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How to zip files that ends with certain extension

I want to get all files in a directory (I reached it after doing several for loops - hence fourth.path) that ends with .npy or with csv and then zip those files.
My code is running putting one file only in the zip file. What am I doing wrong?
I tried to change my indents, but no zip file is being created
import json
import os
import zipfile
import zlib
directory = os.path.join(os.getcwd(), 'recs')
radarfolder = 'RadarIfxAvian'
file = os.listdir(directory)
def r(p, name):
p = os.path.join(p, name)
return p.replace("/", "\\")
#This code will list all json files in e ach file
for first in os.scandir(directory):
if first.is_dir():
for second in os.scandir(first.path):
if second.is_dir():
for third in os.scandir(second.path):
if third.is_dir():
radar_folder_name = ''
list_files = ()
for fourth in os.scandir(third.path):
if fourth.is_dir():
if radarfolder in fourth.path:
radar_folder_name = fourth.path
print(radar_folder_name)
list_files = ()
for file in os.listdir(fourth.path):
if file.endswith(".npy") | file.endswith(".csv"):
list_files = (file)
print(list_files)
with zipfile.ZipFile(radar_folder_name +'\\' +'radar.zip', 'w', compression=zipfile.ZIP_DEFLATED ) as zipMe:
zipMe.write(radar_folder_name +'\\' +list_files)
zipMe.close()
I tried to change my indents either resulting in error: TypeError: can only concatenate str (not "tuple") to str or no zip file being created

As I said in my second comment, your problem comes from the 'w' argument in your zipping statement. It causes the zip to be overwritten every time it's opened, which you do for each file you zip in. You can fix this 2 ways (at least):
Replace 'w' with 'a'; this way the files will be appended to your zip (with the side effect that, if you do this several times, files will be added more than once).
Keep the 'w', but only open the zip once, having listed all the files you want to zip before. See my code below.
I've taken the liberty to rewrite the part of your code where you look for the 'RadarIfxAvian' folder, since embedded for are clumsy (and if your folder structure changes, they might not work), replacing it with a multi-purpose recursive function.
Note that the folder structure will be included in the .zip; if you want to zip only the files themselves, consider doing os.chdir(radar_folder_name) before zipping the files.
# This function recursively looks for the 'filename' file or folder
# under 'start_path' and returns the full path, or an empty string if not found.
def find_file(start_path, filename):
if filename in os.listdir(start_path):
return start_path + '/' + filename
for file in os.scandir(start_path):
if not file.is_dir():
continue
if (deep_path:=find_file(start_path + '/' + file.name, filename)):
return deep_path
return ''
directory = os.path.join(os.getcwd(), 'recs')
radarfolder = 'RadarIfxAvian'
radar_folder_name = find_file(directory, radarfolder)
print(radar_folder_name)
list_files = []
for file in os.listdir(radar_folder_name):
if file.endswith(".npy") or file.endswith(".csv"):
list_files.append(file)
with zipfile.ZipFile(radar_folder_name + '/' + 'radar.zip', 'w', compression=zipfile.ZIP_DEFLATED ) as zipMe:
for file in list_files:
zipMe.write(radar_folder_name + '/' + file)

If I understand your code correctly, you are looking for a folder "RadarIfxAvian" and want to place a .ZIP in that folder containing any .CSV or .NPY files in that directory. This should do the equivalent, using os.walk for the recursive search:
import os
import zipfile
for path, dirs, files in os.walk('recs'):
if os.path.basename(path) == 'RadarIfxAvian':
print(path)
with zipfile.ZipFile(os.path.join(path, 'radar.zip'), 'w', zipfile.ZIP_DEFLATED) as zip:
for file in files:
if file.endswith(".npy") | file.endswith(".csv"):
print(file)
zip.write(file)
break # stop search once the directory is found and processed

