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<p><font size="4">Hi Ivan:</font></p>
<p><font size="4"><br>
</font></p>
<div class="moz-cite-prefix">On 29/12/2023 20:34, Ivan Marchesini
via grass-user wrote:<br>
</div>
<blockquote type="cite"
cite="mid:3daa1189-18b5-40a7-8d69-671837776f2f@gmail.com">
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<p>Hi veronica<br>
</p>
<blockquote type="cite"
cite="mid:CAAMki4EmFest3Czhp+AUHN17Kw-QXUMPftUDQUkHnWGGuykHDQ@mail.gmail.com">
<div dir="ltr">
<div><br>
</div>
<div>I see what you did with creating the days time series. In
that way you acknowledge irregular gaps, right? </div>
</div>
</blockquote>
yes this is the reason<br>
<blockquote type="cite"
cite="mid:CAAMki4EmFest3Czhp+AUHN17Kw-QXUMPftUDQUkHnWGGuykHDQ@mail.gmail.com">
<div dir="ltr">
<div><br>
</div>
<div>However, why do you multiply by days strds? From my
understanding, detrending by subtracting the results of a
model obbeys this rule: value(t) = observed(t) -
predicted(t). Then, this
mystrds-(regression_offset+regression_slope*days) should be
mystrds-(regression_offset+regression_slope*mystrds).</div>
</div>
</blockquote>
<p>hmm <br>
</p>
<p>may be I'm wrong but I used r.regression.series to assess a
relationship between mystrds values and time (days)<br>
</p>
<p>As a consequence the offset and slope maps I obtain are b and a
in the following linear equation</p>
<p>y=ax+b</p>
<p>i.e.<br>
</p>
<p>predicted_mystrds=slope*days+offset</p>
<p>This is why I suppose that for detrending I need to do:<br>
</p>
<p>mystrds-(predicted_mystrds)</p>
<p>i.e.<br>
</p>
<p>mystrds-(slope*days+offset)</p>
<p><br>
</p>
</blockquote>
<p><br>
</p>
<p>I'm not sure about your procedure of regressing days from start
of the time series vs the time series. But I would point out: </p>
<p>The "elephant in the room" for any time series analysis is
temporal autocorrelation. If you have some seasonal variation in
your data, then I don't think that such a regression could work.
You mentioned data from Nov-Feb over several years, so I assume
there is some seasonal effect going on? Normally you have to
first decompose the time series to remove the seasonal variation,
and then examine the trend.</p>
<p><br>
</p>
<p><br>
</p>
<blockquote type="cite"
cite="mid:3daa1189-18b5-40a7-8d69-671837776f2f@gmail.com">
<p> </p>
<p>I'm I wrong?</p>
<p>thank you</p>
<p>Ivan<br>
</p>
<p><br>
</p>
<p><br>
</p>
<blockquote type="cite"
cite="mid:CAAMki4EmFest3Czhp+AUHN17Kw-QXUMPftUDQUkHnWGGuykHDQ@mail.gmail.com">
<div dir="ltr">
<div><br>
</div>
<div>Best,</div>
<div>Vero</div>
</div>
<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">El jue, 28 dic 2023 a las
8:14, Ivan Marchesini (<<a
href="mailto:ivan.marchesini@gmail.com"
moz-do-not-send="true" class="moz-txt-link-freetext">ivan.marchesini@gmail.com</a>>)
escribió:<br>
</div>
<blockquote class="gmail_quote"
style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div>
<p>Dear Veronica</p>
<p>I think I found a simple solution using temporal raster
modules. Here is an example:</p>
<p>#evaluating info of the strds<br>
eval `<a href="http://t.info" target="_blank"
moz-do-not-send="true">t.info</a> mystrds -g`</p>
<p>#getting the starting day (of the year, 0-365) of my
strds<br>
startday=$(date -d "$start_time" "+%j")<br>
</p>
<p>#Creating a new strds where each pixel has the value of
the count of the days starting from the start_day of my
strds (the start day in my dataset is in 2016)<br>
t.rast.mapcalc inputs=mystrds
expression="(start_year()-2016)*365-${startday}
