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 You are using a relative path, which means that the program looks for the file in the working directory. The error is telling you that there is no file of that name in the working directory.
Try using the exact, or absolute, path.
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 '%s': %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") Use the exact path.
import csv with open('C:\path\address.csv','r') as f: reader = csv.reader(f) for row in reader: print row try and remove the '.csv' from the name of the file, or add an extra one in the open function. worked for me.
Lets say we have a script in "c:\script.py" that contain :
result = open("index.html","r") print(result.read()) Lets say that the index.html file is also in the same directory "c:\index.html" when i execute the script from cmd (or shell)
C:\Users\Amine>python c:\script.py You will get error:
FileNotFoundError: [Errno 2] No such file or directory: 'index.html' And that because "index.html" is not in working directory which is "C:\Users\Amine>". so in order to make it work you have to change the working directory
C:\python script.py '' This is why is it preferable to use absolute path.
writing this with open('address.csv','r') as this with open('address.csv',mode='r') worked for me.