I would like to change the file name in the folder, there are jpg file but corrupted with product id number. I tried to rename it and delete the string after ".jpg" by using python, here is the code, but there is no any change.
Do you guys have any suggestion?
import os
path=input('C:\\Users\\pengoul\\Downloads\\Files\\PIC\\')
def rename(path):
for file in os.listdir(path):
if file.endswith(".jpg#v=*"):
print(file)
newfile=file.split("#",1)[0]
newName=file.replace(file,newfile)
os.rename(os.path.join(path,file),os.path.join(path,newName))
rename(path)
print("End")
Rename the file of one folder and delete the string after .jpg by using python.
if you input path from user, python automatically use "\" instead of "\\" so you shouldn't use "\\" while input. So I suggest one of these two ways:
path = input("Input your Path: ")
# And the user entered this: C:\Users\pengoul\Downloads\Files\PIC
or
path = 'C:\\Users\\pengoul\\Downloads\\Files\\PIC'
or
path = r'C:\Users\pengoul\Downloads\Files\PIC'
I hope that works for you. If not, make us aware of that!
Related
I'll try to describe the problem in a simple way.
I have a .txt file that I can not know the full name of it which located under constant path
[for example: the full name is: Hello_stack.txt, I only can give to function the part: 'Hello_']
the input is: Path_to_file/ + 'Hello_'
the expected output is: Path_to_file/Hello_stack.txt
How can i do that?
I tried to give a path and check recursively if part of my file name is exist and if so, to return it's path.
this is my implementation: [of course I'd like to get another way if it works]
def get_CS_R2M_METRO_CALLBACK_FILE_PATH():
directory = 'path_of_file'
file_name = directory + 'part_of_file_name'
const_path = Path(file_name)
for path in [p for p in const_path.rglob("*")]:
if path.is_file():
return path
Thanks for help.
You might retrieve the file list in your path and then select from the list based upon your partial file name. Here is a snippet of code to perform that type of function on a Linux machine.
import os
dir = '/home/craig/Python_Programs/GetFile'
files = os.listdir(dir)
print('Files--> ', files)
for i in files:
myfile = 'Hello_'
if (myfile[0:4] == i[0:4]):
print('File(s) like \"Hello_\"-->', i)
When I executed this simple program over a directory/folder that had various files in the directory, here was the output to the terminal.
Una:~/Python_Programs/GetFile$ python3 GetFile.py
Files--> ['Hello_Stack.txt', 'Okay.txt', 'Hi_Stack.txt', 'GetFile.py', 'Hello_Stack.bak']
File(s) like "Hello_"--> Hello_Stack.txt
File(s) like "Hello_"--> Hello_Stack.bak
The literal value for your path would be different on a Windows machine. I hope this might provide you with a method to achieve your goal.
Regards.
I would like to add a letter at the start of a file name in a file path.
For example change this file path-:
C:\Users\precious\Desktop\hello.txt
To this-:
C:\Users\precious\Desktop\rhello.txt
In other words, is there a way I can input a file path such as-:
C:\Users\precious\Desktop\hello.txt
And the program will remove the extension and path from the string and consolidate it to-:
hello
Then add a letter at the start of that string and make it-:
rhello
Then puts the edited file name back into the file path-:
C:\Users\precious\Desktop\rhello.txt
I have already figured out the first step (consolidating the file name from the path and extension) by using this code-:
file_name = Path(fp).stem
But I still haven't figured out how to take the modified name and put it back into the file path.
My goal is to rename the file to a changed file name using os.rename()
The method I am using to do this probably isn't the best, so you can suggest a better way of changing the file name or help me to continue using my method of doing the same.
I am using windows and python 3.
import pathlib
p = pathlib.Path(r'C:\Users\precious\Desktop\hello.txt')
newname = pathlib.Path(p.parent, 'r' + p.name)
p.rename(newname)
I am trying to access a file from a Box folder as I am working on two different computers. So the file path is pretty much the same except for the username.
