I'm a bit stuck on this one. Could I get one more hint about a way I could get the same thing done without using the illegal indexing?
On Wednesday, July 30, 2014 5:46:58 PM UTC-4, John Myles White wrote: > > Yeah, it’s the combination of (a) the use of i and i+1 indexing with (b) > the use of a loop that goes from i = 1 to i = length(outputarray). > > — John > > On Jul 30, 2014, at 2:44 PM, yaois...@gmail.com <javascript:> wrote: > > To correct the bug, is it this? > > if outputarray[i] < 0 && outputarray[i+1] >= 0 > count += 1 > end > > Factoring in that Julia begins indexing from 1. > > On Wednesday, July 30, 2014 4:41:43 PM UTC-4, John Myles White wrote: >> >> This pseudocode almost works. Just replace Int64[1:len(outputarray)] with >> 1:length(outputarray). >> >> There’s also a bug in your core logic, but I’ll leave fixing that as an >> exercise to the reader. >> >> — John >> >> On Jul 30, 2014, at 1:03 PM, yaois...@gmail.com wrote: >> >> Hi guys, >> >> I asked this in a previous thread, but because that diverged off-topic >> from my existing question, I decided to create a new thread. >> >> Anyhow, say I have an array >> >> outputarray = >> Float64[-1.23423,-3.23423,-2.34234,-2.12342,1.23234,2.23423,-2.23432,5.2341,0.01111,1.23423] >> >> This array lists the output of some function. I want to count the number >> of times that the function passes by or equals 0 while emerging from a >> negative f(x). >> >> In pseudocode, I want to do: >> >> function counter(outputarray) >> count = 0 >> for i in Int64[1:len(outputarray)] >> if outputarray[i] >= 0 && outputarray[i-1] < 0 >> count += 1 >> end >> end >> return count >> end >> >> What would be the most efficient way of doing this in Julia? >> >> Thanks, >> Wally >> >> >> >
