I am using shutil to copy my files to another destination directory , what i want to achieve is
src/red.png
src/blue.png
src/green.png
for example if i select blue and green then while saving i want them to be saved as
dst/blue_1001.png
dst/green_1002.png
Thanks
Kartikey
All you need to do is to construct a destination path in a format that you want and shutil will do rest of the job.
I am using pathlib module here because it provides convenient APIs to work with paths.
>>> import shutil
>>> from pathlib import Path
>>>
>>> files_to_copy = ["src/red.png", "src/blue.png"]
>>>
>>> destination = Path("dst")
>>>
>>> for index, file in enumerate(map(Path, files_to_copy), start=1001):
... destination_filename = f"{file.stem}_{index}{file.suffix}"
... print(f"{destination_filename=}")
... _ = shutil.copy(file, destination / destination_filename)
...
destination_filename='red_1001.png'
destination_filename='blue_1002.png'
Shutils supports setting the name of the target file. It mainly depends on how you splice the name of the target file
import shutil
from pathlib import Path
parent_path = Path(__file__).parent
file_path = parent_path.joinpath("test2.py")
shutil.copy(file_path, parent_path.joinpath("test3.py"))
Related
i want this
>>> from os import listdir
>>> listdir(g:/new folder)
[g:/new folder/file 1, g:/new folder/file 2\]
but i'm getting this
>>> from os import listdir
>>> listdir(g:/new folder)
[file 1, file 2\]
Use os.scandir function to get full path for each entry:
import os
for f in os.scandir('your_path'):
print(f.path)
import os
directory = "g:/new folder"
files = os.listdir(directory)
files_with_path = [directory+"/"+filename for filename in files]
It should do the trick given the doc, you could also look at the glob package
https://docs.python.org/3/library/os.html#os.listdir
https://docs.python.org/3/library/glob.html
I have this code which gives me the filename no problem. It gives it to me without the extension. Any way to get with extension?
from pathlib import Path
file = 'somepath'
path_object = Path(file)
filename = path_object.stem
You can use .name attribute of the Path object.
>>> from pathlib import Path
>>>
>>> p = Path("SomePath/filename.txt")
>>> p.name
'filename.txt'
I have a list of files like this in the images folder.
and How can I create a new folder if there are multiple files with a similar name and move those similar files to that folder?
I am new to python.
Here is my expectation:
Try this:
import glob
from pathlib import Path
for fn in Path("Images").glob("*"):
file_base_name = "_".join(fn.stem.split("_")[:-1])
file_count = len(glob.glob1("Images", f"{file_base_name}*"))
if file_count > 1 or Path(file_base_name).is_dir():
outdir = Path("Images") / file_base_name
outdir.mkdir(exist_ok=True)
fn.rename(outdir / fn.name)
Input:
Output:
Please ignore file names extension. I create those just to test my code
In this case you don't even need re:
from pathlib import Path
for fn in Path("Images").glob("*.jpg"):
outdir = Path("Images") / "_".join(fn.stem.split("_")[:-1])
outdir.mkdir(exist_ok=True)
fn.rename(outdir / fn.name)
What's going on here?
Pathlib is how you want to think of paths if you can. It combines most of the os.path apis. Specifically:
glob gets us all the files matching the glob in the path
mkdir makes the directory (only if it doesn't exist)
rename moves the file there
I am unable to test since I don't have your files. My suggestion would be to comment out the mkdir command and the shutil.move command and replace them with print statements to see what commands would be generated before letting it run for real. But I think it should work.
import pathlib
import os
import re
from itertools import groupby
import shutil
source_dir = 'Images'
files = [os.path.basename(f) for f in pathlib.Path(source_dir).glob('*.jpg')]
def keyfunc(file):
m = re.match('^(.*?)_\d+.jpg$', file)
return m[1]
matched_files = [file for file in files if re.search(r'_\d+.jpg$', file)]
matched_files.sort()
for k, g in groupby(matched_files, keyfunc):
new_dir = os.path.join(source_dir, k)
if not os.path.exists(new_dir):
os.mkdir(new_dir)
for file in g:
shutil.move(os.path.join(source_dir, file), new_dir)
I was messing around just trying to make a script that deletes items by ".zip" extension.
import sys
import os
from os import listdir
test=os.listdir("/Users/ben/downloads/")
for item in test:
if item.endswith(".zip"):
os.remove(item)
Whenever I run the script I get:
OSError: [Errno 2] No such file or directory: 'cities1000.zip'
cities1000.zip is obviously a file in my downloads folder.
