Re: any good way to count no. of specified events

Hi Dave and Dennis:

Thank you very much! Dave's solution is great and Dennis's generalization
makes it even more useful. I guess, I will be able to use ind_resolve
to make it applicable to an multi-dimensional input.

saji

--
* Dennis Shea <shea_at_ucar.edu> [2009-06-04 18:09:58 -0600]:
> Hi Saji
>
> A generalization of Dave Allured's code
>
> Good luck
> D
>
> Saji N. Hameed wrote:
>> Dear NCL-ers,
>>
>>   Suppose I have a 1D array of binaries (/1,1,1,0,0,0,1,0,0,1,1,1,1/)
>>   and I want to calculate number of 3 or more sucessive 1's
>>   (in this example, there are two such events)
>>
>>   is there a good way to do that? FYI, I am trying to make an
>>   index of number of warm spells in a given period. I can write   an 
>> algorithm with some do loops, but was wondering if there is
>>   a simpler way?
>>
>> saji
>>
>> ps: in ruby, i could have done perhaps this way
>>    irb> a=[1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1]
>>    irb> b=a.join.split(/[0]+/) # array to string, split string at multiple 0's
>>      => ["111", "1", "1111"]
>>    irb> b.map! {|num|  (num.to_i>=111)? num=1 : num=nil}
>>      => [1, nil, 1]
>>    irb> b.nitems
>>      => 2
>>
>>
> function numBinOneRuns ( a[*]:integer, nCrit:integer)
> 
> ; Dave Allured: CU/CIRES Climate Diagnostics Center (CDC) 
> ; Return the number of runs [sequences] of 1s
> 
> local inds, ni, deltas, starts, ends, lengths
> begin
>    dim_nCrit  = dimsizes(nCrit)
>    if (dim_nCrit.gt.2) then
>        print("numBinOneRuns: nCrit size must be .le. 2: nCrit="+dim_nCrit)
>        return(-999)
>    end if
> 
>    inds    = ind (a .eq. 1)
>    ni      = dimsizes (inds)
>    deltas  = new (ni+1, integer)
>    deltas  = 0
>    deltas(1:ni-1) = (inds(1:ni-1) - inds(0:ni-2))
>    starts  = inds(ind (deltas(0:ni-1).ne.1 .and. deltas(1:ni).eq.1))
>    ends    = inds(ind (deltas(0:ni-1).eq.1 .and. deltas(1:ni).ne.1))
>    lengths = ends - starts + 1
> 
>    if (dim_nCrit.eq.1) then
>        return(num (lengths.ge.nCrit) )
>    else
>        return(num (lengths.ge.nCrit(0) .and. lengths.le.nCrit(1)) )
>    end if
> end
> 
>    a = (/ 1,1,1,0,0,0,1,0,0,1,1,1,1 /)
> 
>    ncrit = 1
>    nqual = numBinOneRuns ( a, ncrit)
>    print("nqual="+nqual+"    (ncrit="+ncrit+")")
> 
>    ncrit = 3
>    nqual = numBinOneRuns ( a, ncrit)
>    print("nqual="+nqual+"    (ncrit="+ncrit+")")
> 
>    ncrit = 4
>    nqual = numBinOneRuns ( a, ncrit)
>    print("nqual="+nqual+"    (ncrit="+ncrit+")")
> 
>    print(" ")
>    print("============================================")
>    print(" ")
> 
>    x = (/1,1,1,0,0,0,1,0,0,1,1,1,1,0,0,1,1,0,1,1,1,1,1,1,1 \
>         ,0,1,1,0,1,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1/)
>    do ncrit=1,10
>       nqual = numBinOneRuns ( x, ncrit)
>       print("nqual="+nqual+"    (ncrit="+ncrit+")")
>    end do
> 
>    print(" ")
>    print("============================================")
>    print(" ")
> 
>    kcrit = (/2, 5/)      ; number of runs between 2 an 5 inclusive
>    kqual = numBinOneRuns ( x, kcrit)
>    print("kqual="+kqual+"    (kcrit=["+kcrit(0)+","+kcrit(1)+"] )")
> 
>    print(" ")
>    print("============================================")
>    print(" ")
> 
>    kcrit = (/3, 3/)      ; number of runs of length 3
>    kqual = numBinOneRuns ( x, kcrit)
>    print("kqual="+kqual+"    (kcrit=["+kcrit(0)+","+kcrit(1)+"] )")
> 
>    kcrit = (/8, 8/)      ; number of runs of length 8
>    kqual = numBinOneRuns ( x, kcrit)
>    print("kqual="+kqual+"    (kcrit=["+kcrit(0)+","+kcrit(1)+"] )")
-- 
Saji N. Hameed
APEC Climate Center          				
1463 U-dong, Haeundae-gu,                               +82 51 745 3951
BUSAN 612-020, KOREA                    		saji_at_apcc21.net
Fax: +82-51-745-3999
_______________________________________________
ncl-talk mailing list
List instructions, subscriber options, unsubscribe:
http://mailman.ucar.edu/mailman/listinfo/ncl-talk