For some reason my code is having trouble opening a simple file:
This is the code:
file1 = open(\'recentlyUpdated.yaml\')
And the erro
os.listdir() to see the list of files in the current working directoryos.getcwd() (if you launch your code from an IDE, you may well be in a different directory)os.chdir(dir), dir being the folder where the file is
located, then open the file with just its name like you were doing.open call.dir = r'C:\Python32'
'C:\\User\\Bob\\...''C:/Python32' and do not need to be escaped.Let me clarify how Python finds files:
working directory. You can view Python's current working directory by calling os.getcwd(). If you try to do open('sortedLists.yaml'), Python will see that you are passing it a relative path, so it will search for the file inside the current working directory. Calling os.chdir will change the current working directory.
Example: Let's say file.txt is found in C:\Folder.
To open it, you can do:
os.chdir(r'C:\Folder')
open('file.txt') #relative path, looks inside the current working directory
or
open(r'C:\Folder\file.txt') #full path