Archive for the ‘python’ Category

September 22nd, 2019

In many introductions to numpy, one gets taught about np.ones, np.zeros and np.empty. The accepted wisdom is that np.empty will be faster than np.ones because it doesn’t have to waste time doing all that initialisation. A quick test in a Jupyter notebook shows that this seems to be true!

import numpy as np

N = 200
zero_time = %timeit -o some_zeros = np.zeros((N,N))
empty_time = %timeit -o empty_matrix = np.empty((N,N))
print('np.empty is {0} times faster than np.zeros'.format(zero_time.average/empty_time.average))


8.34 µs ± 202 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
436 ns ± 10.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
np.empty is 19.140587654682545 times faster than np.zeros

20 times faster may well be useful in production when using really big matrices. Might even be worth the risk of dealing with uninitialised variables even though they are scary!

However…..on my machine (Windows 10, Microsoft Surface Book 2 with 16Gb RAM), we see the following behaviour with a larger matrix size (1000 x 1000).

import numpy as np

N = 1000
zero_time = %timeit -o some_zeros = np.zeros((N,N))
empty_time = %timeit -o empty_matrix = np.empty((N,N))
print('np.empty is {0} times faster than np.zeros'.format(zero_time.average/empty_time.average))


113 µs ± 2.97 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
112 µs ± 1.01 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
np.empty is 1.0094651980894993 times faster than np.zeros

The speed-up has vanished! A subsequent discussion on twitter suggests that this is probably because the operating system is zeroing all of the memory for security reasons when using numpy.empty on large arrays.

September 2nd, 2019


What is Second Order Cone Programming (SOCP)?

 

Second Order Cone Programming (SOCP) problems are a type of optimisation problem that have applications in many areas of science, finance and engineering. A summary of the type of problems that can make use of SOCP, including things as diverse as designing antenna arrays, finite impulse response (FIR) filters and structural equilibrium problems can be found in the paper ‘Applications of Second Order Cone Programming’ by Lobo et al. There are also a couple of examples of using SOCP for portfolio optimisation in the GitHub repository of the Numerical Algorithms Group (NAG).

A large scale SOCP solver was one of the highlights of the Mark 27 release of the NAG library (See here for a poster about its performance). Those who have used the NAG library for years will expect this solver to have interfaces in Fortran and C and, of course, they are there. In addition to this is the fact that Mark 27 of the NAG Library for Python was released at the same time as the Fortran and C interfaces which reflects the importance of Python in today’s numerical computing landscape.

Here’s a quick demo of how the new SOCP solver works in Python. The code is based on a notebook in NAG’s PythonExamples GitHub repository.

NAG’s handle_solve_socp_ipm function (also known as e04pt) is a solver from the NAG optimization modelling suite for large-scale second-order cone programming (SOCP) problems based on an interior point method (IPM).

$$
\begin{array}{ll}
{\underset{x \in \mathbb{R}^{n}}{minimize}\ } & {c^{T}x} \\
\text{subject to} & {l_{A} \leq Ax \leq u_{A}\text{,}} \\
& {l_{x} \leq x \leq u_{x}\text{,}} \\
& {x \in \mathcal{K}\text{,}} \\
\end{array}
$$

where $\mathcal{K} = \mathcal{K}^{n_{1}} \times \cdots \times \mathcal{K}^{n_{r}} \times \mathbb{R}^{n_{l}}$ is a Cartesian product of quadratic (second-order type) cones and $n_{l}$-dimensional real space, and $n = \sum_{i = 1}^{r}n_{i} + n_{l}$ is the number of decision variables. Here $c$, $x$, $l_x$ and $u_x$ are $n$-dimensional vectors.

$A$ is an $m$ by $n$ sparse matrix, and $l_A$ and $u_A$ and are $m$-dimensional vectors. Note that $x \in \mathcal{K}$ partitions subsets of variables into quadratic cones and each $\mathcal{K}^{n_{i}}$ can be either a quadratic cone or a rotated quadratic cone. These are defined as follows:

  • Quadratic cone:

$$
\mathcal{K}_{q}^{n_{i}} := \left\{ {z = \left\{ {z_{1},z_{2},\ldots,z_{n_{i}}} \right\} \in {\mathbb{R}}^{n_{i}} \quad\quad : \quad\quad z_{1}^{2} \geq \sum\limits_{j = 2}^{n_{i}}z_{j}^{2},\quad\quad\quad z_{1} \geq 0} \right\}\text{.}
$$

