Develop Reference

Searching for an excel file in two Directories and creating a path

I've recently posted a similar question a week about searching through sub directories to find a specific excel file. However this time, I need to find a specific file in either one of the two directories and give a path based on whether the file is located in one folder or the other.
Here is the code I have so far. the work computer i have is running on Python 2.7.18 - there are no errors however when i print out the df as an excel file nothing is shown in the output path
ExcelFilePath = sys.argv[1]
OutputFilePath = sys.argv[2]
# path of excel directory and use glob glob to get all the DigSym files
for (root, subdirs, files) in os.walk(ExcelFilePath):
for f in files:
if '/**/Score_Green_*' in f and '.xlsx' in f:
ScoreGreen_Files = os.path.join(root, f)
for f in ScoreGreen_Files:
df1 = pd.read_excel(f)
df1.to_excel(OutputFilePath)

OutputFilePath is an argument you're passing in. It isn't going to have a value unless you pass one in as a command line argument.
If you want to return the path, the variable you need to return is ScoreGreen_Files. You also don't need to iterate through
ScoreGreen_Files as it should just be the file you're looking for.
ExcelFilePath = sys.argv[1]
OutputFilePath = sys.argv[2]
# path of excel directory and use glob glob to get all the DigSym files
for (root, subdirs, files) in os.walk(ExcelFilePath):
for f in files:
if '/**/Score_Green_*' in f and '.xlsx' in f: # f is a single file
ScoreGreen_File = os.path.join(root, f)
df1 = pd.read_excel(ScoreGreen_File)
df1.to_excel(OutputFilePath)

Using os.walk to find and print names of my files, but unable to open them? Path name issue?

My relevant code block is as follows:
path = "\\Users\\Harmless\\Documents\\untitled"
cellorder = []
cellcont = []
for roots, dirs, files, in os.walk(path):
for file in natural_sort(files):
if "Row" in file:
cellorder.append(file)
with open(file,'r') as txt:
print txt.readlines
#print "file = %s" % file
This would successfully list all the files I want to open (as commented out), but when I try to pass in the filename the same way it was printed in order to read it:
IOError: [Errno 2] No such file or directory: 'Row0Col0Heat.txt'
How can I fix this? Do I need to reference the entire path name and string substitute in each filename? If so, why? Is there a better way to reference/utilize paths?

Try using the absolute path of the file, you can get the absolute path by
abs_file_path = os.path.abspath(file)
since you already have the base path, you can also use:
abs_file_path = os.path.join(path, file)
Hope it helps

Renaming multiple files in a directory using Python

I'm trying to rename multiple files in a directory using this Python script:
import os
path = '/Users/myName/Desktop/directory'
files = os.listdir(path)
i = 1
for file in files:
os.rename(file, str(i)+'.jpg')
i = i+1
When I run this script, I get the following error:
Traceback (most recent call last):
File "rename.py", line 7, in <module>
os.rename(file, str(i)+'.jpg')
OSError: [Errno 2] No such file or directory
Why is that? How can I solve this issue?
Thanks.

You are not giving the whole path while renaming, do it like this:
import os
path = '/Users/myName/Desktop/directory'
files = os.listdir(path)
for index, file in enumerate(files):
os.rename(os.path.join(path, file), os.path.join(path, ''.join([str(index), '.jpg'])))
Edit: Thanks to tavo, The first solution would move the file to the current directory, fixed that.

You have to make this path as a current working directory first.
simple enough.
rest of the code has no errors.
to make it current working directory:
os.chdir(path)

import os
from os import path
import shutil
Source_Path = 'E:\Binayak\deep_learning\Datasets\Class_2'
Destination = 'E:\Binayak\deep_learning\Datasets\Class_2_Dest'
#dst_folder = os.mkdir(Destination)
def main():
for count, filename in enumerate(os.listdir(Source_Path)):
dst = "Class_2_" + str(count) + ".jpg"
# rename all the files
os.rename(os.path.join(Source_Path, filename), os.path.join(Destination, dst))
# Driver Code
if __name__ == '__main__':
main()

As per #daniel's comment, os.listdir() returns just the filenames and not the full path of the file. Use os.path.join(path, file) to get the full path and rename that.
import os
path = 'C:\\Users\\Admin\\Desktop\\Jayesh'
files = os.listdir(path)
for file in files:
os.rename(os.path.join(path, file), os.path.join(path, 'xyz_' + file + '.csv'))