I adjusted my code with the following steps:
Put the if in a function
writing the the zip by looping over each item in the list I appended
import json
import os
import glob
import zipfile
import zlib
directory = os.path.join(os.getcwd(), 'recs')
radarfolder = 'RadarIfxAvian'
file = os.listdir(directory)
list_files = []
def r(p, name):
p = os.path.join(p, name)
return p.replace("/", "\\")
def tozip(path, file):
filestozip = []
if file.endswith(".npy") or file.endswith(".csv"):
filestozip = (path + '\\' + file)
list_files.append(filestozip)
return list_files
#This code will list all json files in each file
for first in os.scandir(directory):
if first.is_dir():
for second in os.scandir(first.path):
if second.is_dir():
for third in os.scandir(second.path):
if third.is_dir():
radar_folder_name = ''
filestozip = []
list_files.clear()
for fourth in os.scandir(third.path):
if fourth.is_dir():
if radarfolder in fourth.path:
radar_folder_name = fourth.path
for file in os.listdir(fourth.path):
filestozip = tozip(radar_folder_name,file)
print(filestozip)
ZipFile = zipfile.ZipFile(r(radar_folder_name,"radar.zip"), "w")
for a in filestozip:
ZipFile.write(a, compress_type= zipfile.ZIP_DEFLATED)
print(radar_folder_name + "added to zip")

Rename multiple files by removing filename prefix

I'm new to Python, I have multiple files in a folder where I need to rename those files as the the given pattern.
Example:
Folder : /Users/Usr1/Documents/FilesFolder and
File's :
0. a101.employee.txt
1. a101.department.txt
2. a101.salary.txt
I want to remove the prefix of the file till a101 and rename to empoloyee.txt/salary.txt.
Expected Output:
My try:
import os
path = '/Users/User1/Documents/FilesFolder'
files = os.listdir(path)
for index, file in enumerate(files):
os.rename(os.path.join(path, file), os.path.join(path,file.removeprefix('a101')))
But unable to get expected result.

You may use regular expression:
import os
import re
path = '/Users/User1/Documents/FilesFolder'
files = os.listdir(path)
p = ".*a101.(.+)"
for file in files:
m = re.match(p, file)
if m is not None:
file_new = m.group(1)
print(file_new)

I think this can solve your problem
import os
import glob
# directory Path
path = "/path/to/dir"
# move to directory
os.chdir(path)
# Getting all files in the directory which contains a101
files = glob.glob("*a101*")
for file in files:
splitted = file.split('.')
filename, ext = splitted[-2], splitted[-1]
new_name = f"{filename}.{ext}"
os.rename(file, new_name)

If your file name is following same pattern with 3 . then you can use this for renaming. removeprefix is introduced in python 3.9.
files = ["0. a101.employee.txt", "1. a101.department.txt" ,"2. a101.salary.txt"]
for file in files:
print(".".join(file.split(".")[-2:]))
output:
employee.txt
department.txt
salary.txt

i can suggest you:
files = ["0. a101.employee.txt", "1. a101.department.txt" ,"2. a101.salary.txt"]
for index, file in enumerate(files):
filename = file.split(".")
print(filename[2]+"."+filename[3])
I got the followingoutput:
employee.txt
department.txt
salary.txt

