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> if c [i] = median (close,60)
That won't work because you're trying to compare a number with an
array of numbers, plus you need the relational equals operator (==)
rather than the assignment one (=). Read the median array into a
variable first and then use indexing on that:
medc = median(close,60);
if (c[i] == medc[i])
However, without seeing the rest of your code, I have a feeling you
still won't end up with what you're expecting. This is comparing the
current bar's close with the median of the previous 60 bars (counting
the current one), so if you only count those, it will give you a count
of how many bars had a close equal to the median of the last 60 bars.
If at each bar you want a count of how many bars in the previous 60
were the same as the median over that range, then the calculation will
be different, probably involving nested loops.
GP
--- In amibroker@xxxxxxxxxxxxxxx, "meenhal" <meenhal@xxx> wrote:
>
> To set an entry point, I am trying to find price deviations from mean
> and median for different intervals.
>
> All is easy except: I can not determine the frequecy of median i.e.
> the number of occurences when the price was median during the inteval.
>
> median (close,260) will give me the median closing price for 260 bars
> but it does not tell whether this prices occured 10 tims or 88 times.
>
> I tried to use the for loop with if c [i] = median (close,60) but
> AFL's handling of arrays does not permit such an attepmt.
>
> Can anyone please help. Thanks in advance
>
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