## venerdì 15 aprile 2016

Can you match the MEPs with their Member States? http://www.thinkingabout.eu/match

## martedì 24 maggio 2011

### A method to exchange keys and values in a python dictionary

Around the web i've found this method to exchange keys and values in a python dictionary:

the_dict = { 'a': 1, 'b': 2, 'c': 3 }
inverted_dict = dict(zip(*zip(*the_dict.items())[::-1]))


output:
{ 1:'a', 2:'b', 3:'c' }


it is nice isn't it?
but how does it work?
The main idea is to split keys and values in a list of two n-tuples $$[ (k_1,...,k_n) , (v_1,...,v_n)] ,$$
switch the tuples and re-create a dictionary.

ingredients are:
1 - operator * (also known as 'splat' operator) that is able to unpack a list
2 - zip function that returns a list of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterable
3 - a_list[::-1] to revert a_list

We can unpack the algorithm in those 5 nested steps:
a - the_dict.items()
b - zip(*the_dict.items())
c - zip(*the_dict.items())[::-1]
d - zip(*zip(*the_dict.items())[::-1])
e - dict(zip(*zip(*the_dict.items())[::-1]))


a - with the_dict.items() we obtain:
 [('a', 1), ('b', 2), ('c', 3)]

b - then, using * operator in the zip function:
zip(*the_dict.items())
we are doing:
zip(('a', 1), ('b', 2), ('c', 3) )
obtaining:
[('a', 'b', 'c'), (1, 2, 3)]


c - with [::-1] we actually switch key and values
zip(*the_dict.items())[::-1] = [('a', 'b', 'c'), (1, 2, 3)][::-1]
=
[('a', 'b', 'c'), (1, 2, 3)]


d - with the second zip * pair we re-unpack the list to recreate the 3 pairs:
zip(*zip(*the_dict.items())[::-1]) =
zip(*[('a', 'b', 'c'), (1, 2, 3)]) =zip(('a', 'b', 'c'), (1, 2, 3) )
=
[('a', 1), ('b', 2), ('c', 3)]


e - now our new dict is almost cooked, we use dict() function to complete the opera:
dict(zip(*zip(*the_dict.items())[::-1]))
=
dict ([('a', 1), ('b', 2), ('c', 3)])
the output:{ 1:'a', 2:'b', 3:'c' }all this stuff in 1 row.



## lunedì 23 maggio 2011

### Happy numbers in python

I'll show you a method in python to check if a number is happy.
When a number is happy? Well, you have to follow this algorithm:

1. Take a number n.
n = 23
2. Dissect it into digits.
2 and 3
3. Square them all and add them up
2^ 2 + 3 ^ 2 = 4 + 9 = 13
4. You get a new number m.
m = 13
5. If m = 1, n is happy; otherwise set n = m and repeat at 1.
1^2 + 3^2 = 1 + 9 = 10
n = 10
1 ^ 2 + 0 ^ 2 = 1
23 is happy!
6. If you run into a loop, n is not a happy number (is sad).

I've used recursion because it is fun (talking about happy numbers).

def happy(n, past = set()):
m = sum(int(i)**2 for i in str(n))
if m == 1:
return True
if m in past:
return False
return happy(m,past)

print [ x for x in range(1,100) if happy(x, set())]


Here 'past' is a set, needed to check if we are in a loop.
The output shows the set of happy numbers below 100.

[1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49, 68, 70, 79, 82, 86, 91, 94, 97]


## giovedì 19 maggio 2011

### logistic map in python

Simple logistic map using python and matplotlib.

import math
import matplotlib.pyplot as plt

def logistic(xa =2.9 , xb=4.0 , imgx = 240 , imgy = 500,
maxit = 200, f=lambda x,r: r * x * (1 - x) ):
xs = []
ys = []
for i in range(imgx):
r = xa + (xb - xa) * float(i)/(imgx - 1)
x = 0.5
for j in range(maxit):
x = f(x,r)
if j > maxit / 2:
xs.append( i )
ys.append(int(x * imgy))
return [xs,ys]

myfunction = lambda x,r: r * (math.sin(x)**2)
points = logistic( xa = 2.1 ,xb = 3.1, imgy=200 , f=myfunction)
ax = plt.subplot(121)
ax.scatter(points[0], points[1], s= 1)

points = logistic()
bx = plt.subplot(122)
bx.scatter(points[0], points[1], s= 1)
plt.show()