I want to move a file based on part of the name to a folder with that name

I have a directory with a large number of files that I want to move into folders based on part of the file name. My list of files looks like this:
001-020-012B-B.nc
001-022-151-A.nc
001-023-022-PY-T1.nc.nc
001-096-016B-A.nc
I want to move the files I have into separate folders based on the first part of the file name (001-096-016B, 001-023-022, 001-022-151). The first parts of the file name always have the same number of numbers and are always in 3 parts separated by an underscore '-'.
The folder names are named like this \oe-xxxx\xxxx\xxxx\001-Disc-PED\020-Rotor-parts-1200.
So for example, this file should be placed in the above folder, based on the folder name (the numbers):
001-020-012B-B.nc
File path divided into column to show where the above file has to be moved to:
(001)-Disc-PED\(020)-Rotor-parts-1200.
Therefore:
(001)-Disc-PED\(020)-Rotor-parts-1200 (001)-(020)-012B-B.nc
This is what I have tried from looking online but it does not work:
My thinking is I want to loop through the folders and look for matches.
import os
import glob
import itertools
import re
#Source file
sourcefile = r'C:\Users\cah\Desktop\000Turning'
destinationPath = r'C:\Users\cah\Desktop\08-CAM'
#Seperation
dirs = glob.glob('*-*')
#Every file with file extension .nc
files = glob.glob('*.nc')
for root, dirs, files in os.walk(sourcefile):
for file in files:
if file.endswith(".nc"):
first3Char = str(file[0:3])
last3Char = str(file[4:7])
for root in os.walk(destinationPath):
first33CharsOfRoot = str(root[0:33])
cleanRoot1 = str(root).replace("[", "")
cleanRoot2 = str(cleanRoot1).replace("]", "")
cleanRoot3 = str(cleanRoot2).replace(")", "")
cleanRoot4 = str(cleanRoot3).replace("'", "")
cleanRoot5 = str(cleanRoot4).replace(",", "")
firstCharOfRoot = re.findall(r'(.{3})\s*$', str(cleanRoot5))
print(firstCharOfRoot==first3Char)
if(firstCharOfRoot == first3Char):
print("Hello")
for root in os.walk(destinationPath):
print(os.path.basename(root))
# if(os.path)
I realized that I should not look for the last 3 chars in the path, because it is the first (001) etc. Numbers that I need to look for in the beginning to find the first path that I need to go to.
EDIT:
import os
import glob
import itertools
import re
#Source file
sourcefile = r'C:\Users\cah\Desktop\000Turning'
destinationPath = r'C:\Users\cah\Desktop\08-CAM'
#Seperation
dirs = glob.glob('*-*')
#Every file with file extension .nc
files = glob.glob('*.nc')
for root, dirs, files in os.walk(sourcefile):
for file in files:
if file.endswith(".nc"):
first3Char = str(file[0:3])
last3Char = str(file[4:7])
for root in os.walk(destinationPath):
cleanRoot1 = str(root).replace("[", "")
cleanRoot2 = str(cleanRoot1).replace("]", "")
cleanRoot3 = str(cleanRoot2).replace(")", "")
cleanRoot4 = str(cleanRoot3).replace("'", "")
cleanRoot5 = str(cleanRoot4).replace(",", "")
firstCharOfRoot = re.findall(r'^(?:[^\\]+\\\\){5}(\d+).*$', str(cleanRoot5))
secondCharOfRoot = re.findall(r'^(?:[^\\]+\\\\){6}(\d+).*$', str(cleanRoot5))
firstCharOfRootCleaned = ''.join(firstCharOfRoot)
secondCharOfRoot = ''.join(secondCharOfRoot)
cleanRoot6 = str(cleanRoot5).replace("(", "")
if(firstCharOfRootCleaned == str(first3Char) & secondCharOfRoot == str(last3Char)):
print("BINGOf")
# for root1 in os.walk(cleanRoot6):