What did I do wrong here? Is the issue that os.remove requires the full path to the file? If this is the issue, than how can I do that in this current script without completely rewriting it.
You can set the path in to a dir_name variable, then use os.path.join for your os.remove.
import os
dir_name = "/Users/ben/downloads/"
test = os.listdir(dir_name)
for item in test:
if item.endswith(".zip"):
os.remove(os.path.join(dir_name, item))
For this operation you need to append the file name on to the file path so the command knows what folder you are looking into.
You can do this correctly and in a portable way in python using the os.path.join command.
For example:
import os
directory = "/Users/ben/downloads/"
test = os.listdir( directory )
for item in test:
if item.endswith(".zip"):
os.remove( os.path.join( directory, item ) )
Alternate approach that avoids join-ing yourself over and over: Use glob module to join once, then let it give you back the paths directly.
import glob
import os
dir = "/Users/ben/downloads/"
for zippath in glob.iglob(os.path.join(dir, '*.zip')):
os.remove(zippath)
I think you could use Pathlib-- a modern way, like the following:
import pathlib
dir = pathlib.Path("/Users/ben/downloads/")
zip_files = dir.glob(dir / "*.zip")
for zf in zip_files:
zf.unlink()
If you want to delete all zip files recursively, just write so:
import pathlib
dir = pathlib.Path("/Users/ben/downloads/")
zip_files = dir.rglob(dir / "*.zip") # recursively
for zf in zip_files:
zf.unlink()
Just leaving my two cents on this issue: if you want to be chic you can use glob or iglob from the glob package, like so:
import glob
import os
files_in_dir = glob.glob('/Users/ben/downloads/*.zip')
# or if you want to be fancy, you can use iglob, which returns an iterator:
files_in_dir = glob.iglob('/Users/ben/downloads/*.zip')
for _file in files_in_dir:
print(_file) # just to be sure, you know how it is...
os.remove(_file)
origfolder = "/Users/ben/downloads/"
test = os.listdir(origfolder)
for item in test:
if item.endswith(".zip"):
os.remove(os.path.join(origfolder, item))
The dirname is not included in the os.listdir output. You have to attach it to reference the file from the list returned by said function.
Prepend the directory to the filename
os.remove("/Users/ben/downloads/" + item)
EDIT: or change the current working directory using os.chdir.
I have 700 files in a single folder. I need to find files that have "h10v03" as part of the name and copy them to a different folder using python.
Heres an example of one of the files: MOD10A1.A2000121.h10v03.005.2007172062725.hdf
I appreciate any help.
Something like this would do the trick.
import os
import shutil
source_dir = "/some/directory/path"
target_dir = "/some/other/directory/path"
part = "h10v03"
files = [file for file in os.listdir(source_dir)
if os.path.isfile(file) and part in file]
for file in files:
shutil.copy2(os.path.join(source_dir, file), target_dir)
Does it need to be python?
A unix shell does that for you quite fine:
cp ./*h10v03* /other/directory/
In python I would suggest you take a look at os.listdir() and shutil.copy()
EDIT:
some untested code:
import os
import shutil
src_dir = "/some/path/"
target_dir = "/some/other/path/"
searchstring = "h10v03"
for f in os.listdir(src_dir):
if searchstring in f and os.path.isfile(os.path.join(src_dir, f)):
shutil.copy2(os.path.join(src_dir, f), target_dir)
print "COPY", f
with the glob module (untested):
import glob
import os
import shutil
for f in glob.glob("/some/path/*2000*h10v03*"):
print f
shutil.copy2(f, os.path.join("/some/target/dir/", os.path.basename(f)))
Firstly, find all the items in that folder with os.listdir. Then you can use the count() method of string to determine if it has your string. Then you can use shutil to copy the file.