You'll probably see a slight speed increase with something like
for a in CustomersToMatch:
for b in Customers:
if a[2] == b[2]:
a[1] = b[1]
break
But a really fast approach is to use a dictionary or other structure
that turns the inner loop into a fast lookup, not a slow loop through
the 'Customers' list. Preparing the dictionary would look like
custmap = {}
for c in Customers:
k = c[2]
if k in custmap: continue
custmap[k] = c
and the loop to update would look like
for a in customerstomatch:
try:
a[1] = custmap[a[2]][1]
except KeyError:
continue(all code is untested) In "big-O" terms, I believe this changes the complexity from O(m*n) to O(m+n). Jeff
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