# Programming oddities

(Difference between revisions)
 Revision as of 08:39, 30 May 2016 (view source)Alex (Talk | contribs) (→datestr)← Older edit Revision as of 12:53, 16 September 2016 (view source)Alex (Talk | contribs) (→Python)Newer edit → Line 117: Line 117: norm2[valid] = numpy.sqrt(norm2[valid]) norm2[valid] = numpy.sqrt(norm2[valid]) - I am wondering if this should not + + == Numpy: arr.max() can return NaN == + + Numpy's max can return a NaN as maximum value: + + In [1]: array([1.,NaN]).max() + Out[1]: nan + + Unlike python building max function: + + In [10]: max([1.,NaN]) + Out[10]: 1.0 + + Also unlike Fortran and Matlab. For a sane behaviour on should use [http://docs.scipy.org/doc/numpy/reference/generated/numpy.nanmax.html arr.nanmax()]. = Shell = = Shell =

# Matlab

## ndims

For scalars and vectors, ndims always returns 2.

>> ndims(0)
ans =
2
>> ndims([1 2 3 4])
ans =
2

>> isvector(1)
ans =
1

## structures as matrices

Even structures are matrices:

s.field = 1;

You can index them and query the dimensions (which is also two):

s(1)
ndims(s)

So every thing in mat(trix)lab is a matrix; Even function handle (tested in octave):

ismatrix (@(x) x)
ans =  1

## datestr

Matlab 7.10.0.499 (R2010a):

>> datestr([1495 1 4  0  0 0])
ans =
03-Feb-0004
01-Jan-0000
04-Jan-0000
00-Jan-0000
00-Jan-0000
00-Jan-0000

>> datestr([1495 1 4  0  0 1])
ans =
04-Jan-1495 00:00:01

What happened the 4th January 1495? But then:

>> datestr([1495 1 5 0 0 0])

ans =
05-Jan-1495

# Appending data to an empty array

For a vector is works as expected:

>> A = [];
>> A(end+1) = 1

A =
1

>> whos A
Name      Size            Bytes  Class     Attributes
A         1x1                 8  double

However, for a 3d-array, we get 2 elements (instead of 1) in matlab R2013a.

>> A = [];
>> A(1,1,end+1) = 1;

A(:,:,1) =
0

A(:,:,2) =
1

>> whos A
Name      Size             Bytes  Class     Attributes
A         1x1x2               16  double

# Python

## sqrt is also calculated on mask value in a Masked array

In numpy version 1.6.1, numpy.sqrt is also applied to masked values and returns a warning if these are negative:

import numpy
norm2 = numpy.ma.array([1, -2, 3], mask = [0, 1, 0])
numpy.sqrt(norm2)
/usr/bin/ipython:1: RuntimeWarning: invalid value encountered in sqrt

This behavior is inconsistent since the ** operators applies only on non-masked value:

b = ma.array([1, -2, 3], mask = [0, 1, 0])
b2 = b**2
b2.data
Out[5]: array([ 1, -2,  9])

Thus numpy.sqrt(b**2) returns this warning.

A work-around:

norm2[valid] = numpy.sqrt(norm2[valid])

## Numpy: arr.max() can return NaN

Numpy's max can return a NaN as maximum value:

In [1]: array([1.,NaN]).max()
Out[1]: nan

Unlike python building max function:

In [10]: max([1.,NaN])
Out[10]: 1.0

Also unlike Fortran and Matlab. For a sane behaviour on should use arr.nanmax().

i=010
let i=\$i+1
echo \$i
9