Develop Reference

How do I select all images from a folder? -MoviePy, Python

I'm pretty new in python and trying to make a python script to put images together into a video(.mp4) using MoviePy.
However, I have multiple files and would like to be more efficient by sort of.... naming the folder and selecting all images within that folder than having to select all images individually.
Here's my Code:
from moviepy.editor import *
import os
clips = []
clip1 = ImageClip('imagesfolder\images0.jpg').set_duration(4)
clip2 = ImageClip('imagesfolder\images1.jpg').set_duration(4)
clip3 = ImageClip('imagesfolder\images2.jpg').set_duration(4)
clip4 = ImageClip('imagesfolder\images3.jpg').set_duration(4)
clip5 = ImageClip('imagesfolder\images4.jpg').set_duration(4)
clip6 = ImageClip('imagesfolder\images5.jpg').set_duration(4)
clip7 = ImageClip('imagesfolder\images6.jpg').set_duration(4)
clip8 = ImageClip('imagesfolder\images7.jpg').set_duration(4)
clip9 = ImageClip('imagesfolder\images8.jpg').set_duration(4)
clip10 = ImageClip('imagesfolder\images9.jpg').set_duration(4)
clips.append(clip1)
clips.append(clip2)
clips.append(clip3)
clips.append(clip4)
clips.append(clip5)
clips.append(clip6)
clips.append(clip7)
clips.append(clip8)
clips.append(clip9)
clips.append(clip10)
video_clip = concatenate_videoclips(clips, method='compose')
video_clip.write_videofile("memes.mp4", fps=24, remove_temp=True, codec="libx264",
audio_codec="aac")

You can use a function called glob to find all files in a directly which match a pattern.
Eg
from glob import glob
clips = [ImageClip(clip).set_duration(4) for clip in glob("imagesfolder\*.gif")]
video_clip = concatenate_videoclips(clips, method="compose")

Python: How do I assign the result of my function to a variable that i can read_csv

The code that I have determines which Operating System is being used. Then it has to search the entire system for my csv file. When it's found I need to be able to read in the csv file (so that its not just inside the function, but useable throughout my code).
So far I am able to locate my file, but I am having trouble to assign the filepath to a variable, so that I can read in that variabel with pd.read_csv()
the code that I have is at follows:
import pandas as pd
import os
import re
import win32api
# https://stackoverflow.com/questions/13067686/search-files-in-all-drives-using-python
def find_file(root_folder, rex):
for root,dirs,files in os.walk(root_folder):
for f in files:
result = rex.search(f)
if result:
print(os.path.join(root, f))
return result
break # if you want to find only one
def find_file_in_all_drives(file_name):
#create a regular expression for the file
rex = re.compile(file_name)
for drive in win32api.GetLogicalDriveStrings().split('\000')[:-1]:
find_file( drive, rex )
return
#file_name = "AB_NYC_2019.csv"
#find_file_in_all_drives(file_name)
df_location = find_file_in_all_drives( "AB_NYC_2019.csv" )
df = pd.read_csv(df_location)
I think that something is not right with the return.
Thank you for your time.
Right now it returns "None"

You haven't returned anything from anywhere.
I'm considering your code to be working and I've placed the necessary return calls but haven't tested it:
def find_file(root_folder, rex):
for root, dirs, files in os.walk(root_folder):
for f in files:
result = rex.search(f)
if result:
file_path = os.path.join(root, f)
return file_path
def find_file_in_all_drives(file_name):
matching_files = list()
# create a regular expression for the file
rex = re.compile(file_name)
for drive in win32api.GetLogicalDriveStrings().split('\000')[:-1]:
file_path = find_file(drive, rex)
if file_path:
matching_files.append(file_path)
return matching_files
df_location = find_file_in_all_drives("AB_NYC_2019.csv")
first_file_df = pd.read_csv(df_location[0])

How to unpack a zip file without changing metadata? [duplicate]

I'm trying to extract files from a zip file using Python 2.7.1 (on Windows, fyi) and each of my attempts shows extracted files with Modified Date = time of extraction (which is incorrect).
import os,zipfile
outDirectory = 'C:\\_TEMP\\'
inFile = 'test.zip'
fh = open(os.path.join(outDirectory,inFile),'rb')
z = zipfile.ZipFile(fh)
for name in z.namelist():
z.extract(name,outDirectory)
fh.close()
I also tried using the .extractall method, with the same results.
import os,zipfile
outDirectory = 'C:\\_TEMP\\'
inFile = 'test.zip'
zFile = zipfile.ZipFile(os.path.join(outDirectory,inFile))
zFile.extractall(outDirectory)
Can anyone tell me what I'm doing wrong?
I'd like to think this is possible without having to post-correct the modified time per How do I change the file creation date of a Windows file?.