  • Rotated quadratic cone:

$$
\mathcal{K}_{r}^{n_{i}} := \left\{ {z = \left\{ {z_{1},z_{2},\ldots,z_{n_{i}}} \right\} \in {\mathbb{R}}^{n_{i}}\quad\quad:\quad \quad\quad 2z_{1}z_{2} \geq \sum\limits_{j = 3}^{n_{i}}z_{j}^{2}, \quad\quad z_{1} \geq 0, \quad\quad z_{2} \geq 0} \right\}\text{.}
$$

For a full explanation of this routine, refer to e04ptc in the NAG Library Manual

Using the NAG SOCP Solver from Python

 

This example, derived from the documentation for the handle_set_group function solves the following SOCP problem

minimize $${10.0x_{1} + 20.0x_{2} + x_{3}}$$

from naginterfaces.base import utils
from naginterfaces.library import opt

# The problem size:
n = 3

# Create the problem handle:
handle = opt.handle_init(nvar=n)

# Set objective function
opt.handle_set_linobj(handle, cvec=[10.0, 20.0, 1.0])

subject to the bounds
$$
\begin{array}{rllll}
{- 2.0} & \leq & x_{1} & \leq & 2.0 \\
{- 2.0} & \leq & x_{2} & \leq & 2.0 \\
\end{array}
$$

# Set box constraints
opt.handle_set_simplebounds(
    handle,
    bl=[-2.0, -2.0, -1.e20],
    bu=[2.0, 2.0, 1.e20]
)

the general linear constraints

\begin{array}{lcrcrcrclcl}
& & {- 0.1x_{1}} & – & {0.1x_{2}} & + & x_{3} & \leq & 1.5 & & \\
1.0 & \leq & {- 0.06x_{1}} & + & x_{2} & + & x_{3} & & & & \\
\end{array}
# Set linear constraints
opt.handle_set_linconstr(
    handle,
    bl=[-1.e20, 1.0],
    bu=[1.5, 1.e20],
    irowb=[1, 1, 1, 2, 2, 2],
    icolb=[1, 2, 3, 1, 2, 3],
    b=[-0.1, -0.1, 1.0, -0.06, 1.0, 1.0]
    );

and the cone constraint

$$\left( {x_{3},x_{1},x_{2}} \right) \in \mathcal{K}_{q}^{3}\text{.}$$

# Set cone constraint
opt.handle_set_group(
    handle,
    gtype='Q',
    group=[ 3,1, 2],
    idgroup=0
);

We set some algorithmic options. For more details on the options available, refer to the routine documentation

# Set some algorithmic options.
for option in [
        'Print Options = NO',
        'Print Level = 1'
]:
    opt.handle_opt_set(handle, option)

# Use an explicit I/O manager for abbreviated iteration output:
iom = utils.FileObjManager(locus_in_output=False)

Finally, we call the solver

# Call SOCP interior point solver
result = opt.handle_solve_socp_ipm(handle, io_manager=iom)
 ------------------------------------------------
  E04PT, Interior point method for SOCP problems
 ------------------------------------------------

 Status: converged, an optimal solution found
 Final primal objective value -1.951817E+01
 Final dual objective value   -1.951817E+01

The optimal solution is

result.x
array([-1.26819151, -0.4084294 ,  1.3323379 ])

and the objective function value is

result.rinfo[0]
-19.51816515094211

Finally, we clean up after ourselves by destroying the handle

# Destroy the handle:
opt.handle_free(handle)

As you can see, the way to use the NAG Library for Python interface follows the mathematics quite closely.
NAG also recently added support for the popular cvxpy modelling language that I’ll discuss another time.

Links

July 1st, 2019

I am a huge user of Anaconda Python and the way I usually get access to the Anaconda Prompt is to start typing ‘Anaconda’ in the Windows search box and click on the link as soon as it pops up. Easy and convenient. Earlier today, however, the Windows 10 menu shortcuts for the Anaconda command line vanished from my machine!

I’m not sure exactly what triggered this but I was heavily messing around with various environments, including the base one, and also installed Visual Studio 2019 Community Edition with all the Python extensions before I noticed that the menu shortcuts had gone missing. No idea what the root cause was.

