On Mon, 12 Apr 2010 13:05:30 -0700, vsoler wrote:
> Hi everyone,
>
> say that 'db' is a list of values
> say 'i' is a list of indexes
> I'd like to get a list where each item is i-th element of db.
>
> For example:
>
> db=[10,20,30,40,50,60,70,80,90] #undefined length
> i=[3,5,7] #undefined length then
> result=[30,50,70] # resulting list
>
> how can I do this?
Others have given a simple solution using list comprehensions. Here's
another few ways.
# The beginner's way
result = []
for index in i: # "i" is a bad name for a list of indexes
x = db[index - 1] # shift from one-based counts to zero-based
result.append(x)
# The advanced functional way
from operator import itemgetter
indexes = map(lambda n: n-1, i) # adjust for one-based counts
result = itemgetter(*indexes)(db)
# The obfuscated, fragile way
map( itemgetter(0), sorted(
zip(db, range(1, len(db)+1)), key=lambda t: t[1] if t[1] in i else -1
)[-len(i):] )
--
Steven
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