I hope extolling the beauty and power of Python on this list is allowed, because I have had a large "WOW!!!" moment tonight. I had a problem I was working on at work this afternoon. I have a list of ~ 10,000 floating point numbers, which run from largest to smallest. There are duplicates scattered throughout, so I might have something like: [5942.6789, 5942.6789, 5941.000003, 5941.01, 5941.01, ... ], etc. I wanted to search the list for a test value, which, depending on the particular list (I can have many such lists, each different from the other.), could conceivably be anywhere within the given list. I needed to return the index where the list values change from being just greater than the test value to just less than the test value at the very next index position. I spent a good chunk of my afternoon writing a binary search function and wondering what theoretically the optimum search algorithm would be, got interrupted (as usual on this project), and decided to look at my books at home to see if a better solution would be staring at me from some book (Like there usually is!).
I haven't studied list comprehensions formally yet, but a snippet of code in a book caught my eye where the author was discussing filtering data in a list. This led me to try: The generalized problem: L = [V0, V1, ..., Vn], where V0 >= V1 >= V2 >= ... >= Vn . Find index i, such that V[i] >= Vt >= V[i + 1], where Vt is the test value being searched for. I need to know the indices i and i + 1, which I need to interpolate based on where Vt falls. The solution (As a sublist, S) I worked out tonight after experimenting with comprehension syntax is: S = [i for i, V in enumerate(L) if L[i] >= Vt >= L[i + 1]] And, of course, the index i I need is: i = S[0] I tested this out with concrete examples in the interpreter, such as with a list, L: L = [item for item in range(10000, 0, -1)] and trying different test values. It was blazingly fast, too! All I can say is: WOW!!! -- boB _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