Fortunately, getting my Anaconda Prompt back was very straightforward:

  • launch Anaconda Navigator
  • Click on Environments
  • Selected base (root)
  • Choose Not installed  from the drop down list
  • Type for console_ in the search box
  • Check the console_shortcut package
  • Click Apply and follow the instructions to install the package

console_shortcut

April 10th, 2019

I recently wrote a blog post for my new employer, The Numerical Algorithms Group, called Exploiting Matrix Structure in the solution of linear systems. It’s a demonstration that shows how choosing the right specialist solver for your problem rather than using a general purpose one can lead to a speed up of well over 100 times!  The example is written in Python but the NAG routines used can be called from a range of languages including C,C++, Fortran, MATLAB etc etc

April 12th, 2018

Update
A discussion on twitter determined that this was an issue with Locales. The practical upshot is that we can make R act the same way as the others by doing

Sys.setlocale("LC_COLLATE", "C")

which may or may not be what you should do!

Original post

While working on a project that involves using multiple languages, I noticed some tests failing in one language and not the other. Further investigation revealed that this was essentially because R's default sort order for strings is different from everyone else's.

I have no idea how to say to R 'Use the sort order that everyone else is using'. Suggestions welcomed.

R 3.3.2

sort(c("#b","-b","-a","#a","a","b"))

[1] "-a" "-b" "#a" "#b" "a" "b"

Python 3.6

sorted({"#b","-b","-a","#a","a","b"})

['#a', '#b', '-a', '-b', 'a', 'b']


MATLAB 2018a

sort([{'#b'},{'-b'},{'-a'},{'#a'},{'a'},{'b'}])

ans =
1×6 cell array
{'#a'} {'#b'} {'-a'} {'-b'} {'a'} {'b'}

C++

int main(){ 

std::string mystrs[] = {"#b","-b","-a","#a","a","b"}; 
std::vector<std::string> stringarray(mystrs,mystrs+6);
std::vector<std::string>::iterator it; 

std::sort(stringarray.begin(),stringarray.end());

for(it=stringarray.begin(); it!=stringarray.end();++it) {
   std::cout << *it << " "; 
} 

return 0;
} 

Result:

#a #b -a -b a b
May 15th, 2017

For a while now, Microsoft have provided a free Jupyter Notebook service on Microsoft Azure. At the moment they provide compute kernels for Python, R and F# providing up to 4Gb of memory per session. Anyone with a Microsoft account can upload their own notebooks, share notebooks with others and start computing or doing data science for free.

They University of Cambridge uses them for teaching, and they’ve also been used by the LIGO people  (gravitational waves) for dissemination purposes.

This got me wondering. How much power does Microsoft provide for free within these notebooks?  Computing is pretty cheap these days what with the Raspberry Pi and so on but what do you get for nothing? The memory limit is 4GB but how about the computational power?

To find out, I created a simple benchmark notebook that finds out how quickly a computer multiplies matrices together of various sizes.

Matrix-Matrix multiplication is often used as a benchmark because it’s a common operation in many scientific domains and it has been optimised to within an inch of it’s life.  I have lost count of the number of times where my contribution to a researcher’s computational workflow has amounted to little more than ‘don’t multiply matrices together like that, do it like this…it’s much faster’

So how do Azure notebooks perform when doing this important operation? It turns out that they max out at 263 Gigaflops! azure_free_notebook

For context, here are some other results:

As you can see, we are getting quite a lot of compute power for nothing from Azure Notebooks. Of course, one of the limiting factors of the free notebook service is that we are limited to 4GB of RAM but that was more than I had on my own laptops until 2011 and I got along just fine.

Another fun fact is that according to https://www.top500.org/statistics/perfdevel/, 263 Gigaflops would have made it the fastest computer in the world until 1994. It would have stayed in the top 500 supercomputers of the world until June 2003 [1].

Not bad for free!

[1] The top 500 list is compiled using a different benchmark called LINPACK  so a direct comparison isn’t strictly valid…I’m using a little poetic license here.