## lunedì 4 aprile 2011

### Caostabile is Online

A little bit of Math and Physics curiosities and amenities ( in italian ) .

http://caostabile.altervista.org/

## martedì 3 agosto 2010

### My drupal site

my drupal site hosted by altervista

## venerdì 26 marzo 2010

### Similarity Matrix in Text Mining

( in this post i'm testing http://www.mathjax.org for latex math formulas )
What is a similarity matrix ( in text mining ) and why is important?

CORPUS

We have to start from a corpus composed by k documents:
$$\left\{ D_i \right\}_{i=1}^k$$
( A corpus is merely collection of documents )

SIMILARITY MATRIX

A way to find semantic structures in the corpus is to study the occurrence and the
co-occurrence for every pair of words in the corpus.
A good tool to find something interesting is a similarity matrix.

DEFINITION

To define a similarity matrix we must define the similarity between two objects ( words )
$$s(w_i, w_j )$$
a similarity matrix becomes simply a matrix that contains the ratio of similarity between
the objects of index i and j for the generic position {i,j}

a good similarity matrix can follow from this definition: $$s(w_i,w_j) = \dfrac {c(w_i,w_j) } { f(w_i) \cdot f(w_j) }$$ where:  $$c(w_i,w_j)$$ is the co-occurrence between two words ( the number of documents containing both
words )
and:  $$f(w_i)$$
is the occurrence of the word.

MATRIX REPRESENTATIONS

- GRAPH

The created matrix is symmetric and could be visualized using a undirected weighted graph.
The nodes represents the words and the similarity between the two words is given by the
weight between two nodes.

this visualization is nearly useless (easily more than 10000 nodes!!!) .

- METRIC SPACE

A way to handle this info is to position k points in an n-dimensional space so that the mutual
distance between a couple of elements previously defined could reflect the weight between
the related pair of words.$$w_i \mapsto p_i | s(w_i,w_j) = \dfrac{1}{||p_i - p_j||} \forall i,j \leq k$$
( higher weight - closer distance )

a problem related with this approach is that is not always operable (matrix could not be
compatible with metrics constraints ) and the requested dimension of the target space
is a $$O(k^2).$$
so we need to use a technique to reduce the dimension preserving the significant information
( reducing the dimension brings a certain loss of information).

SEMANTIC STRUCTURES

Using the representation in an n-dimensional space is important to analyze clusters of points.
A cluster could be defined as a subset of points whose mutual distances are much smaller
than the average distance of the complete set.

A cluster is a reflection of some kind of statistical structure of the corpus.

Structures able to create a cluster can either be:
1. language related rules ( eg: syntactic structures ) or
2. semantic meanings ( eg: topics )

## martedì 29 dicembre 2009

Verlet integration algorithm written in 3 minutes

-- Verlet integration

projOneTwo ( a, b, c , d ,e ) = ( a, b)

nextStepInt (_ ,_ , 0 ,_ ,_ ) = []
nextStepInt ( xt ,vt , n , acc , h ) =
let
atph = acc . ( xt + ) $h vtph = vt + atph * h * 0.5 xtph = xt + h * vt + 0.5 * acc xt * h^2 in ( xtph, vtph , n-1 , acc , h ) : nextStepInt ( xtph, vtph , n-1 , acc , h ) timeInt ( xt ,vt , n , acc , h ) = map ( projOneTwo )$ nextStepInt ( xt ,vt , n , acc , h )

example ( spring ) :
timeInt (0 , 1 , 10000 , (\x -> 0.1 * ( - x ) ) , 0.2 )

