Develop Reference

Python - move all files from one folder to another if their file names contain specified words

I have a folder with many files named like homeXXX_roomXXX_high.csv or homeXXX_roomXXX_low.csv, where the XXX part is replaced with a three-digit number.
I want to use some code to move the files into separate folders based on the number next to "home" in the filename. For example, I want to specify that files with names starting home101, home103, home320, home553, etc. should all be moved into folder A whereas those starting with home555, home431, home105 should go to FolderB.
I have this code so far:
import shutil
import os
source = '/path/to/source_folder'
dest1 = '/path/to/FolderA'
dest2 = '/path/to/FolderB'
files = os.listdir(source)
for f in files:
if (f.startswith("home101") or f.startswith("home103")):
shutil.move(f, dest1)
elif (f.startswith("home431") or f.startswith("home555")):
shutil.move(f, dest2)
However, it's tedious to specify all the if and else cases. I'd like to use some kind of structured data, such as a list, to specify groups of "home" numbers and the corresponding folder paths. How can I do this in Python?

it seems like you can use another for, it would look something like this:
import shutil
import os
source = '/path/to/source_folder'
dest1 = '/path/to/FolderA'
dest2 = '/path/to/FolderB'
list1 = ["home101", "home103"]
list2 = ["home431", "home555"]
files = os.listdir(source)
for f in files:
for home1 in list1:
if f.startswith(home1):
shutil.move(f, dest1)
break
for home2 in list2:
if f.startswith(home2):
shutil.move(f, dest2)
break
You can also create a function:
def check_and_move(file, list_of_patterns, destination):
for pattern in list_of_patterns:
if file.startswith(pattern):
shutil.move(file, destination)
and the code will get cleaner because you avoid repetition :)
for f in files:
check_and_move(f, list1, dest1)
check_and_move(f, list2, dest2)
# etc...

You can make an array for folderA that contains the "home+number"
FolderAGroup = ['home101', 'home103', 'homeXXX', 'homeXXX']
And if they get split like you say with a "_" use this code to filter them
Won't work if they are not split like that.
files = os.listdir(source)
for f in files:
parts = f.split('_')
# Get the first part of the filename before the _
home_number = parts[0]
# Check if the home number is in the FolderA group array
if home_number in FolderAGroup:
shutil.move(f, dest1)
else:
shutil.move(f, dest2)
You can expand with more elif statements if you would want more folders.

If the names homexxx are incremental, you could try something like this:
home_names_list_1 = []
home_names_list_2 = []
for i in range(100):
home_names_list_1.append("home" + str(i))
for i in range(100,200):
home_names_list_2.append("home" + str(i))
for file in files:
moved = False
for name in home_names_list_1:
if file.startswith(name):
print("move somewhere")
moved = True
break
if moved:
break
for name in home_names_list_2:
if file.startswith(name):
print("move somewhere else")
break
print(" did not move because did not match anything")

Renaming a single directory of files with a specific syntax

I'm trying to rename and add pad the names of a few hundred files to the same length.
So far I've managed to correctly rename and pad the file names within my IDE but I'm not sure how I link that to actually rename the files themselves.
Atomic Samurai__________.png
BabyYodatheBased________.png
Baradum_________________.png
bcav____________________.png
This is the code that does the rename and the padding within my IDE:
import glob, os
pad_images = glob.glob(r"C:\Users\test\*.png")
split_images = []
for i in pad_images:
split = i.split("\\")[-1]
split_images.append(split)
longest_file_name = max(split_images, key=len)
longest_int = len(longest_file_name)
new_images = []
for i in split_images:
parts = i.split('.')
new_name = (parts[0]).ljust(longest_int, '_') + "." + parts[1])
I've been trying to get os.rename(old_name, new_name) to work but I'm not sure where I actually get the old name from as I've split things up into different for loops.