January 19th, 2017

There are lots of Widgets in ipywidgets. Here’s how to list them

from ipywidgets import *
widget.Widget.widget_types

At the time of writing, this gave me

{'Jupyter.Accordion': ipywidgets.widgets.widget_selectioncontainer.Accordion,
 'Jupyter.BoundedFloatText': ipywidgets.widgets.widget_float.BoundedFloatText,
 'Jupyter.BoundedIntText': ipywidgets.widgets.widget_int.BoundedIntText,
 'Jupyter.Box': ipywidgets.widgets.widget_box.Box,
 'Jupyter.Button': ipywidgets.widgets.widget_button.Button,
 'Jupyter.Checkbox': ipywidgets.widgets.widget_bool.Checkbox,
 'Jupyter.ColorPicker': ipywidgets.widgets.widget_color.ColorPicker,
 'Jupyter.Controller': ipywidgets.widgets.widget_controller.Controller,
 'Jupyter.ControllerAxis': ipywidgets.widgets.widget_controller.Axis,
 'Jupyter.ControllerButton': ipywidgets.widgets.widget_controller.Button,
 'Jupyter.Dropdown': ipywidgets.widgets.widget_selection.Dropdown,
 'Jupyter.FlexBox': ipywidgets.widgets.widget_box.FlexBox,
 'Jupyter.FloatProgress': ipywidgets.widgets.widget_float.FloatProgress,
 'Jupyter.FloatRangeSlider': ipywidgets.widgets.widget_float.FloatRangeSlider,
 'Jupyter.FloatSlider': ipywidgets.widgets.widget_float.FloatSlider,
 'Jupyter.FloatText': ipywidgets.widgets.widget_float.FloatText,
 'Jupyter.HTML': ipywidgets.widgets.widget_string.HTML,
 'Jupyter.Image': ipywidgets.widgets.widget_image.Image,
 'Jupyter.IntProgress': ipywidgets.widgets.widget_int.IntProgress,
 'Jupyter.IntRangeSlider': ipywidgets.widgets.widget_int.IntRangeSlider,
 'Jupyter.IntSlider': ipywidgets.widgets.widget_int.IntSlider,
 'Jupyter.IntText': ipywidgets.widgets.widget_int.IntText,
 'Jupyter.Label': ipywidgets.widgets.widget_string.Label,
 'Jupyter.PlaceProxy': ipywidgets.widgets.widget_box.PlaceProxy,
 'Jupyter.Play': ipywidgets.widgets.widget_int.Play,
 'Jupyter.Proxy': ipywidgets.widgets.widget_box.Proxy,
 'Jupyter.RadioButtons': ipywidgets.widgets.widget_selection.RadioButtons,
 'Jupyter.Select': ipywidgets.widgets.widget_selection.Select,
 'Jupyter.SelectMultiple': ipywidgets.widgets.widget_selection.SelectMultiple,
 'Jupyter.SelectionSlider': ipywidgets.widgets.widget_selection.SelectionSlider,
 'Jupyter.Tab': ipywidgets.widgets.widget_selectioncontainer.Tab,
 'Jupyter.Text': ipywidgets.widgets.widget_string.Text,
 'Jupyter.Textarea': ipywidgets.widgets.widget_string.Textarea,
 'Jupyter.ToggleButton': ipywidgets.widgets.widget_bool.ToggleButton,
 'Jupyter.ToggleButtons': ipywidgets.widgets.widget_selection.ToggleButtons,
 'Jupyter.Valid': ipywidgets.widgets.widget_bool.Valid,
 'jupyter.DirectionalLink': ipywidgets.widgets.widget_link.DirectionalLink,
 'jupyter.Link': ipywidgets.widgets.widget_link.Link}
August 11th, 2016

This is my rant on import *. There are many like it, but this one is mine.

I tend to work with scientists so I’ll use something from mathematics as my example.  What is the result of executing the following line of Python code?

result = sqrt(-1)

Of course, you have no idea if you don’t know which module sqrt came from. Let’s look at a few possibilities. Perhaps you’ll get an exception:

In [1]: import math
In [2]: math.sqrt(-1)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
in ()
----> 1 math.sqrt(-1)

ValueError: math domain error

Or maybe you’ll just get a warning and a nan

In [3]: import numpy
In [4]: numpy.sqrt(-1)
/Users/walkingrandomly/anaconda/bin/ipython:1: RuntimeWarning: invalid value encountered in sqrt
#!/bin/bash /Users/walkingrandomly/anaconda/bin/python.app
Out[4]: nan

You might get an answer but the datatype of your answer could be all sorts of strange and wonderful stuff.