I am trying to load a numpy array from a .npy file and I could easily change the path each time, but it would be nice if I could make it universal.
Here is what the line of code looks like on my one computer:
y_pred_walking = np.load('C:/Users/Eric/Box/CMU_MBL/Data/Calgary/4_Best_Results/Walking/Knee/bidir_lstm_50_50/predictions/y_pred_test.npy')
And here is what the line of code looks like on the other computer:
y_pred_walking = 'C:/Users/erapp/Box/CMU_MBL/Data/Calgary/4_Best_Results/Walking/Knee/bidir_lstm_50_50/predictions/y_pred_test.npy'
The only difference is that the username on one computer is Eric and the other is erapp, but is there a way where I can make the line universal to all computers where all computers will have the Box folder?
You could either save the file to a path that doesn't depend on the user: e.g. 'C:/Box/CMU_MBL/Data/Calgary/4_Best_Results/Walking/Knee/bidir_lstm_50_50/predictions/y_pred_test.npy'
Or you could do some string formatting. One way would be with an environment or configuration variable that indicates which is the relevant user, and then for your load statement:
import os
current_user = os.environ.get("USERNAME") # assuming you're running on the Windows box as the relevant user
# Now load the formatted string. f-strings are better, but this is more obvious since f-strings are still very new to Python
y_pred_walking = 'C:/Users/{user}/Box/CMU_MBL/Data/Calgary/4_Best_Results/Walking/Knee/bidir_lstm_50_50/predictions/y_pred_test.npy'.format(user=current_user)
Yes, there is a way, at least for the problem as it is right now solution is pretty simple: to use f-strings
user='Eric'
y_pred_walking =np.load(f'C:/Users/{user}/Box/CMU_MBL/Data/Calgary/4_Best_Results/Walking/Knee/bidir_lstm_50_50/predictions/y_pred_test.npy')
or more general
def pred_walking(user):
return np.load(f'C:/Users/{user}/Box/CMU_MBL/Data/Calgary/4_Best_Results/Walking/Knee/bidir_lstm_50_50/predictions/y_pred_test.npy')
so on any machine you just do
y_pred_walking=pred_walking(user)
with defined user before, to receive the result
Simply search the folders recursivly for your file:
filename = 'y_pred_test.npy'
import os
import random
# creates 1000 directories with a 1% chance of having the file as well
for k in range(20):
for i in range(10):
for j in range(5):
os.makedirs(f"./{k}/{i}/{j}")
if random.randint(1,100) == 2:
with open(f"./{k}/{i}/{j}/{filename}","w") as f:
f.write(" ")
# search the directories for your file
found_in = []
# this starts searching in your current folder - you can give it your c:\Users\ instead
for root,dirs,files in os.walk("./"):
if filename in files:
found_in.append(os.path.join(root,filename))
print(*found_in,sep = "\n")
File found in:
./17/3/1/y_pred_test.npy
./3/8/1/y_pred_test.npy
./16/3/4/y_pred_test.npy
./16/5/3/y_pred_test.npy
./14/2/3/y_pred_test.npy
./0/5/4/y_pred_test.npy
./11/9/0/y_pred_test.npy
./9/8/1/y_pred_test.npy
If you get read errors because of missing file/directory permissions you can start directly in the users folder:
# Source: https://stackoverflow.com/a/4028943/7505395
from pathlib import Path
home = str(Path.home())
found_in = []
for root,dirs,files in os.walk(home):
if filename in files:
found_in.append(os.path.join(root,filename))
# use found_in[0] or break as soon as you find first file
You can use the expanduser function in the os.path module to modify a path to start from the home directory of a user
https://docs.python.org/3/library/os.path.html#os.path.expanduser
I'm trying to build a file transfer system with python3 sockets. I have the connection and sending down but my issue right now is that the file being sent has to be in the same directory as the program, and when you receive the file, it just puts the file into the same directory as the program. How can I get a user to input the location of the file to be sent and select the location of the file to be sent to?