Just playing with the accepted answer define the path variable and list:
path = "/Your/path/to/folder/"
files = os.listdir(path)
and then loop over that list:
for index, file in enumerate(files):
#print (file)
os.rename(path+file, path +'file_' + str(index)+ '.jpg')
or loop over same way with one line as python list comprehension :
[os.rename(path+file, path +'jog_' + str(index)+ '.jpg') for index, file in enumerate(files)]
I think the first is more readable, in the second the first part of the loop is just the second part of the list comprehension

If your files are renaming in random manner then you have to sort the files in the directory first. The given code first sort then rename the files.
import os
import re
path = 'target_folder_directory'
files = os.listdir(path)
files.sort(key=lambda var:[int(x) if x.isdigit() else x for x in re.findall(r'[^0-9]|[0-9]+', var)])
for i, file in enumerate(files):
os.rename(path + file, path + "{}".format(i)+".jpg")

I wrote a quick and flexible script for renaming files, if you want a working solution without reinventing the wheel.
It renames files in the current directory by passing replacement functions.
Each function specifies a change you want done to all the matching file names. The code will determine the changes that will be done, and displays the differences it would generate using colors, and asks for confirmation to perform the changes.
You can find the source code here, and place it in the folder of which you want to rename files https://gist.github.com/aljgom/81e8e4ca9584b481523271b8725448b8
It works in pycharm, I haven't tested it in other consoles
The interaction will look something like this, after defining a few replacement functions
when it's running the first one, it would show all the differences from the files matching in the directory, and you can confirm to make the replacements or no, like this

This works for me and by increasing the index by 1 we can number the dataset.
import os
path = '/Users/myName/Desktop/directory'
files = os.listdir(path)
index=1
for index, file in enumerate(files):
os.rename(os.path.join(path, file),os.path.join(path,''.join([str(index),'.jpg'])))
index = index+1
But if your current image name start with a number this will not work.

Python 3x - Remove numbers from file names

import os
def rename_files():
# (1) get file names from a folder
file_list = os.listdir(r"C:\Users\USEER\Desktop\Udacity\Udacity - Programming Foundation with Python\Project\prank\prank")
# print(file_list)
saved_path = os.getcwd()
print("Current Working Directory is " + saved_path)
os.chdir(r"C:\Users\USEER\Desktop\Udacity\Udacity - Programming Foundation with Python\Project\prank\prank")
# (2) for each file, rename file name
for file_name in file_list:
print("Old Name - " + file_name)
print("New Name - " + file_name.translate("0123456789"))
os.rename(file_name, file_name.translate("0123456789"))
os.chdir(saved_path)
rename_files()
The code above doesn't rename the file by removing the integers. Can anyone help? (Python 3x)

import re
new_name = re.sub('[0-9]', '', file_name)

In Python 3 the String.translate is gone. Therefore you need to use the str.translate. It needs 'str.maketrans' which normally creates a translation table with the first two arguments supplied(not needed in this example), the third argument supplies the characters to be stripped.
This line should have the desired effect ...
os.rename(file_name, file_name.translate(str.maketrans('','','0123456789'))
Previous suggestions used .strip() however in this case as the numbers are mixed in with the filenames (not before or after) I believe it would not work, another used Regular Expressions which is perfectly valid, however within the context of this particular Udacity course translate was the suggested solution.
Here are the docs for maketrans :
[https://docs.python.org/3/library/stdtypes.html#str.maketrans][1]

The problem is in your translate function that doesn't do anything. There are better options available, but if your want to use translate then the proper syntax is:
#!/usr/bin/env python2
import string
new_name = string.translate(file_name, None, "0123456789")

Here is one way of renaming files.
Use os.renames to rename the files
Use file_name.strip("0123456789") to remove numbers
Code is given below:
import os
def file_rename():
name_list=os.listdir(r"C:\python\prank")
print(name_list)
saved_path=os.getcwd()
print("Current working directory is"+saved_path)
os.chdir(r"C:\python\prank")
for file_name in name_list:
print("old name"+file_name)
print("new name"+file_name.strip("0123456789"))
os.renames(file_name,file_name.strip("0123456789"))
os.chdir(saved_path)
file_rename()
To read more about os.renames check here.
To read more about the strip function, check here.

