管道或者您甚至可以使用指定的管道fileinput

import fileinput
for line in fileinput.input():
    # do your stuff

And then use it with piping:

然后将其与管道一起使用：

ls -1 | python parse.py

回答by le_vine

you should try using os.walk

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by geekidharsh

You can actually just use os moduleto do both:

你实际上可以只使用os 模块来做这两个：

1. list all files in a folder
2. sort files by file type, file name etc.

1. 列出文件夹中的所有文件
2. 按文件类型、文件名等对文件进行排序。

Here's a simple example:

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

---

---

Now you have not only listed all the files in a folder but also have them (optionally) sorted by starting name, file type and others. Just now iterate over each list and do your stuff.

现在您不仅列出了文件夹中的所有文件，而且还让它们（可选）按起始名称、文件类型和其他排序。现在迭代每个列表并做你的事情。

回答by Andrei

I was looking for this answer:

我正在寻找这个答案：

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))

you can choose as well '*.txt' or other ends of your filename

您也可以选择“*.txt”或文件名的其他结尾

回答by RockwellS

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by SKhendle

The code below reads for any text files available in the directory which contains the script we are running. Then it opens every text file and stores the words of the text line into a list. After store the words we print each word line by line

下面的代码读取包含我们正在运行的脚本的目录中可用的任何文本文件。然后它打开每个文本文件并将文本行的单词存储到一个列表中。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)