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evalf and subs in sympy on single variable expression returns expression instead of expected float value

I'm new to sympy and I'm trying to use it to get the values of higher order Greeks of options (basically higher order derivatives). My goal is to do a Taylor series expansion. The function in question is the first derivative.
f(x) = N(d1)
N(d1) is the P(X <= d1) of a standard normal distribution. d1 in turn is another function of x (x in this case is the price of the stock to anybody who's interested).
d1 = (np.log(x/100) + (0.01 + 0.5*0.11**2)*0.5)/(0.11*np.sqrt(0.5))
As you can see, d1 is a function of only x. This is what I have tried so far.
import sympy as sp
from math import pi
from sympy.stats import Normal,P
x = sp.symbols('x')
u = (sp.log(x/100) + (0.01 + 0.5*0.11**2)*0.5)/(0.11*np.sqrt(0.5))
N = Normal('N',0,1)
f = sp.simplify(P(N <= u))
print(f.evalf(subs={x:100})) # This should be 0.5155
f1 = sp.simplify(sp.diff(f,x))
f1.evalf(subs={x:100}) # This should also return a float value
The last line of code however returns an expression, not a float value as I expected like in the case with f. I feel like I'm making a very simple mistake but I can't find out why. I'd appreciate any help.
Thanks.

If you define x with positive=True (which is implied by the log in the definition of u assuming u is real which is implied by the definition of f) it looks like you get almost the expected result (also using f1.subs({x:100}) in the version without the positive x assumption shows the trouble is with unevaluated polar_lift(0) terms):
import sympy as sp
from sympy.stats import Normal, P
x = sp.symbols('x', positive=True)
u = (sp.log(x/100) + (0.01 + 0.5*0.11**2)*0.5)/(0.11*sp.sqrt(0.5)) # changed np to sp
N = Normal('N',0,1)
f = sp.simplify(P(N <= u))
print(f.evalf(subs={x:100})) # 0.541087287864516
f1 = sp.simplify(sp.diff(f,x))
print(f1.evalf(subs={x:100})) # 0.0510177033783834

Getting an Array/Vector from PARI/GP in Python using Ctypes

I have written a code to compare the solution of sympy and PARI/GP, how ever I am facing a problem to get an array/vector from PARI/GP.
When I try to return the vector res from PARI/GP function nfroots, I get a address like this (see the last line) -
[3, 4]
elements as long (only if of type t_INT):
3
4
<__main__.LP_LP_c_long object at 0x00000000056166C8>
how can I get the res as vector/array from nfroots so I can use that array like normal python vector/array?
The code is given below to download the libpari.dll file, click here-
from ctypes import *
from sympy.solvers import solve
from sympy import Symbol
pari = cdll.LoadLibrary("libpari.dll")
pari.stoi.restype = POINTER(c_long)
pari.cgetg.restype = POINTER(POINTER(c_long))
pari.gtopoly.restype = POINTER(c_long)
pari.nfroots.restype = POINTER(POINTER(c_long))
(t_VEC, t_COL, t_MAT) = (17, 18, 19) # incomplete
pari.pari_init(2 ** 19, 0)
def t_vec(numbers):
l = len(numbers) + 1
p1 = pari.cgetg(c_long(l), c_long(t_VEC))
for i in range(1, l):
#Changed c_long to c_float, but got no output
p1[i] = pari.stoi(c_long(numbers[i - 1]))
return p1
def Quartic_Comparison():
x = Symbol('x')
#a=0;A=0;B=1;C=-7;D=13/12 #PROBLEM 1
a=0;A=0;B=1;C=-7;D=12
#a=0;A=0;B=-1;C=-2;D=1
solution=solve(a*x**4+A*x**3+B*x**2+ C*x + D, x)
print(solution)
V=(A,B,C,D)
P = pari.gtopoly(t_vec(V), c_long(-1))
res = pari.nfroots(None, P)
print("elements as long (only if of type t_INT): ")
for i in range(1, pari.glength(res) + 1):
print(pari.itos(res[i]))
return res #PROBLEM 2
f=Quartic_Comparison()
print(f)

res is an element from the PARI/C world. It is a PARI vector of PARI integers (t_VEC of t_INTs). Python does not know it.
If it is to be processed further on the Python side, it must be converted. This is generally necessary if data needs to be exchanged between Python and the PARI/C world.
So if you have a t_VEC with t_INTs on the PARI/C side, as in this case, you most likely want to convert it to a Python list.
One possible approach might look like this:
...
roots = pari.nfroots(None, P)
result = []
for i in range(1, pari.glength(roots) + 1):
result.append(pari.itos(roots[i]))
return result

SymPy: how can i get the partial derivatives of one equation input beforehand by the user?

Im trying to build a constrained optimization calculator in python using the sympy module. The idea is that a user can enter two functions, "f" and "g", which then are put together to form the equation "L".
I want SymPy to give me the partial derivatives of x, y and lambda in "L", however my code does not seem to be working. When trying to get their partial derivatives i get the following results:
0
0
-x - 4*y + 500
I used x+100*y-y**2 as function 1 and x+4*y-500 and function 2.
Heres the code so far:
import sympy as sp
from sympy.parsing import sympy_parser
x, y = sp.symbols("x y", real=True)
lam = sp.symbols('lambda', real=True)
insert = input("Insert function 1:") #function 1
f = sympy_parser.parse_expr(insert) #transforming the function into a sympy expression
print(f)
insert2 = input("Insert function 2:") #function2
g = sympy_parser.parse_expr(insert2) #transforming function 2
L = f - lam*g #getting the equation "L"
xx = sp.diff(L, x) #partial derivative L'x
yy = sp.diff(L, y) #partial derivative L'y
ll = sp.diff(L, lam) #partial derivative L'lam
print(xx)
print(yy)
print(ll)
I have tried both the "parse_expr" and "simpify" commands to transform the functions input by the user from string to sympy expressions. I might be missing something else.

