Rename files and there store it to csv using Python - python

I want to rename files in subfolders which I can do with this following code.
import os
#This is to rename files
def main():
path = "D:\Proyekan\Data yang udah di extract"
count = 1
for root, dirs, files in os.walk(path):
for i in files:
fullpath = os.path.join(root,i)
os.rename(os.path.join(fullpath), os.path.join(root, "Flight_" + str(1000000+count) + ".mat"))
count += 1
if __name__ == '__main__':
main()
The code works for subfolders so:
Folder 1 the file names are Flight_1-Flight_25 and
Folder 2 the file names are Flight_26-Flight_27.
The problem is, I need to list the old name into new files, so I can track it, the folders too. Ex :
123123 to Flight 1 in Folder 1,
12313 to Flight 2 in Folder 2,
and so on. Any ideas how? I got lost in trying to figure how
I want the results to be like this:

Related

Python Merging files after a name fragment

I have two folders when I have dynamic quantity files.
For example:
In folder "A" I've files:
FileName_1.txt
FileName_2.txt
FileName_3.txt
In folder "B" I've files:
NewFile_1.txt
NewFile_2.txt
NewFile_3.txt
The number of files in both folders will always be the same. Is there any simple way how to merge files by number in filename? As result I want to:
Data from file NewFile_1.txt add to FileName_1.txt
Data from file NewFile_2.txt add to FileName_2.txt, etc.
It doesn't have to be a solution to the problem. Thank you for the tips.
Just something like this. Grab all the file names from the B directory, find the suffix (after the _), and open the A file based on that:
import os
newnames = [k for k in os.listdir( "B") if k[-4:] == '.txt']
for name in newnames:
i = name.find('_')
suffix = name[i:]
olddata = open('B/' + name).read()
open('A/FileName'+suffix, 'a').write(olddata)

Rename specific files in a folder

I am trying to do a program that renames the files that are not named in order for example :
input: in a folder we have these files : ['spam001.txt', 'spam002.txt', 'spam003.txt', 'spam005.txt', 'spam007.txt']
output: i want to end up with the same files named like this : ['spam001.txt', 'spam002.txt', 'spam003.txt', 'spam004.txt', 'spam005.txt']
Here is what i have done until now :
import os,shutil
count = 1
path = '.\\filling in the gaps' #this is the folder where i have the .txt files
files = os.listdir(path)
for i in files:
if(i[6] != str(count)):
os.rename(os.path.join(path,i),'spam00%s.txt' %(count))
count = count+1
My problem is that when I run the program, for the output mentioned above , I get ['spam001.txt', 'spam002.txt', 'spam003.txt'] in the 'filling in the gaps' folder and I get
['spam004.txt', 'spam005.txt'] in the folder that contains 'filling in the gaps' folder.
Basically, my program renames the files, but the files end up in another folder. Any idea why this is happening?
This is where the files go:
Here is a working version of your code, providing your desired results:
import os
count = 1
path = '.\\filling in the gaps' # this is the folder where i have the .txt files
files = os.listdir(path)
for i in files:
if (i[6] != str(count)):
os.rename(os.path.join(path, i),
os.path.join(path, 'spam00%s.txt' % (count))
)
count = count + 1

how can i rename files with different names at once in python

path = '/Users/my/path/tofile'
files = os.listdir(path)
names= ["GAT4", "LO", "sds"]
for filename in files:
if files.startswith("sample" + str[i]):
original_file= os.path.join(path, filename)
new_file= os.path.join(path, names.join([str(i), '.html']))
os.rename(original_file, new_file)
i have many files and i wanna rename all of them using a python code that changes the name depending on a given name from a list:
for example i have a list of x = [sample1, sample236, GAT988] and my files are named like: exp1.html exp2.html exp3.html
how can i make the files names become GAT988.html instead of exp3.html?enter code here
thank you.

