I'm working in Python and I'm confused why I can't open the file I'm trying to. The code is pretty simple. Here it is.
import os
def main():
FILE_NAME = "default_template.csv"
source_path = os.path.join("Documents", FILE_NAME)
file = open(source_path, "r")
At this point I get
IOError: [Errno 2] No such file or directory: 'Documents/FILE_NAME'.
I also decided to try to change directories for whatever reason using os.chdir() and passing just about every high level directory on my computer in seperately and nothing worked. In an attempt to find the fix for opening a file I tried editing the path in a bunch of different ways.
I've tried something like:
os.path.join("/derek/Documents", FILE_NAME)
os.path.join("/Documents", FILE_NAME)
os.path.join("~/derek/Documents", FILE_NAME)
os.path.join("~", FILE_NAME)
If anyone could help me out I would be extremely thankful. I'm still new to using python to navigate and manage files.
Python doesn't expand ~ by default. To make it work try os.path.expanduser:
On Unix and Windows, return the argument with an initial component of ~ or ~user replaced by that user‘s home directory.
source_path = os.path.join("~/Documents", FILE_NAME)
source_path = os.path.expanduser(source_path)
Related
I am trying to run a Streamlit app importing pickle files and a DataFrame. The pathfile for my script is :
/Users/myname/Documents/Master2/Python/Final_Project/streamlit_app.py
And the one for my DataFrame is:
/Users/myname/Documents/Master2/Python/Final_Project/data/metabolic_syndrome.csv
One could reasonably argue that I only need to specify df = pd.read_csv('data/df.csv') yet it does not work as the Streamlit app is unexpectedly not searching in its directory:
FileNotFoundError: [Errno 2] No such file or directory: '/Users/myname/data/metabolic_syndrome.csv'
How can I manage to make the app look for the files in the good directory (the one where it is saved) without having to use absolute pathfiles ?
you can use os.getcwd() to get the Current Working Directory. Read up on what this means exactly, fe here
See the sample script below on how to use it. I'm using os.sep for OS agnostic filepath separators.
import os
print(os.getcwd())
relative_path = "data"
full_path = f"{os.getcwd()}{os.sep}{relative_path}"
print(full_path)
filename = "somefile.csv"
full_file_path = f"{full_path}{os.sep}{filename}"
print(full_file_path)
with open(full_file_path) as infile:
for line in infile.read().splitlines():
print(line)
In which directory are you standing when you are running your code?
From your error message I would assume that you are standing in /Users/myname/ which makes python look for data as a subdirectory of /Users/myname/.
But if you first change directory to /Users/myname/Documents/Master2/Python/Final_Project and then run your code from there I think it would work.
Im tryin to append a short string to every '.png' file. But when I run it, it says cannot find the file. But I know it's there and I can see it in the folder.
Is there anything I need to do?
Here is my script:
import os
for file in os.listdir("./pics"):
if file.endswith(".png"):
newFileName = "{0}_{2}{1}".format(*os.path.splitext(file) + ("z4",))
os.rename(file, newFileName)
Here is the error message I get...02.png is the first file in the folder:
fileNotFoundError: [WinError 2] The system cannot find the file
specified: '02.png' -> '02_z4.png'
It's odd though because it gets the filename, in this case, 02.png. So if it can read the file name, why can't it find it?
Thanks!
I thought my comment might be enough, but for clarity I'll provide a short answer.
02.png doesn't exist relative to your working directory. You need to specify the path to the file for os.rename so you need to include the directory.
import os
for file in os.listdir("./pics"):
if file.endswith(".png"):
newFileName = "/pics/{0}_{2}{1}".format(*os.path.splitext(file) + ("z4",)) # Notice the ./pics
os.rename(os.path.join('pics', file), newFileName)
The name returned from os.listdir() gives the filename, not the full path. So you need to rename pics/02.png to pics/02_zf.png. Right now you don't include the directory name.
I have worked through a number of other threads on this, but not of their solutions seem to work here, that or I am not understanding properly, and would love your help.
I am getting a:
IOError: [Errno 13] Permission denied: 'W:\\test\\Temporary Folder 195\\Sub-fold1
This is the general code i started with.
summary_file = r'W:/test/SDC Analysis Summary.docm'
shutil.copyfile(summary_file, os.getcwd())
I have also varied this a little bit based on other threads, specifically replacing summary_file with the actual text and also adding \ to the end of working directory without success. Really don't know what I'm missing here. I know that the Documentation is looking for complete paths, but I believe I am satisfying that requirement. What am I missing here?
Note: there is a desire to use copyfile over copy due to the speed increase.
From the documentation:
dst must be the complete target file name
You can't just use os.getcwd() as the destination.
you should be the complete target file name for destination
destination = pathdirectory + filename.*
I use this code fir copy wav file with shutil :
# open file with QFileDialog
browse_file = QFileDialog.getOpenFileName(None, 'Open file', 'c:', "wav files (*.wav)")
# get file name
base = os.path.basename(browse_file[0])
os.path.splitext(base)
print(os.path.splitext(base)[1])
# make destination path with file name
destination= "test/" + os.path.splitext(base)[0] + os.path.splitext(base)[1]
shutil.copyfile(browse_file[0], destination)
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.
Could someone shed me some lights on the file path matter in Python?
For example my codes need to read a batch of files, the file names are listed and stored in a .txt file, C:\filelist.txt, which content is:
C:\1stfile.txt
C:\2ndfile.txt
C:\3rdfile.txt
C:\4thfile.txt
C:\5thfile.txt
And the codes start with:
list_open = open('c:\\aaa.txt')
read_list = list_open.read()
line_in_list = read_list.split('\n')
all run fine. But if I want to read files in another path, such as:
C:\WorkingFolder\6thfile.txt
C:\WorkingFolder\7thfile.txt
C:\WorkingFolder\8thfile.txt
C:\WorkingFolder\9thfile.txt
C:\WorkingFolder\10thfile.txt
It doesn’t work. I guess the path here C:\WorkingFolder\ is not properly put so Python cannot recognize it.
So in what way I shall put it? Thanks.
hello all,
sorry that maybe i didn't make my self clear.
the problem is, a text file, c:\aaa.txt contains below:
C:\1stfile.txt
C:\WorkingFolder\1stfile.txt
why only C:\1stfile.txt is readable, but the other one not?
The reason your program isn't working is that you're not changing the directory properly. Use os.chdir() to do so, then open the files as normal:
import os
path = "C:\\WorkingFolder\\"
# Check current working directory.
retval = os.getcwd()
print "Current working directory %s" % retval
# Now change the directory
os.chdir( path )
# Check current working directory.
retval = os.getcwd()
print "Directory changed successfully %s" % retval
REFERENCES:
http://www.tutorialspoint.com/python/os_chdir.htm
import os
BASEDIR = "c:\\WorkingFolder"
list_open = open(os.path.join(BASEDIR, 'aaa.txt'))
Simply try using forward slashes instead.
list_open = open("C:/WorkingFolder/6thfile.txt", "rt")
It works for me.