Print a symbol for product (∏) by using SymPy package
What am I doing wrong with the macro?
from sympy import *
i = Idx("i")
a = IndexedBase("a")
n = symbols("n")
f=Product(a[i], (i,2,n) )
print(f)
g=Product(a[i], (i,1,n-1) )
print(g)
print(simplify(f/g))
# Product(a[i], (i, 2, n))
# Product(a[i], (i, 1, n - 1))
# Product(a[i], (i, 2, n))/Product(a[i], (i, 1, n - 1))
# I want
# a(n) a(n-1)
# a(n-1) a(n-2)
# .
# .
# . a(2)
# a(2) a(1)
# ---->
# a(n)/a(1)
Thank you in advance and sorry for the bad english!
There's nothing wrong - simplify does not support this kind of simplification.
Why not write something yourself?
Related
I'm trying to use the function collect() to simplify mi expression . My desired result is
My code:
from sympy import *
#index
i = symbols('i' , integer = True )
#constants
a = symbols( 'a' )
#variables
alpha = IndexedBase('alpha', positive=True, domain=QQ)
index = (i, 1, 3)
rho = symbols( 'rho')
U = product( alpha[i]**(1/(rho-1)) , index )
U
:
My solution attempt:
U = U.subs(1/(rho-1),a)
collect(U,rho, evaluate=False)[1]
:
What I'm doing wrong?
You must be using a fairly old version of SymPy because in recent versions the form that you wanted arises automatically. In any case you should be able to use powsimp:
In [9]: U
Out[9]:
a a a
alpha[1] ⋅alpha[2] ⋅alpha[3]
In [10]: powsimp(U, force=True)
Out[10]:
a
(alpha[1]⋅alpha[2]⋅alpha[3])
https://docs.sympy.org/latest/tutorials/intro-tutorial/simplification.html#powsimp
from sympy import Sum, Eq
from sympy.abc import n,x
import random
def polynomial(x):
i = 0
def random_value(i):
return random.choice([i for i in range(-10,10) if i not in [0]])
eq = Sum(random_value(i)*x**n, (n,0,random_value(i)))
display(Eq(eq,eq.doit(), evaluate=False))
polynomial(x)
polynomial(x)
With this code, the coefficients are always the same.
Also, I am not sure if the algebra evaluations are correct for b < 0 .
One way is to use IndexedBase to generate symbolic-placeholder coefficients, and then substitute them with numerical coefficients.
from sympy import Sum, Eq, Matrix, IndexedBase
from sympy.abc import n, x
import random
def polynomial(x):
# n will go from zero to this positive value
to = random.randint(0, 10)
# generate random coefficients
# It is important for them to be a sympy Matrix or Tuple,
# otherwise the substitution (later step) won't work
coeff = Matrix([random.randint(-10, 10) for i in range(to + 1)])
c = IndexedBase("c")
eq = Sum(c[n]*x**n, (n, 0, to)).doit()
eq = eq.subs(c, coeff)
return eq
display(polynomial(x))
display(polynomial(x))
Another ways is to avoid using Sum, relying instead on list-comprehension syntax and builtin sum:
def polynomial(x):
to = random.randint(0, 10)
coeff = [random.randint(-10, 10) for i in range(to + 1)]
return sum([c * x**n for c, n in zip(coeff, range(to + 1))])
display(polynomial(x))
display(polynomial(x))
You can pass a list of coefficients (with highest order coefficient first and constant last) directly to Poly and then convert that to an expression:
>>> from sympy import Poly
>>> from sympy.abc import x
>>> Poly([1,2,3,4], x)
Poly(x**3 + 2*x**2 + 3*x + 4, x, domain='ZZ')
>>> _.as_expr()
x**3 + 2*x**2 + 3*x + 4
>>> from random import randint, choice
>>> Poly([choice((-1,1))*randint(1,10) for i in range(randint(0, 10))], x).as_expr()
-3*x**4 + 3*x**3 - x**2 - 6*x + 2
I am trying to use the linprog in python to solve this problem:
# Minimize = (0.035*x1) + (0.015*x2) + (0.025*x3)
# x1+x2+x3=1.2
# 0<=x1<=0.7
# 0<=x2<=0.3
# 0<=x3<=0.5
c = [0.035, 0.015, 0.025] #objective function
A_eq = [[1, 1, 1]]
b = [1.2]
lb = (0, 0, 0)
up = (0.7, 0.3, 0.5)
from scipy.optimize import linprog
linprog(c, A_ub=None, b_ub=None, A_eq=A_eq, b_eq=b, bounds=[lb,up], method='interior-point', callback=None, options=None, x0=None)
However I am getting an error could you help me with that?
thanks a lot!