Try saving the old file names to a list and do all the modifications (split and rename) in a single loop thereafter:
path = "C:/Users/test"
images = [f for f in os.listdir(path) if f.endswith(".png")]
length = len(max(images, key=len))
for file in images:
parts = file.split("\\")[-1].split(".")
new_name = f'{parts[0].ljust(length,"_")}.{parts[1]}'
os.rename(os.path.join(path,file), os.path.join(path,new_name))

Python: moving file to a newly created directory

I've got my script creating a bunch of files (size varies depending on inputs) and I want to be certain files in certain folders based on the filenames.
So far I've got the following but although directories are being created no files are being moved, I'm not sure if the logic in the final for loop makes any sense.
In the below code I'm trying to move all .png files ending in _01 into the sub_frame_0 folder.
Additionally is their someway to increment both the file endings _01 to _02 etc., and the destn folder ie. from sub_frame_0 to sub_frame_1 to sub_frame_2 and so on.
for index, i in enumerate(range(num_sub_frames+10)):
path = os.makedirs('./sub_frame_{}'.format(index))
# Slice layers into sub-frames and add to appropriate directory
list_of_files = glob.glob('*.tif')
for fname in list_of_files:
image_slicer.slice(fname, num_sub_frames) # Slices the .tif frames into .png sub-frames
list_of_sub_frames = glob.glob('*.png')
for i in list_of_sub_frames:
if i == '*_01.png':
shutil.move(os.path.join(os.getcwd(), '*_01.png'), './sub_frame_0/')

As you said, the logic of the final loop does not make sense.
if i == '*_01.ng'
It would evaluate something like 'image_01.png' == '*_01.png' and be always false.
Regexp should be the way to go, but for this simple case you just can slice the number from the file name.
for i in list_of_sub_frames:
frame = int(i[-6:-4]) - 1
shutil.move(os.path.join(os.getcwd(), i), './sub_frame_{}/'.format(frame))
If i = 'image_01.png' then i[-6:-4] would take '01', convert it to integer and then just subtract 1 to follow your schema.

A simple fix would be to check if '*_01.png' is in the file name i and change the shutil.move to include i, the filename. (It's also worth mentioning that iis not a good name for a filepath
list_of_sub_frames = glob.glob('*.png')
for i in list_of_sub_frames:
if '*_01.png' in i:
shutil.move(os.path.join(os.getcwd(), i), './sub_frame_0/')
Additionally is [there some way] to increment both the file endings _01 to _02 etc., and the destn folder ie. from sub_frame_0 to sub_frame_1 to sub_frame_2 and so on.
You could create file names doing something as simple as this:
for i in range(10):
#simple string parsing
file_name = 'sub_frame_'+str(i)
folder_name = 'folder_sub_frame_'+str(i)

Here is a complete example using regular expressions. This also implements the incrementing of file names/destination folders
import os
import glob
import shutil
import re
num_sub_frames = 3
# No need to enumerate range list without start or step
for index in range(num_sub_frames+10):
path = os.makedirs('./sub_frame_{0:02}'.format(index))
# Slice layers into sub-frames and add to appropriate directory
list_of_files = glob.glob('*.tif')
for fname in list_of_files:
image_slicer.slice(fname, num_sub_frames) # Slices the .tif frames into .png sub-frames
list_of_sub_frames = glob.glob('*.png')
for name in list_of_sub_frames:
m = re.search('(?P<fname>.+?)_(?P<num>\d+).png', name)
if m:
num = int(m.group('num'))+1
newname = '{0}_{1:02}.png'.format(m.group('fname'), num)
newpath = os.path.join('./sub_frame_{0:02}/'.format(num), newname)
print m.group() + ' -> ' + newpath
shutil.move(os.path.join(os.getcwd(), m.group()), newpath)

Rename many files sequentially Python

My training on Python is ongoing and I'm currently trying to rename sequentially many files that have this kind of root and extension:
Ite_1_0001.eps
Ite_2_0001.eps
Ite_3_0001.eps
Ite_4_0001.eps
However, I'm trying to rename all these files as follows:
Ite_0001.eps
Ite_0002.eps
Ite_0003.eps
Ite_0004.eps
So I'm proceeding in this way:
for path, subdirs, files in os.walk(newpath):
num = len(os.listdir(newpath))
for filename in files:
basename, extension = os.path.splitext(filename)
for x in range(1, num+1):
new_filename = '_%04d' % x + extension
os.rename(os.path.join(newpath, filename), os.path.join(newpath, new_filename))
It's not working at all because all the files are erased from the directory and when running the script once at a time I have this:
First run: _00004
Second run: _00005
.... and so on.
Could any one have some tips that could help me to achieve this task :).
Thank you very much for your help.

