An integer is required? open()
Question:
I have a very simple python script that should scan a text file, which contains lines formatted as id=’value‘ and put them into a dict. the python module is called chval.py and the input file is in.txt. here’s the code:
import os,sys
from os import *
from sys import *
vals = {}
f = open(sys.argv[1], 'r')
for line in val_f:
t = line.split('=')
t[1].strip(''')
vals.append(t[0], t[1])
print vals
f.close()
when i try to run it i get:
Traceback (most recent call last):
File “chval.py”, line 9, in ?
f = open(sys.argv[1], ‘r’) TypeError: an integer is required
I’m using python 2.4… because i’ve been challenged to not use anything newer, is there something about open() that I don’t know about? Why does it want an integer?
anything after that line is untested. in short: why is it giving me the error and how do i fix it?
Answers:
Because you did from os import *, you are (accidenally) using os.open, which indeed requires an integer flag instead of a textual “r” or “w”. Take out that line and you’ll get past that error.
Don’t do import * from wherever without a good reason (and there aren’t many).
Your code is picking up the os.open() function instead of the built-in open() function. If you really want to use os.open(), do import os then call os.open(....). Whichever open you want to call, read the documentation about what arguments it requires.
Also of note is that starting with Python 2.6 the built-in function open() is now an alias for the io.open() function. It was even considered removing the built-in open() in Python 3 and requiring the usage of io.open, in order to avoid accidental namespace collisions resulting from things such as “from blah import *”. In Python 2.6+ you can write (and can also consider this style to be good practice):
import io
filehandle = io.open(sys.argv[1], 'r')
From http://www.tutorialspoint.com/python/os_open.htm you could also keep your import and use
file = os.open( “foo.txt”, mode )
and the mode could be :
os.O_RDONLY: open for reading only
os.O_WRONLY: open for writing only
os.O_RDWR : open for reading and writing
os.O_NONBLOCK: do not block on open
os.O_APPEND: append on each write
os.O_CREAT: create file if it does not exist
os.O_TRUNC: truncate size to 0
os.O_EXCL: error if create and file exists
os.O_SHLOCK: atomically obtain a shared lock
os.O_EXLOCK: atomically obtain an exclusive lock
os.O_DIRECT: eliminate or reduce cache effects
os.O_FSYNC : synchronous writes
os.O_NOFOLLOW: do not follow symlinks
Providing these parameters resolved my issue:
with open('tomorrow.txt', mode='w', encoding='UTF-8', errors='strict', buffering=1) as file:
file.write(result)
that’s because you should do:
open(sys.argv[2], "w", encoding="utf-8")
or
open(sys.argv[2], "w")
you have from os import * I also got the same error, remove that line and change it to import os and behind the os lib functions, add os.[function]
I have a very simple python script that should scan a text file, which contains lines formatted as id=’value‘ and put them into a dict. the python module is called chval.py and the input file is in.txt. here’s the code:
import os,sys
from os import *
from sys import *
vals = {}
f = open(sys.argv[1], 'r')
for line in val_f:
t = line.split('=')
t[1].strip(''')
vals.append(t[0], t[1])
print vals
f.close()
when i try to run it i get:
Traceback (most recent call last):
File “chval.py”, line 9, in ?
f = open(sys.argv[1], ‘r’) TypeError: an integer is required
I’m using python 2.4… because i’ve been challenged to not use anything newer, is there something about open() that I don’t know about? Why does it want an integer?
anything after that line is untested. in short: why is it giving me the error and how do i fix it?
Because you did from os import *, you are (accidenally) using os.open, which indeed requires an integer flag instead of a textual “r” or “w”. Take out that line and you’ll get past that error.
Don’t do import * from wherever without a good reason (and there aren’t many).
Your code is picking up the os.open() function instead of the built-in open() function. If you really want to use os.open(), do import os then call os.open(....). Whichever open you want to call, read the documentation about what arguments it requires.
Also of note is that starting with Python 2.6 the built-in function open() is now an alias for the io.open() function. It was even considered removing the built-in open() in Python 3 and requiring the usage of io.open, in order to avoid accidental namespace collisions resulting from things such as “from blah import *”. In Python 2.6+ you can write (and can also consider this style to be good practice):
import io
filehandle = io.open(sys.argv[1], 'r')
From http://www.tutorialspoint.com/python/os_open.htm you could also keep your import and use
file = os.open( “foo.txt”, mode )
and the mode could be :
os.O_RDONLY: open for reading only
os.O_WRONLY: open for writing only
os.O_RDWR : open for reading and writing
os.O_NONBLOCK: do not block on open
os.O_APPEND: append on each write
os.O_CREAT: create file if it does not exist
os.O_TRUNC: truncate size to 0
os.O_EXCL: error if create and file exists
os.O_SHLOCK: atomically obtain a shared lock
os.O_EXLOCK: atomically obtain an exclusive lock
os.O_DIRECT: eliminate or reduce cache effects
os.O_FSYNC : synchronous writes
os.O_NOFOLLOW: do not follow symlinks
Providing these parameters resolved my issue:
with open('tomorrow.txt', mode='w', encoding='UTF-8', errors='strict', buffering=1) as file:
file.write(result)
that’s because you should do:
open(sys.argv[2], "w", encoding="utf-8")
or
open(sys.argv[2], "w")
you have from os import * I also got the same error, remove that line and change it to import os and behind the os lib functions, add os.[function]