You sholud define correctly the bounds for each variable in the same order as the coefficients. In this case, they’re between zero and some number:
# Minimize = (0.035*x1) + (0.015*x2) + (0.025*x3)
# x1+x2+x3=1.2
# 0<=x1<=0.7
# 0<=x2<=0.3
# 0<=x3<=0.5
c = [0.035, 0.015, 0.025] #objective function
A_eq = [[1, 1, 1]]
b = [1.2]
x1_b = (0, 0.7)
x2_b = (0, 0.3)
x3_b = (0, 0.5)
from scipy.optimize import linprog
linprog(c, A_ub=None, b_ub=None, A_eq=A_eq, b_eq=b, bounds=[x1_b, x2_b,x3_b], method='interior-point', callback=None, options=None, x0=None)
Thank you in advance and sorry for the bad english!
I want
1.x : how to Determination of Exponentiation , Determination possible?
2.print : base=b ,exponent=n
WrongScript.py
from sympy import *
var('x y z a b n')
x=b**n
y=3**n
z=a
# output 1.---------------------------------
print("x=",x) # x= b**n,**
print("y=",y) # y= 3**n,**
print("z=",z) # z= a, not **
# output 2.---------------------------------
print(MyBaseOut(x),MyExponentOut(x)) # b,n
print(MyBaseOut(y),MyExponentOut(y)) # 3,n
def MyBaseOut(p):
# ans=?
return ans
def MyExponentOut(q):
# ans=?
return ans
2018-11-26------------------------------
FullScript.py
from sympy import *
var('b n')
def MyBaseOut(p):
return p.as_base_exp()[0]
def MyExponentOut(q):
return q.as_base_exp()[1]
x=b**n
y=3**n
print(MyBaseOut(x),MyExponentOut(x))
print(MyBaseOut(y),MyExponentOut(y))
# b n
# 3 n
The attribute is_Pow will tell you if it has an exponent other than 1 and the method as_base_exp() will tell you what the base and exponent are -- select element 0 for the base and element 1 for the exponent:
>>> [(i.is_Pow, i.as_base_exp()) for i in (y,1/y,y**2,y**z)]
[(False, (y, 1)), (True, (y, -1)), (True, (y, 2)), (True, (y, z))]
I wrote the following code for computing character bigrams and the output is right below. My question is, how do I get an output that excludes the last character (ie t)? and is there a quicker and more efficient method for computing character n-grams?
b='student'
>>> y=[]
>>> for x in range(len(b)):
n=b[x:x+2]
y.append(n)
>>> y
['st', 'tu', 'ud', 'de', 'en', 'nt', 't']
Here is the result I would like to get:['st','tu','ud','de','nt]
Thanks in advance for your suggestions.
To generate bigrams:
In [8]: b='student'
In [9]: [b[i:i+2] for i in range(len(b)-1)]
Out[9]: ['st', 'tu', 'ud', 'de', 'en', 'nt']
To generalize to a different n:
In [10]: n=4
In [11]: [b[i:i+n] for i in range(len(b)-n+1)]
Out[11]: ['stud', 'tude', 'uden', 'dent']
Try zip:
>>> def word2ngrams(text, n=3, exact=True):
... """ Convert text into character ngrams. """
... return ["".join(j) for j in zip(*[text[i:] for i in range(n)])]
...
>>> word2ngrams('foobarbarblacksheep')
['foo', 'oob', 'oba', 'bar', 'arb', 'rba', 'bar', 'arb', 'rbl', 'bla', 'lac', 'ack', 'cks', 'ksh', 'she', 'hee', 'eep']
but do note that it's slower:
import string, random, time
def zip_ngrams(text, n=3, exact=True):
return ["".join(j) for j in zip(*[text[i:] for i in range(n)])]
def nozip_ngrams(text, n=3):
return [text[i:i+n] for i in range(len(text)-n+1)]
# Generate 10000 random strings of length 100.
words = [''.join(random.choice(string.ascii_uppercase) for j in range(100)) for i in range(10000)]
start = time.time()
x = [zip_ngrams(w) for w in words]
print time.time() - start
start = time.time()
y = [nozip_ngrams(w) for w in words]
print time.time() - start
print x==y
[out]:
0.314492940903
0.197558879852
True
Although late, NLTK has an inbuilt function that implements ngrams
# python 3
from nltk import ngrams
["".join(k1) for k1 in list(ngrams("hello world",n=3))]
['hel', 'ell', 'llo', 'lo ', 'o w', ' wo', 'wor', 'orl', 'rld']
Ths fucntion gives you ngrams for n = 1 to n:
def getNgrams(sentences, n):
ngrams = []
for sentence in sentences:
_ngrams = []
for _n in range(1,n+1):
for pos in range(1,len(sentence)-_n):
_ngrams.append([sentence[pos:pos+_n]])
ngrams.append(_ngrams)
return ngrams