| Re: Compare two variables [message #61276] |
Mon, 14 July 2008 06:00 |
d.poreh
Messages: 406
Registered: October 2007
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Senior Member |
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On 14 Jul., 04:57, "ben.bighair" <ben.bigh...@gmail.com> wrote:
> On Jul 14, 4:09 am, Joost Aan de Brugh <joost...@gmail.com> wrote:
>
>
>
>
>
>> Hello Dave,
>
>>> hi Joost
>>> but i work with float data. as you said this metheod works for
>>> integer. could we modify it to work for non integer data?
>>> Cheers
>>> Dave
>
>> That is a pity. But isn't that dangerous in any case. If A and B are
>> floats then the expression A eq B is not reliable because of
>> continuous rounding. It may still be reliable if absolutely no
>> arithmetic is involved.
>
>> The compression trick does not work for floats, because of the degree
>> of infinity.
>
>> Maybe a two-step filtering is apropriate
>
>> idx1 = Where(B[2,*] = A[0,j]) ; in for-loop or with the matrix-trick I
>> did with DA and DB.
>> inbetweenresult = B[*,idx1]
>
>> idx2 = Where(inbetweenresult[3,*] = A[1,j]) ; in for-loop or with the
>> matrix-trick I did with DA and DB.
>> result = inbetweenresult[*,idx2]
>
> Hi,
>
> It doesn't seem to me that Dave has provided sufficient information.
> I think the question was not fully fleshed out so it is hard to
> provide helpful answers.
>
> For example, is it possible that the coordinates could be temporarily
> coerced into integers with losing unique pairings? If that is that
> case then he can use the method described by Joost. Or, here is
> another tack, is there a certain granularity (or precision) to the
> coordinates - measured to the nearest tenth or hundreth perhaps? If
> that is the case then he could simply promote the coordinates by
> multiplying by 10 (or 100 or whatever) and then convert to integer.
>
> Cheers,
> Ben- Zitierten Text ausblenden -
>
> - Zitierten Text anzeigen -
Ben
Actually my coordinates are float numbers with 2 decimal. I think your
proposed way is good and I can multiply that numbers whit 100 to take
integers and after finishing I can divide them to 100 to take actual
coordinate.
Cheers
Dave
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| Re: Compare two variables [message #61277 is a reply to message #61276] |
Mon, 14 July 2008 04:57  |
ben.bighair
Messages: 221
Registered: April 2007
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Senior Member |
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On Jul 14, 4:09 am, Joost Aan de Brugh <joost...@gmail.com> wrote:
> Hello Dave,
>
>> hi Joost
>> but i work with float data. as you said this metheod works for
>> integer. could we modify it to work for non integer data?
>> Cheers
>> Dave
>
> That is a pity. But isn't that dangerous in any case. If A and B are
> floats then the expression A eq B is not reliable because of
> continuous rounding. It may still be reliable if absolutely no
> arithmetic is involved.
>
> The compression trick does not work for floats, because of the degree
> of infinity.
>
> Maybe a two-step filtering is apropriate
>
> idx1 = Where(B[2,*] = A[0,j]) ; in for-loop or with the matrix-trick I
> did with DA and DB.
> inbetweenresult = B[*,idx1]
>
> idx2 = Where(inbetweenresult[3,*] = A[1,j]) ; in for-loop or with the
> matrix-trick I did with DA and DB.
> result = inbetweenresult[*,idx2]
Hi,
It doesn't seem to me that Dave has provided sufficient information.
I think the question was not fully fleshed out so it is hard to
provide helpful answers.
For example, is it possible that the coordinates could be temporarily
coerced into integers with losing unique pairings? If that is that
case then he can use the method described by Joost. Or, here is
another tack, is there a certain granularity (or precision) to the
coordinates - measured to the nearest tenth or hundreth perhaps? If
that is the case then he could simply promote the coordinates by
multiplying by 10 (or 100 or whatever) and then convert to integer.
Cheers,
Ben
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| Re: Compare two variables [message #61281 is a reply to message #61277] |
Mon, 14 July 2008 01:09 |
Joost Aan de Brugh
Messages: 16
Registered: July 2008
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Junior Member |
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Hello Dave,
> hi Joost
> but i work with float data. as you said this metheod works for
> integer. could we modify it to work for non integer data?
> Cheers
> Dave
That is a pity. But isn't that dangerous in any case. If A and B are
floats then the expression A eq B is not reliable because of
continuous rounding. It may still be reliable if absolutely no
arithmetic is involved.
The compression trick does not work for floats, because of the degree
of infinity.
Maybe a two-step filtering is apropriate
idx1 = Where(B[2,*] = A[0,j]) ; in for-loop or with the matrix-trick I
did with DA and DB.
inbetweenresult = B[*,idx1]
idx2 = Where(inbetweenresult[3,*] = A[1,j]) ; in for-loop or with the
matrix-trick I did with DA and DB.
result = inbetweenresult[*,idx2]
Cheers,
Joost
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| Re: Compare two variables [message #61283 is a reply to message #61281] |
Sat, 12 July 2008 12:54 |
Sven Geier
Messages: 17
Registered: July 2002
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Junior Member |
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d.poreh@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
I'm probably just misunderstanding something, but what is wrong with
line_1_index = where (b[2,*] eq a[0,0] and b[3,*] eq a[1,0])
line_2_index = ....
...
result = [b[*,line_1_index],b[*,line_2_index],b[... ],... ]
It's not terribly elegant but if it's only 5 lines and has to be typed only
once, what's the point of optimizing the hell out of it?
--
http://www.sgeier.net
My real email address does not contain any "Z"s.
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| Re: Compare two variables [message #61288 is a reply to message #61283] |
Sat, 12 July 2008 06:18 |
d.poreh
Messages: 406
Registered: October 2007
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Senior Member |
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> ; 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.
hi Joost
but i work with float data. as you said this metheod works for
integer. could we modify it to work for non integer data?
Cheers
Dave
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| Re: Compare two variables [message #61301 is a reply to message #61288] |
Fri, 11 July 2008 10:14 |
d.poreh
Messages: 406
Registered: October 2007
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Senior Member |
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On Jul 11, 5:49 pm, "Jean H." <jghas...@DELTHIS.ucalgary.ANDTHIS.ca>
wrote:
> 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
>
> If you only have integers, you may want to re-create the cell index....
> either the "real" cell index, or one of your own (like cell 1;2 ==> 12,
> or 0102 or...). Then a simple intersect will do the trick
>
> Jean
no ny data is not integer
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| Re: Compare two variables [message #61305 is a reply to message #61301] |
Fri, 11 July 2008 08:49 |
Jean H.
Messages: 472
Registered: July 2006
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Senior Member |
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d.poreh@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
If you only have integers, you may want to re-create the cell index....
either the "real" cell index, or one of your own (like cell 1;2 ==> 12,
or 0102 or...). Then a simple intersect will do the trick
Jean
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| Re: Compare two variables [message #61306 is a reply to message #61305] |
Fri, 11 July 2008 08:41 |
Joost Aan de Brugh
Messages: 16
Registered: July 2008
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Junior Member |
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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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