+start_doy()" output=days basename=days nprocs=xxx --o<br>
</p>
<p>#fitting the trend equation<br>
r.regression.series xseries="`t.rast.list in=days
columns=name sep=, format=line`" yseries="` t.rast.list
in=mystrds columns=name sep=, format=line`"
out=regression_offset,regression_slope,regression_rsq,regression_t
meth=offset,slope,rsq,t<br>
</p>
<p>#detrending<br>
t.rast.mapcalc input=mystrds,days
expression="mystrds_detrend
= mystrds-(regression_offset+regression_slope*days)"
output=mystrds_detrend basename=mystrds_detrend
nprocs=xxx method=start --o<br>
</p>
<p><br>
</p>
<p>Best</p>
<p>Ivan<br>
</p>
<p><br>
</p>
<p><br>
</p>
<p><br>
</p>
<div>On 23/12/23 14:53, Ivan Marchesini wrote:<br>
</div>
<blockquote type="cite">
<p>Hi Veronica</p>
<p>Thank you. It goes in the direction of my idea evn
if my problem is exactly trying to take into account
the correct gaps between that data <br>
</p>
<p>I have another idea.</p>
<p>if it works I will come back here to explain how I
did</p>
<p>thank you again</p>
<p>Ivan<br>
</p>
<p><br>
</p>
<div>On 22/12/23 13:45, Veronica Andreo wrote:<br>
</div>
<blockquote type="cite">
<div dir="ltr">Hello Ivan,
<div><br>
</div>
<div>AFAIU you could use the slope and offset maps
from t.rast.series within t.rast.algebra to
detrend the values of the maps within the strds,
something like "detrended_strds = trend_strds -
(trend_strds*map(slope) + map(offset))". Others
suggest, to detrend by subtracting the previous
value, i.e. that would imply using the temporal
algebra with the temporal index, something like
"detrended_strds = trend_strds[1] -
trend_strds[0]". </div>
<div><br>
</div>
<div>I haven't tested any of these, just a couple of
ideas ;-) However, I do not know how this might
interact with seasonality within data, or
irregular gaps. </div>
<div><br>
</div>
<div>hth somehow</div>
<div>Vero</div>
</div>
<br>
<div class="gmail_quote">
<div dir="ltr" class="gmail_attr">El vie, 22 dic
2023 a las 5:10, Ivan Marchesini via grass-user
(<<a href="mailto:grass-user@lists.osgeo.org"
target="_blank" moz-do-not-send="true"
class="moz-txt-link-freetext">grass-user@lists.osgeo.org</a>>)
escribió:<br>
</div>
<blockquote class="gmail_quote"
style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">Dear
colleagues<br>
<br>
I would like to the advantage of the t.* modules
for detrending a strd.<br>
<br>
In the strd I have earth observation data
irregularly sampled (2 or 3 <br>
times per month), in the period November-February,
for 7 years. They are <br>
not equally spaced (i.e gaps have different
duration)<br>
<br>
A simple t.rast.series analysis
(opion=slope,offset) highlights that <br>
probably there is a descending trend when
considering the maps ordered <br>
by id.<br>
<br>
I would like to fit a proper time depending
fitting curve for each pixel <br>
and then subtract the function from the real data.<br>
<br>
any hints on how I can do this task exploiting the
GRASS GIS modules or <br>
some simple bash/python scripting?<br>
<br>
thank you<br>
<br>
Ivan<br>
<br>
<br>
<br>
<br>
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</blockquote>
</div>
</blockquote>
</blockquote>
</div>
</blockquote>
</div>
</blockquote>
<br>
<fieldset class="moz-mime-attachment-header"></fieldset>
<pre class="moz-quote-pre" wrap="">_______________________________________________
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</blockquote>
<pre class="moz-signature" cols="72">--
Micha Silver
Ben Gurion Univ.
Sde Boker, Remote Sensing Lab
cell: +972-523-665918</pre>
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