Well, it does take a little post-processing, but it's not that bad:
import os
import zipfile
import time
outDirectory = 'C:\\TEMP\\'
inFile = 'test.zip'
fh = open(os.path.join(outDirectory,inFile),'rb')
z = zipfile.ZipFile(fh)
for f in z.infolist():
name, date_time = f.filename, f.date_time
name = os.path.join(outDirectory, name)
with open(name, 'wb') as outFile:
outFile.write(z.open(f).read())
date_time = time.mktime(date_time + (0, 0, -1))
os.utime(name, (date_time, date_time))
Okay, maybe it is that bad.

Based on Jia103's answer, I have developed a function (using Python 2.7.14) which preserves directory and file dates AFTER everything has been extracted. This isolates any ugliness in the function, and you can also use zipfile.Zipfile.extractAll() or whatever zip extract method you want:
import time
import zipfile
import os
# Restores the timestamps of zipfile contents.
def RestoreTimestampsOfZipContents(zipname, extract_dir):
for f in zipfile.ZipFile(zipname, 'r').infolist():
# path to this extracted f-item
fullpath = os.path.join(extract_dir, f.filename)
# still need to adjust the dt o/w item will have the current dt
date_time = time.mktime(f.date_time + (0, 0, -1))
# update dt
os.utime(fullpath, (date_time, date_time))
To preserve dates, just call this function after your extract is done.
Here's an example, from a script I wrote to zip/unzip game save directories:
z = zipfile.ZipFile(zipname, 'r')
print 'I have opened zipfile %s, ready to extract into %s' \
% (zipname, gamedir)
try: os.makedirs(gamedir)
except: pass # Most of the time dir already exists
z.extractall(gamedir)
RestoreTimestampsOfZipContents(zipname, gamedir) #<-- USED
print '%s zip extract done' % GameName[game]
Thanks everyone for your previous answers!

Based on Ethan Fuman's answer, I have developed this version (using Python 2.6.6) which is a little more consise:
zf = ZipFile('archive.zip', 'r')
for zi in zf.infolist():
zf.extract(zi)
date_time = time.mktime(zi.date_time + (0, 0, -1))
os.utime(zi.filename, (date_time, date_time))
zf.close()
This extracts to the current working directory and uses the ZipFile.extract() method to write the data instead of creating the file itself.

Based on Ber's answer, I have developed this version (using Python 2.7.11), which also accounts for directory mod dates.
from os import path, utime
from sys import exit
from time import mktime
from zipfile import ZipFile
def unzip(zipfile, outDirectory):
dirs = {}
with ZipFile(zipfile, 'r') as z:
for f in z.infolist():
name, date_time = f.filename, f.date_time
name = path.join(outDirectory, name)
z.extract(f, outDirectory)
# still need to adjust the dt o/w item will have the current dt
date_time = mktime(f.date_time + (0, 0, -1))
if (path.isdir(name)):
# changes to dir dt will have no effect right now since files are
# being created inside of it; hold the dt and apply it later
dirs[name] = date_time
else:
utime(name, (date_time, date_time))
# done creating files, now update dir dt
for name in dirs:
date_time = dirs[name]
utime(name, (date_time, date_time))
if __name__ == "__main__":
unzip('archive.zip', 'out')
exit(0)
Since directories are being modified as the extracted files are being created inside them, there appears to be no point in setting their dates with os.utime until after the extraction has completed, so this version caches the directory names and their timestamps till the very end.

How to extract the file name from a file path?

I have the following code:
os.listdir("staging")
# Seperate filename from extension
sep = os.sep
# Change the casing
for n in os.listdir("staging"):
print(n)
if os.path.isfile("staging" + sep + n):
filename_one, extension = os.path.splitext(n)
os.rename("staging" + sep + n, "staging" + sep + filename_one.lower() + extension)
# Show the new file names
print ('\n--------------------------------\n')
for n in os.listdir("staging"):
print (n)
# Remove the blanks, -, %, and /
for n in os.listdir("staging"):
print (n)
if os.path.isfile("staging" + sep + n):
filename_zero, extension = os.path.splitext(n)
os.rename("staging" + sep + n , "staging" + sep + filename_zero.replace(' ','_').replace('-','_').replace('%','pct').replace('/','_') + extension)
# Show the new file names
print ('\n--------------------------------\n')
for n in os.listdir("staging"):
print (n)
"""
In order to fix all of the column headers and to solve the encoding issues and remove nulls,
first read in all of the CSV's to python as dataframes, then make changes and rewrite the old files
"""
import os
import glob
import pandas as pd
files = glob.glob(os.path.join("staging" + "/*.csv"))
print(files)
# Create an empty dictionary to hold the dataframes from csvs
dict_ = {}
# Write the files into the dictionary
for file in files:
dict_[file] = pd.read_csv(file, header = 0, dtype = str, encoding = 'cp1252').fillna('')
In the dictionary, the dataframes are named as "folder/name(csv)" what I would like to do is remove the prefix "staging/" from the keys in the dictionary.
How can I do this?

