Hi,
I am not sure whether I get you right, but what you are looking for are
those indices where both lists show the same value, right? I guess,
with the line
WHERE(list1-list2(i) EQ min(list1-list2(i))
you try to find those indices where list1 is closest to list2[i]. If
so, you should use
min(abs(list1-list2[i])) instead, as otherwise the minimum value is
where the largest negative difference occurs, not the one closest to
zero.
Apart from that, I'd strongly suggest to follow the suggestion to use
square brackets [] for indexing and round brackets () for function
calls. It makes the code better readable and helps avoiding confusion
if you happen to name a variable after a function.
As for the error message, Ben's comment is perfectly right, generally
WHERE returns an vector, so you have to check its first value against
-1, which luckily does not throw an error if the rreturn value *is* -1,
as you can index scalars with [0]. On the other hand, if you do
something like
index = WHERE(this EQ that, n)
then "n" gives the number of matches, so you could continue with a line
like
IF n gt 0 THEN do_the_rest
And now for your original problem. If you are dealing with integer
values, I'd suggest using histogram instead of the for loop. Something
like this:
Just a few example numbers:
list1 = [0,0,1,1,2,2,3,4,5,5,5]
list2 = [0,1,2,3,4]
We'll use histogram, so make sure to use the same min and max values in
both calls:
min_value = min(list1) < min(list2)
max_value = max(list1) > max(list2)
h1 = histogram(list1, min = min_value, max = max_value, reverse = r1)
h2 = histogram(list2, min = min_value, max = max_value, reverse = r2)
Now, the key is the reverse index, which is just plain confusing when
you first look at it, but it just basically tells you the indices where
the list values do fall in each histogram bin.
If you want to see where list1 and list2 equal, say, 1, you can use
print, list1[r1[r1[1]:r1[2]-1]]
print, list2[r2[r2[1]:r2[2]-1]]
(just printing a lot of "1"s)
which means that
r1[r1[1]:r1[2]-1] gives you the indices where list1 equals 1, while
r2[r2[1]:r2[2]-1] gives the appropriate indices for list2.
Well, you are paying with weird looking code, but you'll gain a lot of
processing speed.
If this approach seems to work for you, then take the time to read JD
Smith's histogram tutorial at
http://www.dfanning.com/tips/histogram_tutorial.html
Cheers,
Peter
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