# Re: [ferret_users] event statistics -- length of contiguous "spell"

Hi Russ,

Thank you for suggetions.

I agree that your 1st suggestion makes the logic generally valid
(though with the particular data I am working with it starts from 1st Jan (which was missing but I filled all missing with value=0) and by defintion of active and break spell they can occur only during Jul and Aug so by design of the dataset 1st time step is always 0)

I tried to implement rest of your suggestions but got following error:
yes? def axis/x=1:`id_good[l=@max]`:1 xax
yes? let id_good_2d = if id_good eq i then id_good + 0*i[gx=xax]
*** NOTE: Ambiguous coordinates on X axis: IF ID_GOOD EQ I THEN ID_GOOD + 0*I[GX=XAX]
*** NOTE: Ambiguous coordinates on X axis: IF ID_GOOD EQ I THEN ID_GOOD + 0*I[GX=XAX]
**ERROR: illegal limits: I on grid (G008) is not in the range X=66.3:98.8
Axis extremes are X=0.5:58.5

What could be the problem?

Thanks,

On Thursday, May 15, 2014 2:38 AM, Russ Fiedler <russell.fiedler@xxxxxxxx> wrote:

I think you also need to consider the case when the first entry of your time series is 1 (i.e. coincides with the start of a spell). In that case your solution (2) would not catch the first event and even numbers would correspond to wet spells rather than dry.

You could either pad the series at the start with a dry day or maybe add one if necessary using the first value in your range of interest.

Something like

def sym first_day 183  ! this is a wet day in your series below
let    ID_GOOD = VAR1[L=@EVNT:0.999] + `VAR1[l=(first_day)]`

list/l=(\$first_day):285 ...

To get the number of days in a spell you could spread the results along a new axis and then count the number of good values.

def axis/x=1:`id_good[l=@max]`:1 xax
let id_good_2d = if id_good eq i then id_good + 0*i[gx=xax] ! Put spell "i" on its own axis.

shade id_good_2d  ! we can see each spell wet or dry. This can be masked using mod(var,2) eq 0 or 1

let length_of_spell = id_good_2d[l=@ngd]

plot/title="Spell length"   length_of_spell

plot/title="Wet spell length"/sym  if mod(id_good,2) eq 1 then length_of_spell

The stats should be easy after this.

Cheers,
Russ

On 15/05/14 05:22, 'jagadish karmacharya' via _OAR PMEL Ferret Users wrote:
Dear steve and Hein,

Thanks for quickly sharing your thoughts.

I have not really gone though all the details you suggestioned but I am glad to report that I have found a solution (in fact 2 different solutions).

First a clearification - What I mean by the spells was: active and break spell has been defined as days when rainfall is above or below a certain threshold provided it last for atleast 3 consequtive days. Previous taking help from the ferret forum I was able implement the criteria such that the result timeseries has value = 1 for active spell (also for break spell but in seperate variable) and missing for other period. That has worked fine as I was then only interested in the total duration of such spell and other composite analysis using those. So my original data and variables looks like following:

yes? sh d
currently SET data sets:
1> ../results/12kmRuns/IMD_active_break_JJASonly.90_05.nc  (default)
name     title                             I         J         K         L         M         N
ATV_JA   ATV                              ...       ...       ...       1:5844    ...       ...
(X=66.3E:98.8E, Y=6.3N:38.8N)
BRK_JA   BRK                              ...       ...       ...       1:5844    ...       ...
(X=66.3E:98.8E, Y=6.3N:38.8N)

yes? list atv_ja[l=@ngd],brk_ja[l=@ngd]
(deleted lines)
ATV_JA  BRK_JA
I / *:     108.0  125.0

Solution 1)
Hein email's yesterday indicating it would be difficult to implement (what I have request for) in ferret prompted me to seek other options which lead to 1st solution in which I avoided the proble mby creating new files for active (and break) period wherein values > 1 during active period and zero for the rest. Next I used:
let ID_atv=atv[l=@evnt:1]
to "tag" the events where successive odd id's corresponds to succesive active spells (& succesive even id's corresponds to succesive gaps between the spells).

Now from that I can easly find duration of each spell. (I then imported those spell lengh in spreadsheet to compute mean, median, max of the spell duration etc, but it should be possible to do it in ferret itself with little more play)

Solution 2)
But quickly going through steve's mail a while ago I though it might be posible to use similar trick:
so what I did is - I filled the missing values in the original data (above) with zero. Then defined the event as before with slight twist:
let id1=var1[l=@evnt:0.999]        ! here any value between 0 and 1 would work
which gave desired result!

i.e.
let VAR1 = IF ATV_JA THEN ATV_JA ELSE 0
let ID_GOOD = VAR1[L=@EVNT:0.999]
let ID_BAD = ATV_JA[L=@EVNT:1]        ! just for illustration, this doesn't work, using value between 0 & 1 also doesn't give desired result

(deleted lines)
30-JUN-1990 00 / 181:   ....   0.00  0.000   0.000
01-JUL-1990 00 / 182:   ....   0.00  0.000   0.000
02-JUL-1990 00 / 183:  1.000   1.00  1.000   1.000
03-JUL-1990 00 / 184:  1.000   2.00  1.000   1.000
04-JUL-1990 00 / 185:  1.000   3.00  1.000   1.000
05-JUL-1990 00 / 186:  1.000   4.00  1.000   1.000
06-JUL-1990 00 / 187:   ....   4.00  0.000   2.000
07-JUL-1990 00 / 188:   ....   4.00  0.000   2.000
(deleted lines)
19-AUG-1990 00 / 231:   ....   4.00  0.000   2.000
20-AUG-1990 00 / 232:   ....   4.00  0.000   2.000
21-AUG-1990 00 / 233:   ....   4.00  0.000   2.000
22-AUG-1990 00 / 234:  1.000   5.00  1.000   3.000
23-AUG-1990 00 / 235:  1.000   6.00  1.000   3.000
24-AUG-1990 00 / 236:  1.000   7.00  1.000   3.000
25-AUG-1990 00 / 237:  1.000   8.00  1.000   3.000
26-AUG-1990 00 / 238:   ....   8.00  0.000   4.000
27-AUG-1990 00 / 239:   ....   8.00  0.000   4.000
28-AUG-1990 00 / 240:   ....   8.00  0.000   4.000
29-AUG-1990 00 / 241:  1.000   9.00  1.000   5.000
30-AUG-1990 00 / 242:  1.000  10.00  1.000   5.000
31-AUG-1990 00 / 243:  1.000  11.00  1.000   5.000
01-SEP-1990 00 / 244:   ....  11.00  0.000   6.000
02-SEP-1990 00 / 245:   ....  11.00  0.000   6.000

I would like to thank your both for sharing your ideas that help me come up with these solutions.

Best regards,