I assume you're opening files with:
open("filename","r")
If you do not provide an absolute path, the open function will always default to a relative path. So, if I wanted to open a file such as /mnt/storage/dnd/5th_edition.txt, I would have to use:
open("/mnt/storage/dnd/4p5_edition","r")
And if I wanted to copy this file to /mnt/storage/trash/ I would have to use the absolute path as well:
open("/mnt/storage/trash/4p5_edition","w")
If instead, I decided to use this:
open("mnt/storage/trash/4p5_edition","w")
Then I would get an IOError if there wasn't a directory named mnt with the directories storage/trash in my present folder. If those folders did exist in my present folder, then it would end up in /whatever/the/path/is/to/my/current/directory/mnt/storage/trash/4p5_edition, rather than /mnt/storage/trash/4p5_edition.
since you said that the file will be placed in the same path where the program is, the following code might work
import os
filename = "name.txt"
f = open(os.path.join(os.path.dirname(__file__),filename))
Its pretty simple just get the path from user
subpath = raw_input("File path = ")
print subpath
file=open(subpath+str(file_name),'w+')
file.write(content)
file.close()
I think thats all you need let me know if you need something else.
like you say, the file should be in the same folder of the project so you have to replace it, or to define a function that return the right file path into your open() function, It's a way that you can use to reduce the time of searching a solution to your problem brother.
It should be something like :
import os
filename = "the_full_path_of_the_fil/name.txt"
f = open(os.path.join(os.path.dirname(__file__),filename))
then you can use the value of the f variable as a path to the directory of where the file is in.
I´m trying to save a file, which I create with the "open" function.
Well I tried nearly everything to change the directory, but nothing works. The file gets always saved in the folder of my file, which I read in before.
file = open(fname[0] + ft, 'w')
file.write("Test")
file.close()
So this is it simple, but what do I have to add, to change the path of creation?
The File Dialog in a individual Function:
global fname
fname = QFileDialog.getOpenFileName(None, 'Please choose your File.',"C:\\Program Files", "Text-Files(*.txt)")
And the File Typ ( in a individual Function too) I set the file type by ticking a check box and ft will set to .py or .pyw
if self.exec_py.isChecked() == True:
global ft
ft = ".py"
I should have mentioned that I already tried os.path.join and os.chdir, but the file will get printed in the file anyway. Any solutions or approaches how to fix it? Here is how i tried it:
tmppath = "C:/temp"
tmp = os.path.join(tmppath,fname[0]+ft)
file = open(tmp, 'w')
Your question is a little short on details, but I am guessing that fname is the tuple returned by QFileDialog, and so fname[0] is the absolute path of the original file. So if you display fname[0], you will see something like this:
>>> fname[0]
'C:\\myfolder\\file.txt'
Now look what happens when you try to use that with os.path.join:
>>> tmppath = 'C:\\temp'
>>> os.path.join(tmppath, fname[0])
'C:\\myfolder\\file.txt'
Nothing! Conclusion: attempting to join two absolute paths will simply return the original path unchanged. What you need to do instead is take the basename of the original path, and join it to the folder where you want to save it:
>>> basename = os.path.basename(fname[0])
>>> basename
'file.txt'
>>> os.path.join(tmppath, basename)
'C:\\tmp\\file.txt'
Now you can use this new path to save your file in the right place.
You need to provide the full filepath
with open(r'C:\entire\path\to\file.txt', 'w') as f:
f.write('test')
If you just provide a file name without a path, it will use the current working directory, which isn't necessarily the directory where the python script your running is located. It will be the directory where you launched the script from.
C:\Users\admin> python C:\path\to\my_script.py
In this instance, the current working directory is C:\Users\admin, not C:\path\to.