Solution
There is an improved solution in the next section. But let's decompose the straightforward solution before.
First, get the complete list of subfolders.
all_folders_splitted = [os.path.split(f)\
for f in glob.iglob(os.path.join(destinationPath, "**"), recursive=True)\
if os.path.isdir(f)]
Then, use a function on each of your file to find its matching folder, or a new filepath if it doesn't exist. I include this function called find_folder() in the rest of the script:
import os
import glob
import shutil
sourcefile= r'C:\Users\cah\Desktop\000Turning'
destinationPath = r'C:\Users\cah\Desktop\08-CAM'
all_folders_splitted = [os.path.split(f)\
for f in glob.iglob(os.path.join(destinationPath , "**"), recursive=True)\
if os.path.isdir(f)]
# It will create and return a new directory if no directory matches
def find_folder(part1, part2):
matching_folders1 = [folder for folder in all_folders_splitted\
if os.path.split(folder[0])[-1].startswith(part1)]
matching_folder2 = None
for matching_folder2 in matching_folders1:
if matching_folder2[-1].startswith(part2):
return os.path.join(*matching_folder2)
# Whole new folder tree
if matching_folder2 is None:
dest = os.path.join(destinationPath, part1, part2)
os.makedirs(dest)
return dest
# Inside the already existing folder part "1"
dest = os.path.join(matching_folder2[0], part2)
os.makedirs(dest)
return dest
# All the files you want to move
files_gen = glob.iglob(os.path.join(source_path, "**", "*-*-*.nc"), recursive=True)
for file in files_gen:
# Split the first two "-"
basename = os.path.basename(file)
splitted = basename.split("-", 2)
# Format the destination folder.
# Creates it if necessary
destination_folder = find_folder(splitted[0], splitted[1])
# Copying the file
shutil.copy2(file, os.path.join(destination_folder, basename))
Improved solution
In case you have a large number of files, it could be detrimental to "split and match" every folder at each iteration.
We can store the folder, found given a pattern, in a dictionary. The dictionary will be updated if a new pattern is given, else it will return the previously found folder.
import os
import glob
import shutil
sourcefile= r'C:\Users\cah\Desktop\000Turning'
destinationPath = r'C:\Users\cah\Desktop\08-CAM'
# Global dictionary to store folder paths, relative to a pattern
found_pattern = dict()
all_folders_splitted = [os.path.split(f)\
for f in glob.iglob(os.path.join(destinationPath , "**"), recursive=True)\
if os.path.isdir(f)]
def find_folder(part1, part2):
current_key = tuple([part1, part2])
if current_key in pattern_match:
# Already found previously.
# We just return the folder path, stored as the value.
return pattern_match[current_key]
matching_folders1 = [folder for folder in all_folders_splitted\
if os.path.split(folder[0])[-1].startswith(part1)]
matching_folder2 = None
for matching_folder2 in matching_folders1:
if matching_folder2[-1].startswith(part2):
dest = os.path.join(*matching_folder2)
# Update the dictionary
pattern_match[current_key] = dest
return dest
if matching_folder2 is None:
dest = os.path.join(destinationPath, part1, part2)
else:
dest = os.path.join(matching_folder2[0], part2)
# Update the dictionary
pattern_match[current_key] = dest
os.makedirs(dest, exist_ok = True)
return dest
# All the files you want to move
files_gen = glob.iglob(os.path.join(source_path, "**", "*-*-*.nc"), recursive=True)
for file in files_gen:
# Split the first two "-"
basename = os.path.basename(file)
splitted = basename.split("-", 2)
# Format the destination folder.
# Creates it if necessary
destination_folder = find_folder(splitted[0], splitted[1])
# Copying the file
shutil.copy2(file, os.path.join(destination_folder, basename))
This updated solution makes it more efficient (especially when many files should share the same folder) and you could also make use of the dictionary later, if you save it.

renaming the extracted file from zipfile

I have lots of zipped files on a Linux server and each file includes multiple text files.
what I want is to extract some of those text files, which have the same name across zipped files and save it a folder; I am creating one folder for each zipped file and extract the text file to it. I need to add the parent zipped folder name to the end of file names and save all text files in one directory. For example, if the zipped folder was March132017.zip and I extracted holding.txt, my filename would be holding_march13207.txt.
My problem is that I am not able to change the extracted file's name.
I would appreciate if you could advise.
import os
import sys
import zipfile
os.chdir("/feeds/lipper/emaxx")
pwkwd = "/feeds/lipper/emaxx"
for item in os.listdir(pwkwd): # loop through items in dir
if item.endswith(".zip"): # check for ".zip" extension
file_name = os.path.abspath(item) # get full path of files
fh = open(file_name, "rb")
zip_ref = zipfile.ZipFile(fh)
filelist = 'ISSUERS.TXT' , 'SECMAST.TXT' , 'FUND.TXT' , 'HOLDING.TXT'
for name in filelist :
try:
outpath = "/SCRATCH/emaxx" + "/" + os.path.splitext(item)[0]
zip_ref.extract(name, outpath)
except KeyError:
{}
fh.close()

import zipfile
zipdata = zipfile.ZipFile('somefile.zip')
zipinfos = zipdata.infolist()
# iterate through each file
for zipinfo in zipinfos:
# This will do the renaming
zipinfo.filename = do_something_to(zipinfo.filename)
zipdata.extract(zipinfo)
Reference:
https://bitdrop.st0w.com/2010/07/23/python-extracting-a-file-from-a-zip-file-with-a-different-name/

