Menu
I am trying to open a CSV file but for some reason python cannot locate it.
Here is my code (it's just a simple code but I cannot solve the problem):
import csv with open('address.csv','r') as f: reader = csv.reader(f) for row in reader: print row 
316
The error "FileNotFoundError: [Errno 2] No such file or directory" is telling you that there is no file of that name in the working directory. So, try using the exact, or absolute path. In the above code, all of the information needed to locate the file is contained in the path string - absolute path.
The error FileNotFoundError Errno 2 no such file or directory occurs when Python cannot find the specified file in the current directory.
The Python FileNotFoundError: [Errno 2] No such file or directory error is often raised by the os library. This error tells you that you are trying to access a file or folder that does not exist. To fix this error, check that you are referring to the right file or folder in your program.
When you open a file with the name address.csv, you are telling the open() function that your file is in the current working directory. This is called a relative path.
To give you an idea of what that means, add this to your code:
import os cwd = os.getcwd() # Get the current working directory (cwd) files = os.listdir(cwd) # Get all the files in that directory print("Files in %r: %s" % (cwd, files)) That will print the current working directory along with all the files in it.
Another way to tell the open() function where your file is located is by using an absolute path, e.g.:
f = open("/Users/foo/address.csv") If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