## giovedì 24 dicembre 2009

### my delicious

now i'm in delicious: http://delicious.com/koteth

## martedì 8 dicembre 2009

Is only a fixed-point iteration over this function:

lgs x r = r * x * exp(- x )

### Dollar $operator in Haskell Do you know what is$ operator in Haskell?
$means simply , 'apply the left function at the right value'. f$ x := f x

it seems really trivial, isn't it ?
But, for example in this kind of situation, is really usefull:

zipWith ( $) ( cycle [ \x -> div (x + 1) 2 , \x -> div x 2 ] ) [1..] here you have a infinite list of function: a = cycle [ \x -> div (x + 1) 2 , \x -> div x 2 ] ( ie: [\x -> div (x + 1) 2 , \x -> div x 2 , \x -> div (x + 1) 2 , \x -> div x 2, ... ] ) and you want to apply every element of the list at the element at the same index in the second list: b = [1..] zipWith, for every index i takes the element a(i) of the left list and b(i) of the right list and execute what is requested inside the parentheses. In this situation is specified$ so:

a(i) $b(i ) := a(i) ( b(i) ) the result must be the following: [ 1 ,1 , 2, 2 , 3 ,3 ... and so on. ### Prime Numbers in haskell well, do you want to know how to find 'prime numbers' in a quick and dirty way using Haskell ? try this! import Data.List nubBy ( \x y -> mod y x == 0 ) [2..] Haskell is so easy and charming... ( ps: if you want to speed up a little bit: nubBy ( \x y -> ( x*x-1 <= y ) && ( mod y x == 0 ) ) [2..] ) ## giovedì 20 agosto 2009 ### re: Mandelbrot Set in Haskell import Data.Complex conv p=(\n->".,:;|!([$O0*%#@?"!!(n-1))(ceiling (p*8)::Int)
gr=map(\y->[(x:+y)|x<-[-2,-1.97..0.7]])[-1.2,-1.13..1.2]
--px w=magnitude(foldl1(\z c ->z^2 +c)(replicate 10 w))
px w=(magnitude.last.take 10.takeWhile(\r->(magnitude r )<6).iterate(\z->z^2+w))w
image=map((map(\el->case(px el<2)of{true->conv(px el);_->' '})))gr
main=mapM_ putStrLn image

better...

import Data.Complex
gr = map(\y-> [( x:+y )|x<-[-3,-2.95..1]])[-2,-1.9..2]
px (a:+b) = magnitude(foldl (\z c ->z^2+c) 0 (take 10([((a*x):+b)|x<-[1,1..]])))
image = map((map(\el->case(px el<2)of{true->'*';_ ->'-'})))gr
main = mapM_ putStrLn ( (map(\el->show el) )image )

A first interpretation ( a little bit unsatisfactory ) of the Mandelbrot set in Haskell

## martedì 18 agosto 2009

The core of numbers post in Haskell:

idens p q x = case ( rem ( x ^ 2 ) p ) of { 1 -> 1 ; _ -> 0 }
numList p x = map ( idens p x ) ( map ( x * ) [ 1..( p-1 ) ] )
matIde p = map ( numList p ) [ 1..(p-1) ]

## domenica 16 agosto 2009

### The Path of St. Augustine

Here you can find info about "Il Cammino di Sant'Agostino ( The Path of St. Augustine )" in Italy.

### Quinta da Regaleira

( Mystic Cristina )

## domenica 19 luglio 2009

### Mandelbrot in Rebol

And this is my interpretation of Mandelbrot fractal using Rebol:

This language allow to write a gui using few rows of code.
There is still a problem with zoom functionality ( I hope to fix it soon ).

## domenica 12 luglio 2009

### Latent Semantic Indexing ( LSI )

LSI is a simple but powerful indexing and retrieval method that is able to identify patterns in the relationships between the terms and concepts contained in a text corpus.
The mathematics of LSI easy to understand cause is based entirely on vector and matrix algebra.
I'll post soon some code ( if someone needs code, my mail is always open ).