如何打开文件夹中的每个文件

Question

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.

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

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

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?

Thanks!

Answer 1

Os

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

import os
for filename in os.listdir(os.getcwd()):
   with open(os.path.join(os.getcwd(), filename), 'r') as f: # open in readonly mode
      # do your stuff

Glob

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

import os, glob
for filename in glob.glob('*.txt'):
   with open(os.path.join(os.getcwd(), 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:

import os, glob
path = '/some/path/to/file'
for filename in glob.glob(os.path.join(path, '*.txt')):
   with open(os.path.join(os.getcwd(), filename), 'r') as f: # open in readonly mode
      # do your stuff

Pipe

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

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

And you can then use it with piping:

ls -1 | python parse.py

Answer 2

You should try using os.walk .

import os

yourpath = 'path'

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

Answer 3

I was looking for this answer:

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

Answer 4

You can actually just use os module to do both:

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

Here's a simple example:

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.

Answer 5

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('')

Answer 6

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

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)

Answer 7

If you would like to open files in a directory and append them into a list, do this:

mylist=[]
for filename in os.listdir('path/here/'):
with open(os.path.join('path/here/', filename), 'r') as f:
    mylist.append(f.read())

Answer 8

you may try another approach of using os.walk and os.path.join which is a little different from the above options:

for root, dirs, files in os.walk(EnterYourPath):
    for name in files:
        with open(os.path.join(root,name))as f:
            text = f.read()

text variable includes all the files in the folder in the directory.