Rename multiple files inside multiple folders

So I have a lot of folders with a certain name. In each folder I have +200 items. The items inside the folders has names like:
CT.34562346.246.dcm
RD.34562346.dcm
RN.34562346.LAO.dcm
And some along that style.
I now wish to rename all files inside all folders so that the number (34562346) is replaced with the name of the folder. So for example in the folder named "1" the files inside should become:
CT.1.246.dcm
RD.1.dcm
RN.1.LAO.dcm
So only the large number is replaced. And yes, all files are similar like this. It would be the number after the first . that should be renamed.
So far I have:
import os
base_dir = "foo/bar/" #In this dir I have all my folders
dir_list = []
for dirname in os.walk(base_dir):
dir_list.append(dirname[0])
This one just lists the entire paths of all folders.
dir_list_split = []
for name in dir_list[1:]: #The 1 is because it lists the base_dir as well
x = name.split('/')[2]
dir_list_split.append(x)
This one extracts the name of each folder.
And then the next thing would be to enter the folders and rename them. And I'm kind of stuck here ?
The pathlib module, which was new in Python 3.4, is often overlooked. I find that it often makes code simpler than it would otherwise be with os.walk.
In this case, .glob('**/*.*') looks recursively through all of the folders and subfolders that I created in a sample folder called example. The *.* part means that it considers all files.
I put path.parts in the loop to show you that pathlib arranges to parse pathnames for you.
I check that the string constant '34562346' is in its correct position in each filename first. If it is then I simply replace it with the items from .parts that is the next level of folder 'up' the folders tree.
Then I can replace the rightmost element of .parts with the newly altered filename to create the new pathname and then do the rename. In each case I display the new pathname, if it was appropriate to create one.
>>> from pathlib import Path
>>> from os import rename
>>> for path in Path('example').glob('**/*.*'):
... path.parts
... if path.parts[-1][3:11]=='34562346':
... new_name = path.parts[-1].replace('34562346', path.parts[-2])
... new_path = '/'.join(list(path.parts[:-1])+[new_name])
... new_path
... ## rename(str(path), new_path)
... else:
... 'no change'
...
('example', 'folder_1', 'id.34562346.6.a.txt')
'example/folder_1/id.folder_1.6.a.txt'
('example', 'folder_1', 'id.34562346.wax.txt')
'example/folder_1/id.folder_1.wax.txt'
('example', 'folder_2', 'subfolder_1', 'ty.34562346.90.py')
'example/folder_2/subfolder_1/ty.subfolder_1.90.py'
('example', 'folder_2', 'subfolder_1', 'tz.34562346.98.py')
'example/folder_2/subfolder_1/tz.subfolder_1.98.py'
('example', 'folder_2', 'subfolder_2', 'doc.34.34562346.implication.rtf')
'no change'
This will rename files in subdirectories too:
import os
rootdir = "foo" + os.sep + "bar"
for subdir, dirs, files in os.walk(rootdir):
for file in files:
filepath = subdir + os.sep + file
foldername = subdir.split(os.sep)[-1]
number = ""
foundnumber = False
for c in filepath:
if c.isdigit():
foundnumber = True
number = number + c
elif foundnumber:
break
if foundnumber:
newfilepath = filepath.replace(number,foldername)
os.rename(filepath, newfilepath)
Split each file name on the . and replace the second item with the file name, then join on .'s again for the new file name. Here's some sample code that demonstrates the concept.
folder_name = ['1', '2']
file_names = ['CT.2345.234.dcm', 'BG.234234.222.dcm', "RA.3342.221.dcm"]
for folder in folder_name:
new_names = []
for x in file_names:
file_name = x.split('.')
file_name[1] = folder
back_together = '.'.join(file_name)
new_names.append(back_together)
print(new_names)
Output
['CT.1.234.dcm', 'BG.1.222.dcm', 'RA.1.221.dcm']
['CT.2.234.dcm', 'BG.2.222.dcm', 'RA.2.221.dcm']

Count and print number of files in subfolders using Python

My folder structure is as follows
Folder A
Folder B1
Folder B2
....
Folder Bn
How can I count the number of files in each of the folders (Folder B1 - Folder Bn), check if the number of files is larger than a given limit and print the folder name and number of files in it on the screen?
Like this:
Folders with too many files:
Folder B3 101
Folder B7 256
Here's what I've tried so far. It goes through every subfolder in each of my Folder B1 etc. I just need file count in one level.
import os, sys ,csv
path = '/Folder A/'
outwriter = csv.writer(open("numFiles.csv", 'w')
dir_count = []
for root, dirs, files in os.walk(path):
for d in dirs:
a = str(d)
count = 0
for fi in files:
count += 1
y = (a, count)
dir_count.append(y)
for i in dir_count:
outwriter.writerow(i)
And then I just printed numFiles.csv. Not quite how I'd like to do it.
Thanks in advance!
As the are all contained in that single folder, you only need to search that directory:
import os
path = '/Folder A/'
mn = 20
folders = ([name for name in os.listdir(path)
if os.path.isdir(os.path.join(path, name)) and name.startswith("B")]) # get all directories
for folder in folders:
contents = os.listdir(os.path.join(path,folder)) # get list of contents
if len(contents) > mn: # if greater than the limit, print folder and number of contents
print(folder,len(contents)
os.walk(path) gives you three tuple for a directory, ie (directory,subdirectory,files).
directory -> list of all directory in current dir, list of subdirectory in current dir, list of files in current dir.
so you can code likes this:
import os
for dir,subdir,files in os.walk(path):
if len(files) > your_limit:
print dir + " has crossed limit, " + "total files: " + len(files)
for x in files:
print x
if you want to walk only one level, you need to code like this:
for x in os.listdir(path):
if os.path.isdir(x):
count = len([ y for y in os.listdir(x) if os.path.isfile(os.path.join(x,y)) ])
if count > your_limit:
print x + " has crossed limit: ", +count

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