I'm learning Python (with Python Crash Course book) and i'm currently working with climate data from the NOAA, but everytime i try to import a file, Python does not find it. I had this error for other programs too and i can't really solve it. Could anyone help me please ?
Here's my code :
import csv
filename = 'new-york-weather_60-20.csv'
try :
with open(filename) as f:
reader = csv.reader(f)
header_row = next(reader)
print(header_row)
except FileNotFoundError:
print(f"Sorry, the file {filename} does not exist.")
The Python interpreter assumes that the file 'new-york-weather_60-20.csv' is in the same directory (folder) as where python currently 'is', i.e., the current working directory.
You can see the current working directory by using the os module.
import os
print(os.getcwd())
This should be the path in which the csv file is located. If it is not, you can either move the file into that same location, or you can move the current working directory to the path where the file is located
import os
os.chdir('/the/location/on/your/computer/wherethefileislocated')
filename = 'new-york-weather_60-20.csv'
# You can check if the file is located in this directory
if os.path.exists(filename): # only if this file exists in this location do we continue
print('The file exists!')
with open(filename) as f:
# your code goes here
else:
print('The file does not exist in this directory')
Related
This is a school program to learn how to use file and directory in Python. So to do my best I create a function to open, set it as a variable and close properly my file.
But I got the error of the title:
FileNotFoundError: [Errno 2] No such file or directory: 'codedata.pkl'
def load_db():
""" load data base properly
And get ready for later use
Return:
-------
cd : (list) list of tuples
"""
file = open('codedata.pkl', 'rb')
codedata = pickle.loads(file)
file.close()
return codedata
From the interpreter, this is the line
file = open('codedata.pkl', 'rb')
Which is the problem, but I don't see where is the source of the problem.
Can anyone help me?
Can you check what is the location of the file?
If your file is located at /Users/abc/Desktop/, then the code to open the file on python would be as shown below
file = open('/Users/abc/Desktop/codedata.pkl', 'rb')
codedata = pickle.load(file)
file.close()
You can also check if the file exists at the desired path by doing something like this
import os
filepath = '/Users/abc/Desktop/codedata.pkl'
if os.path.exists(filepath):
file = open('/Users/abc/Desktop/codedata.pkl', 'rb')
codedata = pickle.load(file)
file.close()
else:
print("File not present at desired location")
it happens when you run the script without determining it's the current working directory (example in vs code if you go to Explorer Tape )
You do not work from the same directory that your data.pkl in that's why No file exists
You can know the current directory from getcwd() usually it will be the C/User/.
print(os.getcwd())
filepath=""
if os.path.exists(r"D:\research\StleGAN\karras2019stylegan-ffhq-1024x1024.pkl"):
print("yes")
else:
print("no")
The solution is to open a directory that contains the script or to add the full path.
I am trying to read content of a file on my work network from my work network. I copy and pasted a code snippet from a google search and modified it to the below. Why might I still be getting [Errno 2] (I have changed some of the path names for this question board)
The file path in my file explorer shows that "> This PC > word common common" and I don't have "This PC" in my path. I tried adding that into the place I would think it goes in the string. That didn't solve it.
I tried making sure I have matching capitalization. That didn't solve it.
I tried renaming the file to have a *.txt on the end. That didn't solve it.
I tried the different variations of // and / and \ with and without the r predecessor and while that did eliminate the first error I was getting. It didn't help this error.
(Looking at the code errors in the right gutter is says my line length is greater than the PEP8 standard. While I doubt that is the root of my problem, if you can throw in the 'right' wrap method for a file path that long that would be helpful.)
myfile = open("z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246", "rt") # open lorem.txt for reading text
contents = myfile.read() # read the entire file into a string
myfile.close() # close the file
print(contents) # print contents
Full Error Copy:
C:\Users\e087680\PycharmProjects\FailureCompiling\venv\Scripts\python.exe C:/Users/e087680/PycharmProjects/FailureCompiling/FirstScriptAttempt.py
Traceback (most recent call last):
File "C:/Users/e087680/PycharmProjects/FailureCompiling/FirstScriptAttempt.py", line 1, in
myfile = open("z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246", "rt") # open lorem.txt for reading text
FileNotFoundError: [Errno 2] No such file or directory: 'z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246'
EDIT
DEBUG EFFORTS
working to figure out how to change directory. Just in case that is the problem. Tested this code bit
import os
path = "z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246"
os.chdir(path)
isExist = os.path.exists(path)
print(isExist)
Received this error
C:\Users\e087680\PycharmProjects\FailureCompiling\venv\Scripts\python.exe C:/Users/e087680/PycharmProjects/FailureCompiling/ScriptDebugJunkFile.py
Traceback (most recent call last):
File "C:/Users/e087680/PycharmProjects/FailureCompiling/ScriptDebugJunkFile.py", line 5, in <module>
os.chdir(path)
FileNotFoundError: [WinError 3] The system cannot find the path specified: 'z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246'
My intention for adding the picture below is to show how File Explorer displays the file path for my file
FileExplorerPathScreenShot
EDIT
I think this confirms that my 'OS' doesn't have my file.
from os import path
path.exists("PCC_ESS_FC_Full_Report_65000122-1_R0016_962019_9246")
def main():
print ("File exists:" + str(path.exists('PCC_ESS_FC_Full_Report_65000122-1_R0016_962019_9246')))
if __name__== "__main__":
main()
Output
File exists: False
I thought OS was a standard variable for Operating system. Now I'm not sure.
EDIT
Using Cmd in DOS, I confirmed that my path for the z: is correct
EDIT - Success
I ran
import os
print( os.listdir("z:/"))
Confirmed I don't need the monster string of folders.
Confirmed, although explorer doesn't show it, it is a *.txt file
Once I implemented these two items the first code worked fine.
Thank you #Furas
To open and read a file specify the filename in your path:
myfile = open("U:/matrix_neo/word common common/hello world.txt", "rt") # open file
contents = myfile.read() # read the entire file into a string
myfile.close() # close the file
print(contents) # print contents
The U: is a mapped drive in my network.
I did not find any issue with your change dir example. I used a path on my U: path again and it returned True.
import os
path = "U:/matrix_neo/word common common"
os.chdir(path)
isExist = os.path.exists(path)
print(isExist)
The check the attributes on the directory that you are trying to read from. Also try to copy the file to a local drive for a test and see if you can read the file and also check if it exists.
This is an alternative to the above and uses your path to make sure that the long file path works:
import os
mypath = "z:/abcdefg/abc123_proj2/word_general/word common common/Users/Mariee/Python/abc_abc_ab_Full_Report_12345-1_R9999_962019_9246"
myfile = 'whatever is your filename.txt'
if not os.path.isdir(mypath):
os.makedirs (mypath)
file_path = os.path.join(mypath, myfile)
print(file_path)
if os.path.exists(file_path) is True:
with open(file_path) as filein:
contents = filein.read()
print(contents)
I tested this code using a long csv file.,Replace the variable myfile with whatever is your file name.
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'))
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()