f = open("demofile3.txt", "w")
f.writelines(["\nSee you soon!", "\nOver and out."])
f.close()
`
I have been trying to create or open a file and write something in it but it does not seem to do anything.
it's about pwd.
But here is a method: use full path.
For example:
from pathlib import Path
folder = Path(__file__).parent
your_txt = folder / "demofile3.txt"
with open(your_txt, "w") as f:
f.writelines(["\nSee you soon!", "\nOver and out."])
Related
I am having an issue of trying to load CSV files in a python script when trying to run the file from outside the directory of the script called main.py and the CSV files. (CSV files and main are in the same directory) I think the same issues as this SO post which doesn't appear to contain a solution.
If I run this from terminal:
$ python /home/bbartling/Desktop/Load-Shredder/Load-Shift/main.py
I get some CSV file loading errors [Errno 2] No such file or directory: 'addresses_jci.csv'
But if I run the script in the directory of Load-Shift $ python main.py it works just fine.
How do I fix my script to accommodate this? I added this to the top of my script:
script_dir = os.path.abspath( os.path.dirname( __file__ ) )
print("script directory: ",script_dir)
which prints:
script directory: /home/bbartling/Desktop/Load-Shredder/Load-Shift
But still no luck. Any ideas to try?
Edit
CSV file loading function in main.py
dir_path = os.path.dirname(os.path.realpath(__file__))
filename = os.path.join(dir_path, f'log_{dt_string}.log')
script_dir = os.path.abspath( os.path.dirname( __file__ ) )
print("script directory: ",script_dir)
def load_addresses(csv_file):
try:
print(f"Loading csv file: {csv_file}!")
os.path.join(script_dir, f'{csv_file}.csv')
with open(f'{csv_file}', newline='') as f:
reader = csv.reader(f)
data = list(reader)
# flattens list of lists
csv_file = sum(data, [])
print(f"{csv_file} addresses loaded: ",csv_file)
except Exception as error:
print(f"CSV file load error: {error}")
csv_file = [] # errors out
return csv_file
The problem in your code, I think you did not update the csv_file variable update:
csv_file = os.path.join(script_dir, f'{csv_file}.csv')
Though it can be more simpler:
import os
script_path = os.path.dirname(__file__)
filename = 'sample'
file_path = os.path.join(script_path, f'{filename}.csv')
print(file_path)
filepaths in general are a bit confusing in programming languages, so this is a very common problem for beginners.
I typically have a fixed working directory for my projects, and tend to access all files relative to that working directory. If you can implement it, this is the simplest solution.
$ cd /home/bbartling/Desktop/Load-Shredder/Load-Shift/
$ python main.py
can give the desired solution, but depending upon your project this may or may not be feasible. This is by no means a hard-code or a hot fix, and perfectly valid for my projects. However, if you're making a reusable shell script, you probably don't want this.
I've used Path(__file__).parent before to get the directory the file is running in, but IDK how the speed compares to the way you compute it.
I've used this a couple times before V
import os
HERE = os.path.dirname(os.path.realpath(__file__))
with open(os.path.join(HERE, 'mkdocs.yml')) as fl:
print(fl.read())
You can change the filename here to your script's path, and that will work. If you know where it will be stored relative to your main.py dir, you can hard-code it. otherwise get it from the command line or console, whichever you prefer. Use the relative path from your main.py directory to your csv here.
This is working! in the function to load a file:
THIS WORKS with specifing object to reference for the os.path.join:
full_csv_path = os.path.join(dir_path, f'{csv_file}')
THIS DOESNT WORK without specifying an object for the program to reference:
os.path.join(dir_path, f'{csv_file}')
Example below that also includes logging:
import csv, os, logging
# datetime object containing current date and time
now = datetime.now()
dt_string = now.strftime("%m_%d_%Y %H_%M_%S")
dir_path = os.path.dirname(os.path.realpath(__file__))
filename = os.path.join(dir_path, f'log_{dt_string}.log')
# Logger
logger = logging.getLogger(__name__)
logger.setLevel(logging.INFO)
file_handler = logging.FileHandler(filename)
file_handler.setLevel(logging.INFO)
file_handler.setFormatter(logging.Formatter('%(asctime)s - %(levelname)s - %(message)s'))
logger.addHandler(file_handler)
print("dir_path is: ",dir_path)
def load_addresses(csv_file):
try:
print(f"Loading csv file: {csv_file}!")