You could test the approach with a list of strings. So you do not run the risk of deleting the files. ;-)
files = ["Ite_1_0001.eps", "Ite_2_0001.eps", "Ite_3_0001.eps", "Ite_4_0001.eps",]
for f in files:
# Get the value between underscores. This is the index.
index = int(f[4:f.index('_', 4)])
new_name = '_%04d' % index
# Join the prefix, index and sufix file
print ''.join([f[:3], new_name, f[-4:]])
Ite_0001.eps
Ite_0002.eps
Ite_0003.eps
Ite_0004.eps

You can dynamically change the thing you're substituting in within your loop, like so
import os, re
n = 1
for i in os.listdir('.'):
os.rename(i, re.sub(r'\(\d{4}\)', '(Ite_) ({n})'.format(n=n), i))
n += 1

I write a function that if you give in input your basename it returns the correct name.
def newname(old_name):
num = old_name[4]
return (old_name[0:3] + old_name[5:-1] + num)

Python - Error when opening two files [duplicate]

I'm creating a program that will create a file and save it to the directory with the filename sample.xml. Once the file is saved when i try to run the program again it overwrites the old file into the new one because they do have the same file name. How do I increment the file names so that whenever I try to run the code again it will going to increment the file name. and will not overwrite the existing one. I am thinking of checking the filename first on the directory and if they are the same the code will generate a new filename:
fh = open("sample.xml", "w")
rs = [blockresult]
fh.writelines(rs)
fh.close()

I would iterate through sample[int].xml for example and grab the next available name that is not used by a file or directory.
import os
i = 0
while os.path.exists("sample%s.xml" % i):
i += 1
fh = open("sample%s.xml" % i, "w")
....
That should give you sample0.xml initially, then sample1.xml, etc.
Note that the relative file notation by default relates to the file directory/folder you run the code from. Use absolute paths if necessary. Use os.getcwd() to read your current dir and os.chdir(path_to_dir) to set a new current dir.

Sequentially checking each file name to find the next available one works fine with small numbers of files, but quickly becomes slower as the number of files increases.
Here is a version that finds the next available file name in log(n) time:
import os
def next_path(path_pattern):
"""
Finds the next free path in an sequentially named list of files
e.g. path_pattern = 'file-%s.txt':
file-1.txt
file-2.txt
file-3.txt
Runs in log(n) time where n is the number of existing files in sequence
"""
i = 1
# First do an exponential search
while os.path.exists(path_pattern % i):
i = i * 2
# Result lies somewhere in the interval (i/2..i]
# We call this interval (a..b] and narrow it down until a + 1 = b
a, b = (i // 2, i)
while a + 1 < b:
c = (a + b) // 2 # interval midpoint
a, b = (c, b) if os.path.exists(path_pattern % c) else (a, c)
return path_pattern % b
To measure the speed improvement I wrote a small test function that creates 10,000 files:
for i in range(1,10000):
with open(next_path('file-%s.foo'), 'w'):
pass
And implemented the naive approach:
def next_path_naive(path_pattern):
"""
Naive (slow) version of next_path
"""
i = 1
while os.path.exists(path_pattern % i):
i += 1
return path_pattern % i
And here are the results:
Fast version:
real 0m2.132s
user 0m0.773s
sys 0m1.312s
Naive version:
real 2m36.480s
user 1m12.671s
sys 1m22.425s
Finally, note that either approach is susceptible to race conditions if multiple actors are trying to create files in the sequence at the same time.

def get_nonexistant_path(fname_path):
"""
Get the path to a filename which does not exist by incrementing path.
Examples
--------
>>> get_nonexistant_path('/etc/issue')
'/etc/issue-1'
>>> get_nonexistant_path('whatever/1337bla.py')
'whatever/1337bla.py'
"""
if not os.path.exists(fname_path):
return fname_path
filename, file_extension = os.path.splitext(fname_path)
i = 1
new_fname = "{}-{}{}".format(filename, i, file_extension)
while os.path.exists(new_fname):
i += 1
new_fname = "{}-{}{}".format(filename, i, file_extension)
return new_fname
Before you open the file, call
fname = get_nonexistant_path("sample.xml")
This will either give you 'sample.xml' or - if this alreay exists - 'sample-i.xml' where i is the lowest positive integer such that the file does not already exist.
I recommend using os.path.abspath("sample.xml"). If you have ~ as home directory, you might need to expand it first.
Please note that race conditions might occur with this simple code if you have multiple instances running at the same time. If this might be a problem, please check this question.

