I follow some articles regarding on how to list files in a directory using the below code :
import os, sys
path = "/python 3.6.4"
dirs = os.list(path)
for file in dirs:
print file
Note: The .py file in which the above code is on the same directory as the path.
Im running it using a command prompt. Any idea would be much appreciated. Thanks!
Related
I'm using Windows 10
Anyway I used os.rename and forgot to add a directory path to the beginning. Where did my files end up?
The file could be sent to the current working directory. To find the current working directory, use
import os
os.getcwd()
To find out where the file you are executing is located, use
import os
os.path.dirname(os.path.realpath(__file__))
This question already has an answer here:
Could not open resource file, pygame error: "FileNotFoundError: No such file or directory."
(1 answer)
Closed 2 years ago.
I am somewhat new to programming, I don't like it that much but as it is school work I have to do it, good thing I find it interesting.
Here is my code, note I'm using VS code (some things are in french but that shouldn't affect anything):
import pygame
screen = pygame.display.set_mode((800,600))
pygame.display.set_caption("Space vaders")
fond = pygame.image.load (testingfield.png)
and I get:
Traceback (most recent call last):
File "c:/Users/faver/Desktop/Space_Invaders-NSI/space_invaders_1.py", line 11, in <module>
fond = pygame.image.load ('testingfield.png')
FileNotFoundError: No such file or directory.
the file is indeed in the folder, and if I put fond = pygame.image.load ((r'C:\Users\faver\Desktop\Space_Invaders-NSI\testingfield.png'))it works perfectly, the problem now is that it'd be hard to actually send it since the other person would lack the folders I place there.
I've also tried import os a = os.path.lexists("/Users/suyeon/Documents/pythonWorkspace/pygame_basic/character.png") print(a) and it returned false, so I'm pretty much confused, I'd appreciate help if possible ^^'
The image file path has to be relative to the current working directory. The working directory is possibly different to the directory of the python file.
The name and path of the file can be get by __file__. The current working directory can be get by os.getcwd() and can be changed by os.chdir(path).
One solution is to change the working directory:
import os
os.chdir(os.path.dirname(os.path.abspath(__file__)))
An alternative solution is to find the absolute path.
If the image is relative to the folder of the python file (or even in the same folder), then you can get the directory of the file and join (os.path.join()) the image filename. e.g.:
import pygame
import os
# get the directory of this file
sourceFileDir = os.path.dirname(os.path.abspath(__file__))
# [...]
# join the filepath and the filename
fondImgPath = os.path.join(sourceFileDir, 'testingfield.png')
fond = pygame.image.load(fondImgPath)
How to extract a zip file using python when zip file present in different directory where script file present.
I try this ,but i got error because source path is not accepted ,try to solve me this problem.
from zipfile import ZipFile
def func(source, target):
with ZipFile('source', 'target'):
ZipFile.Extractall('target')
Use this code. To move through directories you could either hard code the directory where your script is present or you could toggle through the directories using simple commands such as "../" to move out of the given directory or "/" to move inside a folder in the directory. For example - "../script.py" or "/folder/script.py". Similarly you can use this to find your .zip file.
import zipfile
with zipfile.ZipFile("file.zip","r") as zip_ref:
zip_ref.extractall("targetdir")
For just unpacking, shutil should suffice:
import shutil
shutil.unpack_archive('path-to-zipfile')
You'll have to check for source path of the zip file which is relative to your current working directory. To know your current working directory you can try
import os
print(os.getcwd())
zip - Unzipping files in python
relative-paths-in-python
I have downloaded a csv file(ABC) using python and selenium and need to change the name of the file.
Input= ABC.csv #downloaded file with name ABC
Output= DEF.csv #New file with name DEF.
Any help is appreciated.
try this
import os
os.rename('a.txt', 'b.kml')
From:
How to rename a file using Python
Thanks All.. Used the below code and it is working fine. All others kept throwing the error as File not found. Used complete file path.
os.rename('C:\\Users\\pathname\\ABC.csv', 'C:\\Users\\pathname\\DEF.csv')
You can use os module to rename the file:
import os
os.rename('path to ABC.csv', 'path to DEF.csv')
Explanation:
rename first argument is the original file which needs to be renamed and second argument is the new name which it should be renamed. If the files are in current directory it will work by simply using the file names. Otherwise include the path where these files are there
More information at docs **https://docs.python.org/3/library/os.html#os.rename
I have one application developed in python 3.2, which has inbuilt modules(ex: Tkinter, matplotlib, openpyxl), user defined modules & classes(ex: draw_graph, generate_report), icon files, log file, .csv, .docx etc. I am running this application from script(ex: testapplication.py)
I have setup file as
import sys
from cx_Freeze import setup, Executable
exe = Executable(
script=r"C:\Python32\testapplication.py",
base="Win32GUI",
)
setup(
name = "TESTApp",
version = "0.1",
description = "An example",
executables = [exe]
)
Now I want to create a exe file of this application. can anyone please suggest me a way to do this?
So this is what you need to do. For starters, change script=r"C:\Python32\testapplication.py" to script=r"testapplication.py"
Then, put ALL the files to need to convert into C/python32 including the setup file. Then what you wan to do is get your command line up, and type the following commands: (assuming that you're cx_freeze file is named setup.py):
cd
cd python32
python setup.py build
And then you should have a build folder in that directory containing the exe file.