In [5]: import cmath
In [6]: cmath.sqrt(-1)
Out[6]: 1j
In [7]: type(cmath.sqrt(-1))
Out[7]: complex

In [8]: import scipy
In [9]: scipy.sqrt(-1)
Out[9]: 1j
In [10]: type(scipy.sqrt(-1))
Out[10]: numpy.complex128

In [11]: import sympy
In [12]: sympy.sqrt(-1)
Out[12]: I
In [13]: type(sympy.sqrt(-1))
Out[13]: sympy.core.numbers.ImaginaryUnit

Even the humble square root function behaves very differently when imported from different modules! There are probably other sqrt functions, with yet more behaviours that I’ve missed.

Sometimes, they seem to behave in very similar ways:-

In [16]: math.sqrt(2)
Out[16]: 1.4142135623730951

In [17]: numpy.sqrt(2)
Out[17]: 1.4142135623730951

In [18]: scipy.sqrt(2)
Out[18]: 1.4142135623730951

Let’s invent some trivial code.

from scipy import sqrt

x = float(input('enter a number\n'))
y = sqrt(x)

# important things happen after here. Complex numbers are fine!

I can input -1 just fine. Then, someone comes along and decides that they need a function from math in the ‘important bit’. They use import *

from scipy import sqrt
from math import *

x = float(input('enter a number\n'))
y = sqrt(x)

# important things happen after here. Complex numbers are fine!

They test using inputs like 2 and 4 and everything works (we don’t have automated tests — we suck!). Of course it breaks for -1 now though. This is easy to diagnose when you’ve got a few lines of code but it causes a lot of grief when there’s hundreds…or, horror of horrors, if the ‘from math import *’ was done somewhere in the middle of the source file!

I’m sometimes accused of being obsessive and maybe I’m labouring the point a little but I see this stuff, in various guises, all the time!

So, yeah, don’t use import *.

February 8th, 2016

While waiting for the rain to stop before heading home, I started messing around with the heart equation described in an old WalkingRandomly post. Playing code golf with myself, I worked to get the code tweetable. In Python:

In R:

I liked the look of the default plot in R so animated it by turning 200 into a parameter that ranged from 1 to 200. The result was this animation:

The code for the above isn’t quite tweetable:

options(warn=-1)
for(num in seq(1,200,1))
{
    filename = paste("rplot" ,sprintf("%03d", num),'.jpg',sep='')
    jpeg(filename)
    x=seq(-2,2,0.001)
    y=Re((sqrt(cos(x))*cos(num*x)+sqrt(abs(x))-0.7)*(4-x*x)^0.01)
    plot(x,y,axes=FALSE,ann=FALSE)
    dev.off()
}

This produces a lot of .jpg files which I turned into the animated gif with ImageMagick:

convert -delay 12 -layers OptimizeTransparency -colors 8 -loop 0 *.jpg animated.gif 
September 5th, 2015

The test suite of a project I’m working on is poking around at the extreme edges of the range of double precision numbers. I noticed a difference between Windows and other platforms that I can’t yet fully explain. On Windows, the test suite was pumping out RuntimeWarnings that we don’t see in Linux or Mac. I’ve distilled the issue down to a single numpy command:

np.log1p(1.7976931348622732e+308)

On Windows 7 Anaconda Python 2.3, this gives a RuntimeWarning and returns inf whereas on Linux and Mac OS X it evaluates to 709.78-ish

Numpy version is 1.9.2 in all cases.

64 bit Windows 7

Python 2.7.10 |Continuum Analytics, Inc.| (default, May 28 2015, 16:44:52) [MSC
v.1500 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
Anaconda is brought to you by Continuum Analytics.
Please check out: http://continuum.io/thanks and https://binstar.org
>>> import numpy as np
>>> np.log1p(1.7976931348622732e+308)
__main__:1: RuntimeWarning: overflow encountered in log1p
inf

64 bit Linux

Python 2.7.9 (default, Apr  2 2015, 15:33:21) 
[GCC 4.9.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import numpy as np
>>> np.log1p(1.7976931348622732e+308)
709.78271289338397

Mac OS X

Python 2.7.10 |Anaconda 2.3.0 (x86_64)| (default, May 28 2015, 17:04:42) 
[GCC 4.2.1 (Apple Inc. build 5577)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
Anaconda is brought to you by Continuum Analytics.
Please check out: http://continuum.io/thanks and https://binstar.org
>>> import numpy as np
>>> np.log1p(1.7976931348622732e+308)
709.78271289338397

The argument to log1p is getting close to the largest double precision number:

>>> sys.float_info.max
1.7976931348623157e+308