Still some other things wrong with my code, but I think I figured out my
way around this issue. Thanks for the heads ups, John and Leah.
This is what I did:
function spikecounter(Varray)
Varraycut = splice!(Varray,1:spliceend)
push!(Varraycut,Varraycut[length(Varraycut)])
spikecount = 0
for i in 1:(length(Varraycut)-1)
if Varraycut[i] < 0 && Varraycut[i+1] >= 0
spikecount += 1
end
end
return spikecount
end
I added a duplicate entry at the end to pad the array and make the [i-1] or
[i+1] possible.
On Wednesday, July 30, 2014 6:32:26 PM UTC-4, John Myles White wrote:
>
> This looks like you’re doing implicit if-clauses, which aren’t currently
> possible with Julia’s list comprehensions.
>
> — John
>
> On Jul 30, 2014, at 3:30 PM, [email protected] <javascript:> wrote:
>
> So in Python, I would do something like this:
>
> count = sum([a<0 and b>=0 for a,b in zip(countlist,countlist[1:])])
>
> I am having a bit of trouble Googling for this, since I do not know the
> right keywords to use, but is there a means of doing something similar with
> a list comprehension in Julia?
>
> On Wednesday, July 30, 2014 6:17:18 PM UTC-4, Leah Hanson wrote:
>>
>> Your problem is with the first index (`i == 1`). You can't check if the
>> previous element is < 0. You could adjust your range (the
>> `1:length(outputarray)`) to only run through elements for which your
>> if-condition makes sense.
>>
>> -- Leah
>>
>>
>> On Wed, Jul 30, 2014 at 5:12 PM, <[email protected]> wrote:
>>
>>> 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, [email protected] 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, [email protected] 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
>>>>>
>>>>>
>>>>>
>>>>
>>
>
