[SciPy-User] Order of numpy orperations is not equal to logic (and also octave)?
Oz Nahum Tiram
nahumoz@gmail....
Sun Oct 31 03:34:02 CDT 2010
Hi Everyone,
So here is something which I discovered lately and is making me wonder.
I want to define a scalar which I call Net Absolute Mass Balance Error or in
short NAMBE. This NAMBE is the absolute difference between a base vector and
another vector, divided by the base vector and multiplied by a hundred, in
pseudo-code notation:
NAMBE=sum(abs(a-b)/a)*100
When I do it in python, I decided to break the line into two lines so the
code is more readable:
>>> a=np.array([0.1,0.1,0.1,0.1,0.1])
>>> b=np.array([0.1,0.1,0.1,0.1,0.1])*2
>>> b
array([ 0.2, 0.2, 0.2, 0.2, 0.2])
>>> a-b
array([-0.1, -0.1, -0.1, -0.1, -0.1])
>>> s=np.sum(abs(a-b))
>>> s
0.5
>>> s/np.sum(a)
1.0
I thought the numpy does everything element wise so if I do it one line, I
noticed the the result is different:
>>> s=np.sum(abs(a-b)/a)
>>> s
5.0
Now If I check myself on the data I have with a octave, I get different
results:
octave:1> a=[0.1,0.1,0.1,0.1,0.1]
a =
0.10000 0.10000 0.10000 0.10000 0.10000
octave:2> b=a*2
b =
0.20000 0.20000 0.20000 0.20000 0.20000
octave:3> sum(a)
ans = 0.50000
octave:4> sum(b)
ans = 1
octave:5> sum(a-b)
ans = -0.50000
octave:6> sum(abs(a-b))
ans = 0.50000
octave:7> s=sum(abs(a-b))
s = 0.50000
octave:8> s/sum(a)
ans = 1
octave:9> s=sum(abs(a-b)/a)
s = 1.0000
octave:10> s=sum(abs(a-b)/sum(a))
s = 1
Note that the is no difference in the output of 9 and 10 in Octave, although
there is in Python ... So, my question is: Why is python is behaving like
that ? Which one is right ? Octave or Python ?
Thanks in advance,
--
Oz Nahum
Graduate Student
Zentrum für Angewandte Geologie
Universität Tübingen
---
Imagine there's no countries
it isn't hard to do
Nothing to kill or die for
And no religion too
Imagine all the people
Living life in peace
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