<div dir="ltr">Yes ... 12 values is not nearly enough to fit a distribution. As noted in the documentation for the function:<br><br><a href="http://www.ncl.ucar.edu/Document/Functions/Built-in/dim_spi_n.shtml">http://www.ncl.ucar.edu/Document/Functions/Built-in/dim_spi_n.shtml</a><br>
<div><em>x</em>
<p class=""><i>Monthly precipitation</i> of type 'float' or 'double' and any dimensionality.
The size of the specified <em>dims</em> must be divisible by 12.
Since a distribution is being fit, there should be a 'reasonably' large sample size.</p><p class="">
<b>At least 30 years of monthly data (360=12*30) is recommended.</b></p><p class=""><b>^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^<br></b>
</p><br></div></div><div class="gmail_extra"><br><br><div class="gmail_quote">On Fri, Aug 1, 2014 at 5:14 PM, debasish mazumder <span dir="ltr"><<a href="mailto:debasish@ucar.edu" target="_blank">debasish@ucar.edu</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div bgcolor="#FFFFFF" text="#000000">
Hi All,<br>
I am trying to use function <strong>dim_spi_n.<br>
</strong>My precipitation values varies between 0 - 334 mm/month
and precipitation time series is very small (12 months).<br>
I am getting strange values for SPI3. I am wondering is it because
of length of my time series or am I missing something?<br>
<br>
with regards<span class="HOEnZb"><font color="#888888"><br>
-Debasish<br>
<br>
<br>
</font></span></div>
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