[Tutor] Zipfile and File manipulation questions.

[Luke Paireepinart]

Chris Hengge wrote:
> I chose the way I used the names because to me...
>
> outFile = open(aFile.lower(), 'w') # Open output buffer for writing.
> = open a file with lowercase name for writing.
> it is implied that aFile is from the zip, since it is created in the 
> loop to read the zip..
>
> outFile.write(zFile.read(insideZip)) # Write the file.
> = write what is read from inside the zip file.
>
> I guess for declaration it isn't very clear, but thats what comments 
> are for?
No!!!!
Consider the following 2 examples:

#temp.py
def aa(a):
    for b in range(2,a):
        if a % b == 0:
            return False
    return True

a = int(raw_input('Please enter a prime number: '))
if not aa(a):
    print "That's not a prime."
    raise SystemExit
a = (2**a)-1
if aa(a):
    print "It's a mersenne prime."
else:
    print "It's not a mersenne prime."
#----




#Check_Mersenne_Numbers.py
def isprime(candidate):
    for number in range(2,candidate):
       if number % candidate == 0:
          return False
    return True

power = int(raw_input('Please enter a prime number: '))
if not isprime(power):
    print "That's not a prime."
    raise SystemExit

mersenne_number = (2**power) - 1

if not isprime(mersenne_number):
    print "It's not a mersenne prime."
else:
    print "It is a mersenne prime."
# ----

Obviously, the whole aa(a) thing was designed to look ridiculous,
but you have to remember that other people looking at your code don't 
know anything about it, like you do.
You can comment the example #1 to the point where people can figure out 
what's going on,
but example #2 is more or less completely clear as to what is happening 
at every step without any need for documentation.
If you find yourself commenting "Remember here that file is no longer 
the input file but is now the output file"
chances are you'd be better off having 2 variables, infile and outfile.
If you fall into this pattern of willy-nilly naming variables you might 
someday have a script that looks like this:

import os, sys
print sys.argv
sys = "Intel Core 2 Duo"
os = "Windows XP Professional"
print "This computer is running %s on a(n) %s" % (os,sys) #remember os 
and sys are strings and not modules

#we need to use the os and sys modules again so we need to reimport them,
#but we'll go ahead and import them as something else so we don't 
overwrite our string vars.
import os as Os
import sys as Sys
if sys in Sys.argv: print "They passed ",sys," as an argument!"
#we need an alternate configuration now
SYs = "AMD ATHLON 64"
OS = "Ubuntu"
#let's get the name of this script
SYs = Os.path.split(Sys.argv[0])[-1]
sys = "Error creating directory."
print "this script is called %s" % SYs
#subdirectories for our oses
try:
    Os.mkdir(OS)
    Os.mkdir(os)
except:
    print sys

#------
Haha.
Just be forewarned!  Coming up with good variable names is extremely 
important!

> My naming was purely for my ease of mind.. I personally care less 
> about what I call it when I declare, as to how it logically flows when 
> I go to use it. I'm sure this is considered poor method, but once I 
> declare a method I tend to never need to change the declaration, just 
> how I use the info... I hope that makes sense.
No, I don't know what you mean here.
>
> On 10/16/06, *Kent Johnson* <kent37 at tds.net <mailto:kent37 at tds.net>> 
> wrote:
>
>     Chris Hengge wrote:
>     > Here is my solution, completed with (I think) all your
>     suggestions...
>     >
>     >
>     #########################################################################
>     > def extractZip(filePathName):
>     >     """
>     >     This method recieves the zip file name for decompression,
>     placing the
>     >     contents of the zip file appropriately.
>     >     """
>     >     if filePathName == "":
>     >         print "No file provided...\n"
>     >     else:
>     >         try: # Attempt to unzip file.
>     >             zFile = zipfile.ZipFile(filePathName.strip('"'), "r")
>     >             for aFile in zFile.namelist(): # For every file in
>     the zip.
>     >                 # If the file ends with a needed extension,
>     extract it.
>     >                 for ext in ['.cap', '.hex', '.fru', '.cfg', '.sdr']:
>     >                     if aFile.lower().endswith(ext):
>     >                         insideZip = aFile # Copy of Filename.
>     >                         if "/" in aFile: # Split the filename if
>     '/'.
>     >                           aFile = aFile.rsplit('/', 1)[-1]
>     >                         elif  "\\" in aFile: # Split the
>     filename if '\'.
>     >                           aFile = aFile.rsplit('\\',
>     > 1)[-1]
>     >                         outfile = open( aFile.lower(), 'w') # Open
>     > output buffer for writing.
>     >                         outfile.write(zFile.read(insideZip)) #
>     Write the
>     > file.
>     >                         outfile.close() # Close the output file
>     buffer.
>     >             print "Resource extraction completed successfully!\n"
>     >         except IOerror, message: # If file creation fails, let
>     the user
>     > know.
>     >             print "File could not be written: \n"
>     >             print message
>     >
>     >
>     #########################################################################
>     > Definatly an improvement! Thanks Kent.
>
>     Yes, that is what I meant. One minor quibble, I think I would keep
>     aFile
>       as the name in the zip, since that is what it starts as, and use
>     a new
>     name for the external file name. Maybe you could use better names, for
>     example zipPath and fileName. I think that would make the code a
>     little
>     clearer but it is a very minor point.
>
>     Kent
>
>
>
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