I have a file in my site-packages directory called wordlist.py that consists of just one line:
f = open("words.txt")
There is a file called words.txt in the same directory. When I run wordlist.py it works fine. However, whenever I use import wordlist, I get an error:
FileNotFoundError: [Errno 2] No such file or directory: 'words.txt'
I am using IDLE for Python 3.4
if you use relative paths for file or directory names python will look for them (or create them) in your current working directory (the $PWD variable in bash).
if you want to have them relative to the current python file, you can use (python 3.4)
import pathlib
HERE = Path(__file__).parent.absolute()
WORDS_PATH = HERE / '../path/to/words.txt'
with WORDS_PATH.open() as file_pointer:
'do something with file_pointer...'
Related
So I currently use Path to check if a file exists
from pathlib import Path
if Path("main.conf").is_file():
pass
else:
setup_config()
While this works as long as I'm in the directory where I'm running the script, I'd like to make it work whatever directory I'm at and just run the script. I know it doesn't work because it's expecting the main.conf to be in the directory I'm currently in but how do I tell path to only check the in the folder where the script is located in?
You can resolve the absolute path of the script by using sys.argv[0] and then replace the name of that script with the config file to check, eg:
import sys
import pathlib
path = path.Pathlib(sys.argv[0]).resolve()
if path.with_name('main.conf').is_file():
# ...
else:
# ...
Although it seems like you should probably not worry about that check and structure your setup_config so it takes a filename as an argument, eg:
def setup_config(filename):
# use with to open file here
with open(filename) as fin:
# do whatever for config setup
Then wrap your main in a try/except (which'll also cover file not exists/can't open file for other reasons), eg:
path = pathlib.Path(sys.argv[0]).resolve()
try:
setup_config(path.with_name('main.conf'))
except IOError:
pass
using
import sys
folder = sys.argv[1]
for i in folder:
for file in i:
if file == "test.txt":
print (file)
would this access a file in the folder of a subfolder? For Example 1 main folder, with 20 subfolders, and each subfolder has 35 files. I want to pass the folder in commandline and access the first subfolder and the second file in it
Neither. This doesn't look at files or folders.
sys.argv[1] is just a string. i is the characters of that string. for file in i shouldn't work because you cannot iterate a character.
Maybe you want to glob or walk a directory instead?
Here's a short example using the os.walk method.
import os
import sys
input_path = sys.argv[1]
filters = ["test.txt"]
print(f"Searching input path '{input_path}' for matches in {filters}...")
for root, dirs, files in os.walk(input_path):
for file in files:
if file in filters:
print("Found a match!")
match_path = os.path.join(root, file)
print(f"The path is: {match_path}")
If the above file was named file_finder.py, and you wanted to search the directory my_folder, you would call python file_finder.py my_folder from the command line. Note that if my_folder is not in the same directory as file_finder.py, then you have to provide the full path.
No, this won't work, because folder will be a string, so you'll be iterating through the characters of the string. You could use the os module (e.g., the os.listdir() method). I don't know what exactly are you passing to the script, but probably it would be easiest by passing an absolute path. Look at some other methods in the module used for path manipulation.
I making a script in python3. this script takes an input file. depends on who is running the script every time the location of this input file is different but always would be in the same directory as the script is. so I want to give the location of the input file to the script but basically the script should find it. my input file always would have the same name (infile.txt). to do so, I am using this way in python3:
path = os.path.join(os.getcwd())
input = path/infile.txt
but it does not return anything. do you know how to fix it?
os.getcwd() return the working directory which can be different to the directory where the script is. The working directory corresponds from where python is executed.
In order to know where is the scipt you should use
`
import os
input = os.path.join(os.path.dirname(os.path.realpath(__file__)), infile.txt)
# and you should use os.path.join
`
If i understand your question properly;
You have python script (sample.py) and a input file (sample_input_file.txt) in a directory say; D:\stackoverflow\sample.y and D:\stackoverflow\sample_input_file.txt respectively.
import os
stackoverflow_dir = os.getcwd()
sample_txt_file_path = os.path.join(stackoverflow_dir, 'sample_input_file.txt')
print(sample_txt_file_path)
os.path.join() takes *args as second argument which must have been your file path to join.
I am trying to run multiple experiments, which are in different folder. I want to save the results in the main folder. Something like this:
Main folder
Main_run_file.py
Results.txt
Experiment_1
Run_file.py
Experiment_2
Run_file.py
Experiment_3
Run_file.py
I already tried with the following code:
import os
mydir = os.getcwd() # would be the MAIN folder
mydir_tmp = mydir + "\Experiment_1 - 40kmh" # add the first experiment folder name
mydir_new = os.chdir(mydir_tmp) # change the current working directory
mydir = os.getcwd() # set the main directory again
import Run_file
mydir = os.getcwd() # would be the MAIN folder
mydir_tmp = mydir + "/Experiment_1 - 60kmh" # add the second experiment folder name
mydir_new = os.chdir(mydir_tmp) # change the current working directory
mydir = os.getcwd() # set the main directory again
import Run_file
However, this only runs the first Run_file and not the second one. Can someone help me with this?
The second import Run_file is ignored, since python considers this module as already imported.
Either you replace those import statements by statements like this: import Experiment_1.Run_file, without forgetting to add the __init__.py files in your subdirectories,
Or you call your python scripts with subprocess as you would do from a command-line;
from subprocess import call
call(["python", "./path/to/script.py"])
You are also missing the point about current directories:
In the second mydir = os.getcwd() # would be the MAIN folder, mydir is still the previous subfolder
Python importing system does not care about you to change the working directory: imports are managed using the python path which depends on your python installation and the directory where you run your main script.
More here: https://docs.python.org/3/reference/import.html
try out subprocess :
import subprocess
with open("Results.txt", "w+") as output:
subprocess.call(["python", "./Experiment_1/Run_file.py"], stdout=output);
subprocess.call(["python", "./Experiment_2/Run_file.py"], stdout=output);
...
if you need to pass arguments to your Run_file.py, just add it as so :
subprocess.call(["python", "./Experiment_2/Run_file.py arg1"], stdout=output);
Does any one know why i got an error of "FileNotFoundError: [Errno 2] No such file or directory: 'bcs.xlsx'" when i'm loading this file of size 2mb it has around 60,000 rows and 4 columns.
i tried using csv instead of xlsx but i get the same error and i've checked hundreds times that the script and the file are at he same directory.
This is because Python does not find your file, errors are not lying.
But there's a misunderstanding in your question, you checked that the file is in the same directory as your script, but that's not the check you have to do. You have to check the file is in the current working directory of your python script.
To see your current working directory, use:
import os
print(os.getcwd())
And as we're at it you can list this directory:
print(os.listdir())
I don't know how you execute your script, but if you're using a terminal emulator, a typical way to give a file name to a program is by argument, not hardcoding its name, like by using argparse. And if you do this way, your shell completion may help you naming your file properly, like:
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('file', type=argparse.FileType('r'))
args = parser.parse_args()
print(args.file.read())
Now on a shell if you type:
python3 ./thescript.py ./th[TAB]
your shell will autocomplete "./th" to "./thescript.py" (if and only if it exists), highly reducing the probablity of having a typo. Typically if there's a space in the filename like "the script.py", your shell should properly autocomplete the\ script.py.
Also if you use argparse with the argparse.FileType as I did, you'll have a verbose error in case the file does not exist:
thescript.py: error: argument file: can't open 'foo': [Errno 2] No such file or directory: 'foo'
But… you already have a verbose error.