full_csv_path = os.path.join(dir_path, f'{csv_file}')
print("full_csv_path: ",full_csv_path)
with open(full_csv_path, newline='') as f:
reader = csv.reader(f)
data = list(reader)
# flattens list of lists
csv_file = sum(data, [])
print(f"{csv_file} addresses loaded: ",csv_file)
except Exception as error:
print(f"CSV file load error: {error}")
csv_file = [] # errors out
return csv_file
You can use pandas , os and fnmatch for this purpose as given below,
import pandas as pd
import os,fnmatch
# change path to the folder from another folder
os.chdir("/path_to_the_folder_having_files_from_another_folder")
# print the current working directory to check if you have arrived
# in the right folder/directory
print(os.getcwd())
# then use fnmatch package to get the CSV files only
# or you can make some other pattern using some prefix or suffix
# like *this.csv or *that.csv
files = fnmatch.filter(os.listdir('.'), '*.csv')
# print the files to check that you have captured the right files
files
# read your files using pandas
# be careful with `sep`, you need to check specifically for your
# files
csv_files = [pd.read_csv(f, low_memory=False,sep = "\t",header=0 for f in files]
I am currently accessing a script that opens a file in the directory it's located in. I am accessing this file from both the main.py file located in the same directory, as well as a testfile which is located in a "Test" subdirectory. Trying to use a file from the Test subdirectory to call the function that opens the file causes the script to try and open it from the Test directory instead of the super directory, since I am opening the file simply by calling it as following:
with open(filename,"w") as f:
Is there a way to define the location of the file in a way that makes it possible for the script opening it to be called from anywhere?
Use __file__ to get the path to the current script file, then find the file relative to that:
# In main.py: find the file in the same directory as this script
import os.path
open(os.path.join(os.path.dirname(__file__), 'file.txt'))
# In Test/test.py: find the file one directory up from this script
import os.path
open(os.path.join(os.path.dirname(__file__), '..', 'file.txt'))
just give the absolute file path instead of giving a relative one
for eg
abs_path = '/home/user/project/file'
with open(abs_path, 'r') as f:
f.write(data)
Try specifying the path:
import os
path = 'Your path'
path = os.path.abspath(path)
with open(path, 'w') as f:
f.write(data)
From what I understood your file is in a directory parent_directory/file_name.txt
and in another folder parent_directory/sub_directory/file_name.txt. All you have to do is paste the below code in both parent and sub directories.
import os
file_name = 'your_file_name'
# if the file is in current directory set the path to file_name
if file_name in os.listdir(os.getcwd()):
path = file_name
# if the path is not in current directory go back to parent directory set the path to parent directory
else:
path = os.path.abspath(os.path.join(os.getcwd(), os.pardir))
print('from',os.getcwd())
with open(path, 'r') as filename:
print(filename.read())
is it possible to create a file with a specific extension, name and python with the sys module or any other module?
Thanks!
Using
f = open('myfile.myextension', 'w')
f.write('hello')
f.close()
You can create file with any extension you would like. If you want another software to read it, for example, if you want to create an excel file (xlsx) or pdf you might need additional packages.
This should work with the path and the file name:
import os
PATH = '/home/user/somepath'
FILE = 'somefile.someext'
os.makedirs( PATH, exist_ok=True) # create the PATH if not exists
full_name = os.path.join( PATH, FILE )
with open( full_name ) as fout :
print >>fout, 'some message'
I'm struggling with reading a file in python, the py file and CSV file are in the same folder but the VSCode makes an error and can't find the file:
import csv
with open('file.csv','r') as f:
reader = reader(f)
...
how can I fix this??
and the error is:
Exception has occurred: FileNotFoundError
[Errno 2] No such file or directory: 'file.csv'
If you run:
import os
os.getcwd()
You'll find out your current working directory which I assume is not the one you were expecting. If you're running the python script through VS code it could be using it could be the directory which you have open on the left hand side.
So either run the python using the correct working directory or use an absolute path like this:
import csv
with open('pathname/file.csv','r') as f:
reader = reader(f)
There might be an issue with your relative path settings.
Try this:
import os
import csv
dir = os.path.dirname(__file__)
filename = os.path.join(dir, 'file.csv')
with open(filename,'r') as f:
reader = reader(f)
Are you using spyder?
If so, please check if the current working path is the path your py file locates.
import csv
with open('file.csv','r') as f:
reader = csv.reader(f)
in this case your file.csv should be in folder where is your python script (current working folder)
or, instead of 'file.csv' you can put absolute path
Normally, executing the following code will pickle an object to a file in my current directory:
fp = open('somefile.txt', 'wb')
pickle.dump(object, fp)
How do I re-direct the output from pickle.dump to a different directory?
with open('/full/path/to/file', 'wb') as f:
pickle.dump(object, f)
How about combination of pathlib and with, I think it more flexible and safer.
# from python 3.4
from pathlib import Path
my_path = Path("{path to you want to set root like . or ~}") / "path" / "to" / "file"
with my_path.open('wb') as fp:
pickle.dump(object, fp)
If you wish to save the file to a sub-folder located inside the folder containing your code you can use the pathlib module. This will allow the code to work even if its location is moved on your computer, or you code is added to a different machine.
import the module:
from pathlib import Path
Set root equal to your current folder:
root = Path(".")
Create a path to your sub-folder and file name:
my_path = root / "my_sub_folder" / "my_file_name"
Open your file, dump data to your file, close your file:
my_file = open(my_path, 'wb')
my_file = pickle.dump("data_to_save", my_file)
my_file.close()
Note if your my_file doesn't currently exist you will want to create it before running this code. *
You can just try
import pickle
filepath = r'<FolderName>/<FileName>.pkl'
pickle.dump(<model>, open(filepath, 'wb'))
so final code look likes
If your Folder name is Models
and you wanted to pickle randomForest Model
import pickle
filepath = r'Models/rfPickle.pkl'
pickle.dump(randomForest, open(filepath, 'wb'))