I have a folder that stores a json file in my django application folder, ie, test_data/data.json.
In my tests.py, I am trying to read this file using the following code:
with open('/test_data/data.json', 'r') as f:
self.response_data = json.load(f)
However, I keep on getting the following error:
FileNotFoundError: [Errno 2] No such file or directory: '/test_data/data.json'
What am I doing wrong? Thanks.
Edit: I tried removing the leading slash, yet I still get the same error.
import os
try this
with open(os.getcwd() + '/test_data/data.json', 'r') as f:
self.response_data = json.load(f)
If you're opening files in directories close to where your code is, it is common to place
import os
DIRNAME = os.path.dirname(__file__) # the directory of this file
at the top of the file.
Then you can open files in a test_data subdirectory with
with open(os.path.join(DIRNAME, 'test_data', 'data.json'), 'rb') as fp:
self.response_data = json.load(fp)
you probably want to open json files, which should be utf-8 encoded, in 'rb' (read-binary) mode.
Related
I am using the OS module to open a file for reading, but I'm getting a FileNotFoundError.
I am trying to
find all the files in a given sub-directory that contain the word "mda"
for each of those files, grab the string in the filename just after two "_"s (indicates a specific code called an SIC)
open that file for reading
will write to a master file for some Mapreduce processing later
When I try to do the opening, I get the following error:
File "parse_mda_SIC.py", line 16, in <module>
f = open(file, 'r')
FileNotFoundError: [Errno 2] No such file or directory:
'mda_3357_2017-03-08_1000230_000143774917004005__3357.txt'
I am suspicious the issue is either with the "file" variable or the fact that it is one directory down, but confused why this would occur when I am using OS to address that lower directory.
I have the following code :
working_dir = "data/"
for file in os.listdir(working_dir):
if (file.find("mda") != -1):
SIC = re.findall("__(\d+)", file)
f = open(file, 'r')
I would expect to be able to open the file without issue and then create my list from the data. Thanks for your help.
This should work for you. You need to append the directory because it sees it as just the file name at the top of your code and will look only in the directory where your code is located for that file name.
for file in os.listdir(working_dir):
if (file.find("mda") != -1):
SIC = re.findall("__(\d+)", file)
f = open(os.path.join(working_dir, file), 'r')
Also it's a good practice to open files using a context manager of with as it will handle closing your file when it is no longer needed:
for file in os.listdir(working_dir):
if (file.find("mda") != -1):
SIC = re.findall("__(\d+)", file)
with open(os.path.join(working_dir, file), 'r') as f:
# do stuff with f here
You need to append the directory, like this:
f = open(os.path.join(working_dir, file, 'r'))
for a pattern recognition application, I want to read and operate on jpeg files from another folder using the os module.
I tried to use str(file) and file.encode('latin-1') but they both give me errors
I tried :
allLines = []
path = 'results/'
fileList = os.listdir(path)
for file in fileList:
file = open(os.path.join('results/'+ str(file.encode('latin-1'))), 'r')
allLines.append(file.read())
print(allLines)
but I get an error saying:
No such file or directory "results/b'thefilename"
when I expect a list with the desired file names that are accessible
If you can use Python 3.4 or newer, you can use the pathlib module to handle the paths.
from pathlib import Path
all_lines = []
path = Path('results/')
for file in path.iterdir():
with file.open() as f:
all_lines.append(f.read())
print(all_lines)
By using the with statement, you don't have to close the file descriptor by hand (what is currently missing), even if an exception is raised at some point.
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
I'm having an issue trying to open a file that is definitely saved to my computer ('NYT-bestsellers.txt'), but whenever I try opening it with my code I get the error
FileNotFoundError: [Errno 2] No such file or directory: 'NYT-bestsellers.txt'
I thought about using the method where you use the full path to open the file… but this is part of an assignment that I'll be submitting later this week. If I open the file using a specific path from my laptop, I'm worried that it won't open for the marker. Please advise!
with open('NYT-bestsellers.txt', 'r') as file:
file = file.splitlines()
As Ryan said, every time you open a file by a relative name, you need to make clear for the current work path.
import sys
import os
current_work_directory = os.getcwd() # Return a string representing the current working directory.
print('Current work directory: {}'.format(current_work_directory))
# Make sure it's an absolute path.
abs_work_directory = os.path.abspath(current_work_directory)
print('Current work directory (full path): {}'.format(abs_work_directory))
print()
filename = 'NYT-bestsellers.txt'
# Check whether file exists.
if not os.path.isfile(filename):
# Stop with leaving a note to the user.
print('It seems file "{}" not exists in directory: "{}"'.format(filename, current_work_directory))
sys.exit(1)
# File exists, go on!
with open(filename, 'r') as file:
file = file.splitlines()
If you confirm that the file will be along with your python script file, you can do some preparatory work before opening the file:
script_directory = os.path.split(os.path.abspath(__file__))[0]
print(script_directory)
abs_filename = os.path.join(script_directory, filename)
print(abs_filename)
with open(abs_filename, 'r') as file:
file = file.splitlines()
I am trying to process every files inside a folder line by line. I need to check for a particular string and write into an excel sheet. Using my code, if i explicitly give the file name, the code will work. If I try to get all the files, then it throws an IOError. The code which I wrote is as below.
import os
def test_extract_programid():
folder = 'C://Work//Scripts//CMDC_Analysis//logs'
for filename in os.listdir(folder):
print filename
with open(filename, 'r') as fo:
strings = ("/uri")
<conditions>
for line in fo:
if strings in line:
<conditions>
I think the error is that the file is already opened when the for loop started but i am not sure. printing the file name prints the file name correctly.
The error shown is IOError: [Errno 2] No such file or directory:
if your working directory is not the same as folder, then you need to give open the path the the file as well:
with open(folder+'/'+filename, 'r') as fo
Alternatively, you can use glob
import glob
for filename in glob.glob(folder+'/*'):
print filename
It can't open the path. You should do
for filename in os.listdir(folder):
print folder+os.sep()+filename