Why not just read the file in question and save it yourself instead of extracting? Something like:
import os
import zipfile
source_dir = "/feeds/lipper/emaxx" # folder with zip files
target_dir = "/SCRATCH/emaxx" # folder to save the extracted files
# Are you sure your files names are capitalized in your zip files?
filelist = ['ISSUERS.TXT', 'SECMAST.TXT', 'FUND.TXT', 'HOLDING.TXT']
for item in os.listdir(source_dir): # loop through items in dir
if item.endswith(".zip"): # check for ".zip" extension
file_path = os.path.join(source_dir, item) # get zip file path
with zipfile.ZipFile(file_path) as zf: # open the zip file
for target_file in filelist: # loop through the list of files to extract
if target_file in zf.namelist(): # check if the file exists in the archive
# generate the desired output name:
target_name = os.path.splitext(target_file)[0] + "_" + os.path.splitext(file_path)[0] + ".txt"
target_path = os.path.join(target_dir, target_name) # output path
with open(target_path, "w") as f: # open the output path for writing
f.write(zf.read(target_file)) # save the contents of the file in it
# next file from the list...
# next zip file...

You could simply run a rename after each file is extracted right? os.rename should do the trick.
zip_ref.extract(name, outpath)
parent_zip = os.path.basename(os.path.dirname(outpath)) + ".zip"
new_file_name = os.path.splitext(os.path.basename(name))[0] # just the filename
new_name_path = os.path.dirname(outpath) + os.sep + new_file_name + "_" + parent_zip
os.rename(outpath, new_namepath)
For the filename, if you want it to be incremental, simply start a count and for each file, go up by on.
count = 0
for file in files:
count += 1
# ... Do our file actions
new_file_name = original_file_name + "_" + str(count)
# ...
Or if you don't care about the end name you could always use something like a uuid.
import uuid
random_name = uuid.uuid4()

outpath = '/SCRATCH/emaxx'
suffix = os.path.splitext(item)[0]
for name in filelist :
index = zip_ref.namelist().find(name)
if index != -1: # check the file exists in the zipfile
filename, ext = os.path.splitext(name)
zip_ref.filelist[index].filename = f'{filename}_{suffix}.{ext}' # rename the extracting file to the suffix file name
zip_ref.extract(zip_ref.filelist[index], outpath) # use the renamed file descriptor to extract the file

I doubt this is possible to rename file during their extraction.
What about renaming files once they are extracted ?
Relying on linux bash, you can achieve it in a one line :
os.system("find "+outpath+" -name '*.txt' -exec echo mv {} `echo {} | sed s/.txt/"+zipName+".txt/` \;")
So, first we search all txt files in the specified folder, then exec the renaming command, with the new name computed by sed.
Code not tested, i'm on windows now ^^'

Python Want to move/Create a text file into a folder with the same name

I have a series of text files that are named with a series of numbers (20040719.txt) that I need edited and placed into folders with the same name as the text file (but without the .txt in the folder name). I am able to do my edits and create the folders with the correct names, but can't seem to get the edited files into their corresponding folders. There are no errors, so my question is how can i do this type of file move.
here is what I have so far
import glob
import os
import shutil
list_of_files = glob.glob("f:/Python scripts/Tests2/*.txt")
root_path = 'f:/Python scripts/Tests2/'
for file_name in list_of_files:
folders = [file_name.replace('.txt', 'D')]
for folder in folders:
os.mkdir(os.path.join(root_path,folder))
input = open(file_name, 'r')
output = open(file_name.replace('.txt', 't2.txt'), "w")
for line in input:
str = line.strip(" dd/mm/yyyy hh:mm:ss kA\t")
str = str.replace("date", "ddmmyyyy_hhmmss")
str = str.replace("lat. long. amp.", " lat long ka")
output.write(str)
input.close()
output.close()
list_of_folders = glob.glob("f:/Python scripts/Tests2/*D")
list_of_t2txt = glob.glob("f:/Python scripts/Tests2/*t2.txt")
for Folder_Name in list_of_folders:
for t2txt_Name in list_of_t2txt:
if t2txt_Name.replace('*t2.txt', '*D') == Folder_Name:
shutil.move(t2txt_Name, Folder_Name)
the end 'if' statement was a trial to see if I could do it that way