Try setting a count variable, and then incrementing that variable nested inside the same loop you write your file in. Include the count loop inside the name of the file with an escape character, so every loop ticks +1 and so does the number in the file.
Some code from a project I just finished:
numberLoops = #some limit determined by the user
currentLoop = 1
while currentLoop < numberLoops:
currentLoop = currentLoop + 1
fileName = ("log%d_%d.txt" % (currentLoop, str(now())))
For reference:
from time import mktime, gmtime
def now():
return mktime(gmtime())
which is probably irrelevant in your case but i was running multiple instances of this program and making tons of files. Hope this helps!

The two ways to do it are:
Check for the existence of the old file and if it exists try the next file name +1
save state data somewhere
an easy way to do it off the bat would be:
import os.path as pth
filename = "myfile"
filenum = 1
while (pth.exists(pth.abspath(filename+str(filenum)+".py")):
filenum+=1
my_next_file = open(filename+str(filenum)+".py",'w')
as a design thing, while True slows things down and isn't a great thing for code readability
edited: #EOL contributions/ thoughts
so I think not having .format is more readable at first glance - but using .format is better for generality and convention so.
import os.path as pth
filename = "myfile"
filenum = 1
while (pth.exists(pth.abspath(filename+str(filenum)+".py")):
filenum+=1
my_next_file = open("{}{}.py".format(filename, filenum),'w')
# or
my_next_file = open(filename + "{}.py".format(filenum),'w')
and you don't have to use abspath - you can use relative paths if you prefer, I prefer abs path sometimes because it helps to normalize the paths passed :).
import os.path as pth
filename = "myfile"
filenum = 1
while (pth.exists(filename+str(filenum)+".py"):
filenum+=1
##removed for conciseness

Another solution that avoids the use of while loop is to use os.listdir() function which returns a list of all the files and directories contained in a directory whose path is taken as an argument.
To answer the example in the question, supposing that the directory you are working in only contains "sample_i.xlm" files indexed starting at 0, you can easily obtain the next index for the new file with the following code.
import os
new_index = len(os.listdir('path_to_file_containing_only_sample_i_files'))
new_file = open('path_to_file_containing_only_sample_i_files/sample_%s.xml' % new_index, 'w')

You can use a while loop with a counter which checks if a file with a name and the counter's value exists if it does then move on else break and make a file.
I have done it in this way for one of my projects:`
from os import path
import os
i = 0
flnm = "Directory\\Filename" + str(i) + ".txt"
while path.exists(flnm) :
flnm = "Directory\\Filename" + str(i) + ".txt"
i += 1
f = open(flnm, "w") #do what you want to with that file...
f.write(str(var))
f.close() # make sure to close it.
`
Here the counter i starts from 0 and a while loop checks everytime if the file exists, if it does it moves on else it breaks out and creates a file from then you can customize. Also make sure to close it else it will result in the file being open which can cause problems while deleting it.
I used path.exists() to check if a file exists.
Don't do from os import * it can cause problem when we use open() method as there is another os.open() method too and it can give the error. TypeError: Integer expected. (got str)
Else wish u a Happy New Year and to all.

Without storing state data in an extra file, a quicker solution to the ones presented here would be to do the following:
from glob import glob
import os
files = glob("somedir/sample*.xml")
files = files.sorted()
cur_num = int(os.path.basename(files[-1])[6:-4])
cur_num += 1
fh = open("somedir/sample%s.xml" % cur_num, 'w')
rs = [blockresult]
fh.writelines(rs)
fh.close()
This will also keep incrementing, even if some of the lower numbered files disappear.
The other solution here that I like (pointed out by Eiyrioü) is the idea of keeping a temporary file that contains your most recent number:
temp_fh = open('somedir/curr_num.txt', 'r')
curr_num = int(temp_fh.readline().strip())
curr_num += 1
fh = open("somedir/sample%s.xml" % cur_num, 'w')
rs = [blockresult]
fh.writelines(rs)
fh.close()

