LBP histogram dimension??
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Goutham
May 19, 2015, 1:02:24 PM5/19/15
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Why the LBP histogram in mahotas gives 36 bins instead of 59.Sorry,I can't follow the previous post regarding this.Can anybody explain me further?
Luis Pedro Coelho
May 19, 2015, 3:52:36 PM5/19/15
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Hi,
For 8 bit codes, 36 is just what comes out from the rotation invariant
condition.
The rule is that the binary codes are invariant to rotation. One way of
implementing this is to normalize all the codes to the minimal value of
all rotations (this value is called the pivot and aggregates all values
it represents). That is, you take each value, you try all possible
binary rolls and you take the minimal value. So, code 1 will actually
represent 1, 2, 4, 8, 16...; code 3 (0000 0011) will also represent 6
(0000 0110), 12 (0000 1100)...; code 5(000 0101) will also represent 10
(0000 1010)...
Now, we can just count how many of these pivots there are.
For 8 bit codes, a 1 bit roll can be implemented like this:
def roll1(i):
return ((i << 1) & 0xff) + bool(i & 128)
Now, we can check whether a binary value is a pivot using this basic
loop:
def is_min(v):
i = v
for _ in range(8):
i = roll1(i)
if i > v:
return False
return True
Ok, so how many pivots are there?
n_pivots = sum([is_min(i) for i in range(256)])
It's 36.
HTH
--
Luis Pedro Coelho | EMBL | http://luispedro.org
My blog: http://metarabbit.wordpress.com
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For 8 bit codes, 36 is just what comes out from the rotation invariant
condition.
The rule is that the binary codes are invariant to rotation. One way of
implementing this is to normalize all the codes to the minimal value of
all rotations (this value is called the pivot and aggregates all values
it represents). That is, you take each value, you try all possible
binary rolls and you take the minimal value. So, code 1 will actually
represent 1, 2, 4, 8, 16...; code 3 (0000 0011) will also represent 6
(0000 0110), 12 (0000 1100)...; code 5(000 0101) will also represent 10
(0000 1010)...
Now, we can just count how many of these pivots there are.
For 8 bit codes, a 1 bit roll can be implemented like this:
def roll1(i):
return ((i << 1) & 0xff) + bool(i & 128)
Now, we can check whether a binary value is a pivot using this basic
loop:
def is_min(v):
i = v
for _ in range(8):
i = roll1(i)
if i > v:
return False
return True
Ok, so how many pivots are there?
n_pivots = sum([is_min(i) for i in range(256)])
It's 36.
HTH
--
Luis Pedro Coelho | EMBL | http://luispedro.org
My blog: http://metarabbit.wordpress.com
> You received this message because you are subscribed to the Google Groups
> "pythonvision" group.
> To unsubscribe from this group and stop receiving emails from it, send an
> email to pythonvision...@googlegroups.com.
> For more options, visit https://groups.google.com/d/optout.
Luis Pedro Coelho
May 19, 2015, 3:58:50 PM5/19/15
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There is a bug in the previous code (the < comparison should have read
>). As written, it uses the maximum as the pivot. Of course, it doesn't
really matter as you end up with the same number of pivots (different
ones, but the same number of equivalent classes).
>). As written, it uses the maximum as the pivot. Of course, it doesn't
really matter as you end up with the same number of pivots (different
ones, but the same number of equivalent classes).
Goutham
May 20, 2015, 10:16:28 AM5/20/15
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Thanks, :)
Goutham
May 21, 2015, 1:54:55 PM5/21/15
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So the 36 bins corresponds to the 36 pivot values,like 0,1,3,5,7.....127,255..Right??
Luis Pedro Coelho
May 21, 2015, 2:11:18 PM5/21/15
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Yes, exactly.
Goutham
May 22, 2015, 2:41:44 AM5/22/15
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Thank you :)
Goutham
May 22, 2015, 3:39:34 AM5/22/15
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For 16 neighbours the histogram bins in mahotas implementation is 4116 instead of 243 !!! I think this implementation take more computing time. !!!
On Thursday, May 21, 2015 at 11:41:18 PM UTC+5:30, Luis Pedro Coelho wrote:
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