get an array variable in python

Question

can I do this in a loop, by producing the file name from the name of the array to store ?

ab = array.array('B', map( operator.xor, a, b ) )
f1 = open('ab', 'wb')
ab.tofile(f1)
f1.close
ac = array.array('B', map( operator.xor, a, c ) )
f1 = open('ac', 'wb')
ac.tofile(f1)
f1.close
ad = array.array('B', map( operator.xor, a, d ) )
f1 = open('ad', 'wb')
ad.tofile(f1)
f1.close
ae = array.array('B', map( operator.xor, a, e ) )
f1 = open('ae', 'wb')
ae.tofile(f1)
f1.close
af = array.array('B', map( operator.xor, a, f ) )
f1 = open('af', 'wb')
af.tofile(f1)
f1.close

thank you for any help!

Answer 1

Assuming you are storing all the intermediate arrays for a reason.

A={}
for v,x in zip((b,c,d,e,f),'bcdef'):
 fname = 'a'+x
 A[fname] = (array.array('B', map( operator.xor, a, v ) ))
 f1 = open(fname, 'wb')
 A[fname].tofile(f1)
 f1.close

Or something like this should work too

A={}
for x in 'bcdef':
 fname = 'a'+x
 A[fname] = (array.array('B', map(a.__xor__, vars()[x] ) ))
 f1 = open(fname, 'wb')
 A[fname].tofile(f1)
 f1.close

• Mark Dickinson: Doesn't this just compute map(operator.xor, a, b) repeatedly?
• John La Rooy: @Mark Dickinson, yes it was. I fixed it now :)

Answer 2

One way is to have a,b,c,d,e,f in a dict. Then you'd just do something like:

for x in 'bcdef':
 t = array.array('B', map( operator.xor, mydict['a'], mydict[x] ) )
 f1 = open(''.join('a',x),'wb')
 t.tofile(f1)
 f1.close()

• Ian Clelland: They are in a dict -- it's called locals() :)
• Justin Peel: @Ian Yes, but some people don't like to use that because it is kind of a hack.
• Justin Peel: The way I was doing it was to have mydict['a'] return the array a. You can use locals() by just replacing mydict with locals().