On Jul 11, 2:05 pm, d.po...@gmail.com wrote:
> Hi everyone
> I have got a problem and I don’t know how to solve it:
> I have two variables like this:
> A, which col*row=2*4
> B, which col*row=4*100.
> All A’s couples (x,y)’s are somewhere in the 2th and 3th column of
> Result. I want to take this rows (I want to extract that rows in
> result which 2th and 3th columns are settled in the B’s ).for
> example:
> A=[[1,2], [3,4] ,[5,6], [7,8]]
>
> B=[….[11,22,1,2,],….. [44,55,3,4]….. ,[22,99,5,6], ….[77,66,7,8]…]
>
> …..=means there are some other data
> I just want to extract these rows and put them in a new variable
> Any help?
> Cheers
There are two problems. One is that you have two variables you have to
compare and one is that you have a whole vector of values they can be.
Probably you are familiar with the Where-function
idx = Where(array <condition considering array as a scalar>)
idx will be a vector of indices where the array has values that meet
the condition. If there are none, idx will be -1.
The problem that you have to check both the 2nd and the 3rd column
will force constructions like
idx1 = Where(vector1 eq value1)
idx2 = Where(vector2 eq value2)
if idx1[0] eq -1 or idx2[0] eq -1 then screw it
...
It is better to create another vector. As you work with integers, you
compress two integers into one.
totalB = B[2,*] + B[3,*]
DB = (total*(total+1))/2 + B[2,*] ; The value in DB[j] determines
exactly what B[2,j] and B[3,j] are
totalA = A[0,*] + A[1,*]
DA = (total*(total+1))/2 + A[0,*] ; Same joke, so if DA[i] eq DB[j]
then A[0,i] eq B[2,j] and A[1,i] eq B[3,j]
And now we are left with two vectors. As long as DA only contains four
values (like your example), you can use a forloop
for j=0,4-1 do begin
idx = Where(DB eq DA[j])
if idx[0] eq -1 then continue ; Subscripting [0] is required if idx
can have more than one value.
if N_Elements(idxs) eq 0 then idxs = idx else idxx = [idxx,idx] ;
Add the index to the result vector.
end
result = B[*,idxs]
There must be a more efficient way to do the last part, but Where(DB
eq DA) does not work.
You can also do it at the array way, create one array like. Watch out
where to use Transposes (Check it out first)
MA =
DA[0] DA[1] DA[2] DA[3]
DA[0] DA[1] DA[2] DA[3]
DA[0] DA[1] DA[2] DA[3]
DA[0] DA[1] DA[2] DA[3]
DA[0] DA[1] DA[2] DA[3]
...
and one like
MB =
DB[0] DB[0] DB[0] DB[0]
DB[1] DB[1] DB[1] DB[1]
DB[2] DB[2] DB[2] DB[2]
DB[3] DB[3] DB[3] DB[3]
DB[4] DB[4] DB[4] DB[4]
...
MC = LonArr(4,100) ; Comparison
idx = Where(MB eq MA)
MC[idx] = 1
IDL> print,MC
0 0 0 0 ; Drop this B-index
0 1 0 0 ; There is an A-index that fits
0 0 0 0
0 0 1` 0
1 0 0 0
...
idxs = total(MC,1)
IDL> print,idxs
0,1,0,1,1,...
result = B[*,idxs]
; This is more work, but probably faster than a for-loop if we have
more than 4 rows in A.
I hope this helps you. Probably there are better ways, but this should
work.
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