Add minimum, maximum functions

[FrancescAlted]

From [email protected] on June 07, 2012 16:43:59

It would be useful to add np.mininum and np.maximum functions. 'maximum(x-a, 0)' would be faster and clearer than 'where(x-a>0,x-a,0)

Original issue: http://code.google.com/p/numexpr/issues/detail?id=86

[FrancescAlted]

From [email protected] on August 30, 2013 13:04:32

Interested in a numpy.min numpy.max equiv.

[mythsmith]

This would be extremely useful, especially when searching a "nearest value", eg: abs(x-val)==min(abs(x-val)).

[Ablinne]

In the context of min and max I would also suggest to implement clip. While clip(a,b,c) is equivalent to min(max(a,b),c), the former is easier to read.

[vallsv]

+1 for minimum and clip

But at least i am happy to read that i can use this where function.

[wilko339]

Just found this thread still open. Any progress? Interested in a np.minimum() and np.maximum() style element wise min / max!

[robbmcleod]

Min and max were implemented awhile ago. There is no room left for more opcodes, so clip cannot be implemented in NumExpr 2.7. I do believe numexpr3 could have clip implemented.

[vallsv]

Really! That's super nice. How can we use it?

I tried all this sentences:

a = numpy.array([2,3,4])
b = numpy.array([3,2,4])

> numexpr.evaluate("min(a,2)")
ValueError: reduction axis is out of bounds

> numexpr.evaluate("min(a,b)")
TypeError: You can't use Python's standard boolean operators in NumExpr expressions.
You should use their bitwise counterparts instead: '&' instead of 'and', '|' instead of 'or',
and '~' instead of 'not'.

> numexpr.evaluate("minimum(a,2)")
TypeError: 'VariableNode' object is not callable

> numexpr.evaluate("a.min(b)")
AttributeError: 'VariableNode' object has no attribute 'min'

> numexpr.version.version                                                        
'2.7.1'

[robbmcleod]

Oh, sorry, they are reductions.

Element-wise could be done for float32 and float64 but not for integers.

[sebbelese]

Hi,

Sorry if I misunderstood: does it mean that element-wise min and max should work with float32 and float64 arrays? I tried @vallsv expressions with a and b as float arrays, but got the same issue as for integer.

Are element-wise min and max implemented, and do you have an example?

Thanks

[robbmcleod]

Element-wise are not implemented at this time. However, you can use the where function instead. It's just uglier.

[shubham2941]

Really! That's super nice. How can we use it?

I tried all this sentences:

a = numpy.array([2,3,4])
b = numpy.array([3,2,4])

> numexpr.evaluate("min(a,2)")
ValueError: reduction axis is out of bounds

> numexpr.evaluate("min(a,b)")
TypeError: You can't use Python's standard boolean operators in NumExpr expressions.
You should use their bitwise counterparts instead: '&' instead of 'and', '|' instead of 'or',
and '~' instead of 'not'.

> numexpr.evaluate("minimum(a,2)")
TypeError: 'VariableNode' object is not callable

> numexpr.evaluate("a.min(b)")
AttributeError: 'VariableNode' object has no attribute 'min'

> numexpr.version.version                                                        
'2.7.1'

Hi, Is there a way to use all these in the current version of numexpr? I tried these and same error appears.

[robbmcleod]

Such as this:

import numexpr as ne
import numpy as np

a = np.array([2,3,4])
b = np.array([3,2,4])

out1 = ne.evaluate('where(a < 2, a, 2)')
out2 = ne.evaluate('where(a < b, a, b)')

[shubham2941]

Thanks, @robbmcleod. I somehow missed that function earlier.

Is there a way to define your own operator in numexpr? For eg., if you want to do multiple operations on two parameters in a user-defined function. If there is a post somewhere or some documentation, can someone point to it? Any suggestions is welcome.

[robbmcleod]

No, I'd suggest you look at numba for such a task.

[shubham2941]

I am not sure how to use numba with numexpr.evaluate and user-defined function.

Let me explain my issue with numexpr.evaluate in detail:

I have a string function in the form with data in variables A and B in data dictionary form: def ufunc(A,B): ............ return var The evaluation function goes like this: numexpr.evaluate( 'A + B**2 + A*B + sin(A) + ufunc(A,B)' , data_dict) The problem is I don't how to send the 'ufunc' information to 'numexpr.evaluate'. I am not sure how 'numba' will help with that.

[josedaniel-escribano-clarity]

I was evaluating this library and not having min or max operator is a big negative aspect... is there any plan to implement them? It's very strange not using min or max in the formula and have to use the where function.

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