I am trying to identify all .kml in a specific directory and save them into a new directory. Is this possible? I'm able to print the file path but I would like to use Python to copy those files to a new directory.
Here is my code so far:
import os
# traverse whole directory
for root, dirs, files in os.walk(r'C:\Users\file_path_here'):
# select file name
for file in files:
# check the extension of files
if file.endswith('.kml'):
# print whole path of files
print(os.path.join(root, file))
Try this:
import os
# traverse whole directory
for root, dirs, files in os.walk(r'C:\Users\file_path_here'):
# select file name
for each_file in files:
# check the extension of files
if each_file.endswith('.kml'):
# print whole path of files
print(os.path.join(root, file))
kml_file = open(each_file, "r")
content = kml_file.read()
file.close()
with open('newfile.kml', 'w') as f:
f.write(content)
Related
I have following script which is supposed to list all picture etc from my disk drives, but it failes to find any file. Any Idea what I may be doing wrong. I am doing in PyCharm environment:
files = []
for root, dirs, files in os.walk("."):
for file in files:
if file.endswith((".jpg", ".jpeg", ".mp3", ".mp4")):
files.append(os.path.join(root, file))
# Sort the files by their current folder name
files.sort(key=lambda x: os.path.dirname(x))
# Display a list of the files
for file in files:
print(file)
Your for-loop will never end as you add the found file name into files which you are looping through. Use another variable to store the list.
This works fine:
import os
results = []
for root, dirs, files in os.walk("."):
print(f"Got: {len(results)} files, current dir: {root}")
for file in files:
if file.endswith((".jpg", ".jpeg", ".mp3", ".mp4")):
results.append(os.path.join(root, file))
# Sort the files by their current folder name
results.sort(key=lambda x: os.path.dirname(x))
# Display a list of the files
for file in results:
print(file)
I am trying to use the os.walk() module to go through a number of directories and move the contents of each directory into a single "folder" (dir).
In this particular example I have hundreds of .txt files that need to be moved. I tried using shutil.move() and os.rename(), but it did not work.
import os
import shutil
current_wkd = os.getcwd()
print(current_wkd)
# make sure that these directories exist
dir_src = current_wkd
dir_dst = '.../Merged/out'
for root, dir, files in os.walk(top=current_wkd):
for file in files:
if file.endswith(".txt"): #match files that match this extension
print(file)
#need to move files (1.txt, 2.txt, etc) to 'dir_dst'
#tried: shutil.move(file, dir_dst) = error
If there is a way to move all the contents of the directories, I would be interested in how to do that as well.
Your help is much appreciated! Thanks.
Here is the file directory and contents
current_wk == ".../Merged
In current_wkthere is:
Dir1
Dir2
Dir3..
combine.py # python script file to be executed
In each directory there are hundreds of .txtfiles.
Simple path math is required to find source files and destination files precisely.
import os
import shutil
src_dir = os.getcwd()
dst_dir = src_dir + " COMBINED"
for root, _, files in os.walk(current_cwd):
for f in files:
if f.endswith(".txt"):
full_src_path = os.path.join(src_dir, root, f)
full_dst_path = os.path.join(dst_dir, f)
os.rename(full_src_path, full_dst_path)
You have to prepare the complete path of source file, and make sure dir_dst exists.
for root, dir, files in os.walk(top=current_wkd):
for file in files:
if file.endswith(".txt"): #match files that match this extension
shutil.move(os.path.join(root, file), dir_dst)
Lets say my python script is in a folder "/main". I have a bunch of text files inside subfolders in main. I want to be able to open a file just by specifying its name, not the subdirectory its in.
So open_file('test1.csv') should open test1.csv even if its full path is /main/test/test1.csv.
I don't have duplicated file names so it should no be a problem.
I using windows.
you could use os.walk to find your filename in a subfolder structure
import os
def find_and_open(filename):
for root_f, folders, files in os.walk('.'):
if filename in files:
# here you can either open the file
# or just return the full path and process file
# somewhere else
with open(root_f + '/' + filename) as f:
f.read()
# do something
if you have a very deep folder structure you might want to limit the depth of the search
import os
def get_file_path(file):
for (root, dirs, files) in os.walk('.'):
if file in files:
return os.path.join(root, file)
This should work. It'll return the path, so you should handle opening the file, in your code.
import os
def open_file(filename):
f = open(os.path.join('/path/to/main/', filename))
return f
I'm trying to make a script that copies all the files other than the zipped files from a source folder to another destination folder and extract zipped files from the source folder to the destination, this is what i have come till this far:
import os
import zipfile
import shutil
myPath = "Source dir"
for root, dirs, files in os.walk(myPath):
for file in files:
if file.endswith('.zip'):
fh = open(file, 'rb')
z = zipfile.ZipFile(fh)
for name in z.namelist():
outpath = "destination dir"#Put the name of the destination folder
z.extract(name, outpath)
fh.close()
else:
fileList = os.listdir('source dir')
for f in fileList:
shutil.copy2(f, 'destination directory')
The code shows no error but there is no output too.
From Python Standard Library To get a full path (which begins with top) to a file or directory in dirpath, do os.path.join(dirpath, name) so you should write
fh = open(so.path.join(root,file))
to have the correct path.
Im trying to create a zipfile with the content of a folder ( some dirs and files ) using the code belllow:
import zip,os
path = "c:\\tester\\folderToZip"
zip = zipfile.ZipFile("zippedFolder.zip", "w")
for root, dirs, files in os.walk(path):
for file in files:
zip.write(os.path.join(root, file))
zip.close()
But after the code runs, when i check the zip file, the zip file, instead having the content of the folder "folderToZip" ( ex: f1,f2,f2, a.txt,b.txt,c.txt ) it have the full path of the variable path.
So, the question is, how can i create a zipfile based on a folder content, but not his fullpath ?
write takes a second optional parameter arcname, which when specified will provide the name for the file stored. Use that to specify the name you want.
If you only want the filename:
import zip,os
path = "c:\\tester\\folderToZip"
zip = zipfile.ZipFile("zippedFolder.zip", "w")
for root, dirs, files in os.walk(path):
for file in files:
filename = os.path.join(root, file)
zip.write(filename, filename)
zip.close()