# Re: trend problem

Date: Tue Aug 02 2011 - 16:52:01 MDT

Hi Shinn,
Your coding is somewhat correct. There is no need to go through a double
do loop nor do you need to use the y-intercept. Your coding could be
condensed to:
trend = data(:,:,0)
temp = dtrend_msg(ispan(0,29,1),data,False,True)
trend = (/ ndtooned(temp@slope,(/dimsizes(data&lat),dimsizes(data&lon))*30/)

To calculate the significance of the trend, you should use the betainc
function:
http://www.ncl.ucar.edu/Document/Functions/Built-in/betainc.shtml

continuing from the coding above:
tval =new((/dimsizes(data&lat),dimsizes(data&lon/),typeof(data))
beta_b = tval
df = new((/dimsizes(data&lat),dimsizes(data&lon/),"integer")
rc = regcoef(ispan(0,29,1),data,tval,df)
df = df - 2 ; regcoef returns N, need N-2
beta_b = 0.5
signif = trend
signif = (/ 1-betainc(df/(df+tval^2),df/2.0,beta_b))*100. /)

Note that in the above coding regcoef sets the degrees of freedom to
simply the number of timesteps. If your data is highly auto-correlated,
then it is suggested that you use equiv_sample_size to compute the degrees
of freedom:
http://www.ncl.ucar.edu/Document/Functions/Built-in/equiv_sample_size.shtml

> Dear all,
> I would like to get the 30-year linear trend of my data (lat, lon, time)
> in
> each grid box. I have tried to write a script but didn't know whether it
> is
> correct. Could anyone help me to check with that?Also, I would like to
> know
> how can I test the significance of the trend using ncl. Thank in advance
> for
> any help given regarding on that.
> Shinn
>
> trend = new((/dimsizes(data&lat), dimsizes(data&lon)/), "float")
> trend!0 = "lat"
> trend&lat = data&lat
> trend!1 = "lon"
> trend&lon = data&lon
> trend&lat@units = "degrees north"
> trend&lon@units = "degrees east"
>
> do i =0, dimsizes(data&lat)-1
> do j = 0, dimsizes(data&lon)-1
> temp = dtrend_msg(ispan(0,29,1),data(i,j,:),False,True)
> trend(i,j) = (29*(temp@slope) + temp@y_intercept) ; is the trend
> here
> represent the 30-year linear trend?
> end do
> end do
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