Another example using recursion
import os
def checkFilePath(testString, extension, currentCount):
if os.path.exists(testString + str(currentCount) +extension):
return checkFilePath(testString, extension, currentCount+1)
else:
return testString + str(currentCount) +extension
Use:
checkFilePath("myfile", ".txt" , 0)

I needed to do something similar, but for output directories in a data processing pipeline. I was inspired by Vorticity's answer, but added use of regex to grab the trailing number. This method continues to increment the last directory, even if intermediate numbered output directories are deleted. It also adds leading zeros so the names will sort alphabetically (i.e. width 3 gives 001 etc.)
def get_unique_dir(path, width=3):
# if it doesn't exist, create
if not os.path.isdir(path):
log.debug("Creating new directory - {}".format(path))
os.makedirs(path)
return path
# if it's empty, use
if not os.listdir(path):
log.debug("Using empty directory - {}".format(path))
return path
# otherwise, increment the highest number folder in the series
def get_trailing_number(search_text):
serch_obj = re.search(r"([0-9]+)$", search_text)
if not serch_obj:
return 0
else:
return int(serch_obj.group(1))
dirs = glob(path + "*")
num_list = sorted([get_trailing_number(d) for d in dirs])
highest_num = num_list[-1]
next_num = highest_num + 1
new_path = "{0}_{1:0>{2}}".format(path, next_num, width)
log.debug("Creating new incremented directory - {}".format(new_path))
os.makedirs(new_path)
return new_path
get_unique_dir("output")

Here is one more example. Code tests whether a file exists in the directory or not if exist it does increment in the last index of the file name and saves
The typical file name is: Three letters of month_date_lastindex.txt ie.e.g.May10_1.txt
import time
import datetime
import shutil
import os
import os.path
da=datetime.datetime.now()
data_id =1
ts = time.time()
st = datetime.datetime.fromtimestamp(ts).strftime("%b%d")
data_id=str(data_id)
filename = st+'_'+data_id+'.dat'
while (os.path.isfile(str(filename))):
data_id=int(data_id)
data_id=data_id+1
print(data_id)
filename = st+'_'+str(data_id)+'.dat'
print(filename)
shutil.copyfile('Autonamingscript1.py',filename)
f = open(filename,'a+')
f.write("\n\n\n")
f.write("Data comments: \n")
f.close()

Continues sequence numbering from the given filename with or without the appended sequence number.
The given filename will be used if it doesn't exist, otherwise a sequence number is applied, and gaps between numbers will be candidates.
This version is quick if the given filename is not already sequenced or is the sequentially highest numbered pre-existing file.
for example the provided filename can be
sample.xml
sample-1.xml
sample-23.xml
import os
import re
def get_incremented_filename(filename):
name, ext = os.path.splitext(filename)
seq = 0
# continue from existing sequence number if any
rex = re.search(r"^(.*)-(\d+)$", name)
if rex:
name = rex[1]
seq = int(rex[2])
while os.path.exists(filename):
seq += 1
filename = f"{name}-{seq}{ext}"
return filename

My 2 cents: an always increasing, macOS-style incremental naming procedure
get_increased_path("./some_new_dir").mkdir() creates ./some_new_dir ; then
get_increased_path("./some_new_dir").mkdir() creates ./some_new_dir (1) ; then
get_increased_path("./some_new_dir").mkdir() creates ./some_new_dir (2) ; etc.
If ./some_new_dir (2) exists but not ./some_new_dir (1), then get_increased_path("./some_new_dir").mkdir() creates ./some_new_dir (3) anyways, so that indexes always increase and you always know which is the latest
from pathlib import Path
import re
def get_increased_path(file_path):
fp = Path(file_path).resolve()
f = str(fp)
vals = []
for n in fp.parent.glob("{}*".format(fp.name)):
ms = list(re.finditer(r"^{} \(\d+\)$".format(f), str(n)))
if ms:
m = list(re.finditer(r"\(\d+\)$", str(n)))[0].group()
vals.append(int(m.replace("(", "").replace(")", "")))
if vals:
ext = " ({})".format(max(vals) + 1)
elif fp.exists():
ext = " (1)"
else:
ext = ""
return fp.parent / (fp.name + ext + fp.suffix)