Thank you Olivier and Robert for your replies!
Some remarks about the dictionnary solution:
from numpy import *
def f(arr): return arr + 100.
arrs = {} arrs['a'] = array( [1,1,1] ) arrs['b'] = array( [2,2,2] ) arrs['c'] = array( [3,3,3] ) arrs['d'] = array( [4,4,4] )
for key,value in arrs.iteritems(): arrs[key] = f(value)
1. about the memory Memory is first allocated with the array functions: arrs['a'] = array( [1,1,1] ) arrs['b'] = array( [2,2,2] ) arrs['c'] = array( [3,3,3] ) arrs['d'] = array( [4,4,4] )
Are there others memory allocations with this assignemnt: arrs[key] = f(value) or is the already allocated memory used to store the result of f(value)?
In other words, if I have N arrays of the same shape, each of them costing nbytes of memory, does it use N*nbytes memory, or 2*N*bytes?
I think this is well documented on the web and I can find it....
2. about individual array The problem is that now, if one want to use a individual array, one have now to use: arrs['a'] instead of just: a So I'm sometime tempted to use locals() instead of arrs...
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