If all you want to do is truncate the file paths to just the filename, you can use os.path.basename:
for file in files:
fname = os.path.basename(file)
dict_[fname] = (pd.read_csv(file, header=0, dtype=str, encoding='cp1252')
.fillna(''))
Example:
os.path.basename('Desktop/test.txt')
# 'test.txt'

import os
pathname ='c:\\hello\\dickins\\myfile.py'
head, tail = os.path.split(pathname)
print head
print tail

This article here worked out just fine for me
import os
inputFilepath = 'path/to/file/foobar.txt'
filename_w_ext = os.path.basename(inputFilepath)
filename, file_extension = os.path.splitext(filename_w_ext)
#filename = foobar
#file_extension = .txt
path, filename = os.path.split(path/to/file/foobar.txt)
# path = path/to/file
# filename = foobar.txt
Hope it helps someone searching for this answer

In the same spirt as truncate the file paths, use pathlib in python standard library. It will turn the path into an easy to use class.
from pathlib import Path
path = Path('Desktop/folder/test.txt')
path.name # test.txt
path.stem # test
path.suffix # .txt
path.parent.name # folder
path.parent.name.name # Desktop

As ColdSpeed said, you can use "os.path.basename" to truncate a file to its name, but I think what you are refering to is the ability to pycache the data?
For Example here is my Directory:
You see the pycache folder? that initializes it as a module.
Then, you can import a file from that module (for example the staging.txt file and operate on it.)
I use the IpConfig.txt File from the assets folder level (or should be) and take a line of information out of it.
import pygame as pyg
import sys
import os
import math
import ssl
import socket as sock
import ipaddress as ipad
import threading
import random
print("Modules Installed!")
class two:
# Find out how to refer to class super construct
def main(Display, SecSock, ipadd, clock):
# I have code here that has nothing to do with the question...
def __init__():
print("Initializing[2]...")
# Initialization of Pygame and SSL Socket goes here
searchQuery = open("IpConfig.txt", 'r') #Opening the File IpConfig(Which now should open on the top level of the game files)
step2 = searchQuery.readlines()# read the file
ipadd = step2[6] # This is what you should have or something similar where you reference the line you want to copy or manipulate.
main(gameDisplay, SSLSock, ipadd, clock)# Im having issues here myself - (main() is not defined it says)
print(ipadd)
print("Server Certificate Configuration Enabled...")
__init__() # Start up the procedure

How do I apply my python code to all of the files in a folder at once, and how do I create a new name for each subsequent output file?

The code I am working with takes in a .pdf file, and outputs a .txt file. My question is, how do I create a loop (probably a for loop) which runs the code over and over again on all files in a folder which end in ".pdf"? Furthermore, how do I change the output each time the loop runs so that I can write a new file each time, that has the same name as the input file (ie. 1_pet.pdf > 1_pet.txt, 2_pet.pdf > 2_pet.txt, etc.)
Here is the code so far:
path="2_pet.pdf"
content = getPDFContent(path)
encoded = content.encode("utf-8")
text_file = open("Output.txt", "w")
text_file.write(encoded)
text_file.close()

The following script solve your problem:
import os
sourcedir = 'pdfdir'
dl = os.listdir('pdfdir')
for f in dl:
fs = f.split(".")
if fs[1] == "pdf":
path_in = os.path.join(dl,f)
content = getPDFContent(path_in)
encoded = content.encode("utf-8")
path_out = os.path.join(dl,fs[0] + ".txt")
text_file = open(path_out, 'w')
text_file.write(encoded)
text_file.close()

Create a function that encapsulates what you want to do to each file.
import os.path
def parse_pdf(filename):
"Parse a pdf into text"
content = getPDFContent(filename)
encoded = content.encode("utf-8")
## split of the pdf extension to add .txt instead.
(root, _) = os.path.splitext(filename)
text_file = open(root + ".txt", "w")
text_file.write(encoded)
text_file.close()
Then apply this function to a list of filenames, like so:
for f in files:
parse_pdf(f)

One way to operate on all PDF files in a directory is to invoke glob.glob() and iterate over the results:
import glob
for path in glob.glob('*.pdf')
content = getPDFContent(path)
encoded = content.encode("utf-8")
text_file = open("Output.txt", "w")
text_file.write(encoded)
text_file.close()
Another way is to allow the user to specify the files:
import sys
for path in sys.argv[1:]:
...
Then the user runs your script like python foo.py *.pdf.

You could use a recursive function to search the folders and all subfolders for files that end with pdf. Than take those files and then create a text file for it.
It could be something like:
import os
def convert_PDF(path, func):
d = os.path.basename(path)
if os.path.isdir(path):
[convert_PDF(os.path.join(path,x), func) for x in os.listdir(path)]
elif d[-4:] == '.pdf':
funct(path)
# based entirely on your example code
def convert_to_txt(path):
content = getPDFContent(path)
encoded = content.encode("utf-8")
file_path = os.path.dirname(path)
# replace pdf with txt extension
file_name = os.path.basename(path)[:-4]+'.txt'
text_file = open(file_path +'/'+file_name, "w")
text_file.write(encoded)
text_file.close()
convert_PDF('path/to/files', convert_to_txt)
Because the actual operation is changeable, you can replace the function with whatever operation you need to perform (like using a different library, converting to a different type, etc.)