Another way to do this without import re. Instead of utilizing the .translate, use the .strip.
os.rename(file_name, file_name.strip('0123456789'))
Another observation is that your code wont read the new file name after changing it. At the top of your code you are reading file names and saving those name in file_list
# (1) get file names from a folder
file_list = os.listdir(r"C:\Users\USEER\Desktop\Udacity\Udacity - Programming Foundation with Python\Project\prank\prank")
In the for loop, where you are changing the name of each file, YOU ARE NOT reading the new file's name. You need to do something like this.
# (2) for each file, rename file name
for file_name in file_list:
print("Old Name - " + file_name)
os.rename(file_name, file_name.strip("0123456789"))
# (3) read file's name again... 'file_list' has old names
new_file_list = os.listdir(r"C:\Users\USEER\Desktop\Udacity\Udacity - Programming Foundation with Python\Project\prank\prank")
for file_name in new_file_list:
print("New file's name: " + new_file_name)
os.chdir(saved_path)

import os
def rename_files():
#1 Get file names from the folder
file = os.listdir(r"C:\Web\Python\prank")
print(file)
saved_path = os.getcwd()
print("Current Working Directory is"+saved_path)
os.chdir(r"C:\Web\Python\prank")
#2 For each file name rename file names
for file_name in file:
print("Old Name - " + file_name)
os.rename(file_name,file_name.strip("0123456789"))
print("New Name - " + file_name)
os.chdir(saved_path)
rename_files()

enter code here
import os
dir="/home/lucidvis/myPythonHome/prank/"
def rename_files():
# get file names from a folder
filenames = os.listdir(dir)
# print(filenames)
for file in filenames:
#print(file)
try:
#with os.open(filename for filename in filenames,"r+"):
#read_file = filename.read()
# new_name = file.translate(str.maketrans('','', "0123456789"))
new_filname = (file.translate(str.maketrans('','', "0123456789"))).replace(" ","")
#print(dir+new_file_name)
os.rename(dir+file,dir+new_file_name)
except SyntaxError as e:
print(e)
continue
# for each file, rename filename
rename_files()
import os
dir="/home/lucidvis/myPythonHome/prank/"
def rename_files():
# get file names from a folder
filenames = os.listdir(dir)
# iterate through the list of filenames
for file in filenames:
#print(file)
try:
#assign a variable to new names for easy manipulation
new_filname = (file.translate(str.maketrans('','', "0123456789"))).replace(" ","")
#concatenating the directory name to old and new file names
os.rename(dir+file,dir+new_file_name)
#just to manage errors
except SyntaxError as e:
print(e)
continue
#file renaming function call
rename_files()

will guys i was trying to solve this problem because i see udacity online courses and it require to rename file without numbers thanks to simon for his replay i have to figured it out
this is my code to rename files without numbers, hope it jhelp anyone who stuck
import os
import re
def rename():
#get the list of the photo name
plist = os.listdir(r"D:\pay\prank")
print(plist)
#removing the numbers from the photo names
os.chdir(r"D:\pay\prank")
for pname in plist :
os.rename(pname, re.sub('[0-9]', '' , pname))
print(pname)
rename()

import os
import re
from string import digits
#Get file names
file_list = os.listdir(r"C:\Users\703305981\Downloads\prank\prank")
print(file_list)
#chenage directory.
os.chdir(r"C:\Users\703305981\Downloads\prank\prank")
print (os.getcwd())
#Change the Name.
for file_name in file_list:
os.rename(file_name, re.sub(r'[0-9]+', '', file_name))

Opening multiple CSV files

I am trying to open multiple excel files. My program throws error message "FileNotFoundError". The file is present in the directory.
Here is the code:
import os
import pandas as pd
path = "C:\\GPA Calculations for CSM\\twentyfourteen"
files = os.listdir(path)
print (files)
df = pd.DataFrame()
for f in files:
df = pd.read_excel(f,'Internal', skiprows = 7)
print ("file name is " + f)
print (df.loc[0][1])
print (df.loc[1][1])
print (df.loc[2][1])
Program gives error on df = pd.read_excel(f,'Internal', skiprows = 7).
I opened the same file on another program (which opens single file) and that worked fine. Any suggestions or advice would be highly appreciated.

os.listdir lists the filenames relative to the directory (path) you're giving as argument. Thus, you need to join the path and filename together to get the absolute path for each file. Thus, in your loop:
for filename in files:
abspath = os.path.join(path, filename)
<etc, replace f by abspath>