Your local x and y are real but those that the parser returns are vanilla, they have not assumptions. Since symbols match by name and assumptions, your input functions don't have (the same) x and y:
>>> f.has(x)
False
So either don't make your local symbols real
>>> var('x')
x
>>> f = sympy_parser.parse_expr(insert)
>>> f.has(x)
True
Or pass your local symbols to the parser so it can use them to build your functions:
>>> f = sympy_parser.parse_expr(insert, dict(x=x,y=y))
>>> f.has(x)
True
And once you are using the same symbols, the rest of your issues should move to a new level :-)

Writing a function for x * sin(3/x) in python

I have to write a function, s(x) = x * sin(3/x) in python that is capable of taking single values or vectors/arrays, but I'm having a little trouble handling the cases when x is zero (or has an element that's zero). This is what I have so far:
def s(x):
result = zeros(size(x))
for a in range(0,size(x)):
if (x[a] == 0):
result[a] = 0
else:
result[a] = float(x[a] * sin(3.0/x[a]))
return result
Which...doesn't work for x = 0. And it's kinda messy. Even worse, I'm unable to use sympy's integrate function on it, or use it in my own simpson/trapezoidal rule code. Any ideas?
When I use integrate() on this function, I get the following error message: "Symbol" object does not support indexing.

This takes about 30 seconds per integrate call:
import sympy as sp
x = sp.Symbol('x')
int2 = sp.integrate(x*sp.sin(3./x),(x,0.000001,2)).evalf(8)
print int2
int1 = sp.integrate(x*sp.sin(3./x),(x,0,2)).evalf(8)
print int1
The results are:
1.0996940
-4.5*Si(zoo) + 8.1682775
Clearly you want to start the integration from a small positive number to avoid the problem at x = 0.
You can also assign x*sin(3./x) to a variable, e.g.:
s = x*sin(3./x)
int1 = sp.integrate(s, (x, 0.00001, 2))
My original answer using scipy to compute the integral:
import scipy.integrate
import math
def s(x):
if abs(x) < 0.00001:
return 0
else:
return x*math.sin(3.0/x)
s_exact = scipy.integrate.quad(s, 0, 2)
print s_exact
See the scipy docs for more integration options.

If you want to use SymPy's integrate, you need a symbolic function. A wrong value at a point doesn't really matter for integration (at least mathematically), so you shouldn't worry about it.
It seems there is a bug in SymPy that gives an answer in terms of zoo at 0, because it isn't using limit correctly. You'll need to compute the limits manually. For example, the integral from 0 to 1:
In [14]: res = integrate(x*sin(3/x), x)
In [15]: ans = limit(res, x, 1) - limit(res, x, 0)
In [16]: ans
Out[16]:
9⋅π 3⋅cos(3) sin(3) 9⋅Si(3)
- ─── + ──────── + ────── + ───────
4 2 2 2
In [17]: ans.evalf()
Out[17]: -0.164075835450162

How do you make this code more pythonic?

Could you guys please tell me how I can make the following code more pythonic?
The code is correct. Full disclosure - it's problem 1b in Handout #4 of this machine learning course. I'm supposed to use newton's algorithm on the two data sets for fitting a logistic hypothesis. But they use matlab & I'm using scipy
Eg one question i have is the matrixes kept rounding to integers until I initialized one value to 0.0. Is there a better way?
Thanks
import os.path
import math
from numpy import matrix
from scipy.linalg import inv #, det, eig
x = matrix( '0.0;0;1' )
y = 11
grad = matrix( '0.0;0;0' )
hess = matrix('0.0,0,0;0,0,0;0,0,0')
theta = matrix( '0.0;0;0' )
# run until convergence=6or7
for i in range(1, 6):
#reset
grad = matrix( '0.0;0;0' )
hess = matrix('0.0,0,0;0,0,0;0,0,0')
xfile = open("q1x.dat", "r")
yfile = open("q1y.dat", "r")
#over whole set=99 items
for i in range(1, 100):
xline = xfile.readline()
s= xline.split(" ")
x[0] = float(s[1])
x[1] = float(s[2])
y = float(yfile.readline())
hypoth = 1/ (1+ math.exp(-(theta.transpose() * x)))
for j in range(0,3):
grad[j] = grad[j] + (y-hypoth)* x[j]
for k in range(0,3):
hess[j,k] = hess[j,k] - (hypoth *(1-hypoth)*x[j]*x[k])
theta = theta - inv(hess)*grad #update theta after construction
xfile.close()
yfile.close()
print "done"
print theta

One obvious change is to get rid of the "for i in range(1, 100):" and just iterate over the file lines. To iterate over both files (xfile and yfile), zip them. ie replace that block with something like:
import itertools
for xline, yline in itertools.izip(xfile, yfile):
s= xline.split(" ")
x[0] = float(s[1])
x[1] = float(s[2])
y = float(yline)
...
(This is assuming the file is 100 lines, (ie. you want the whole file). If you're deliberately restricting to the first 100 lines, you could use something like:
for i, xline, yline in itertools.izip(range(100), xfile, yfile):
However, its also inefficient to iterate over the same file 6 times - better to load it into memory in advance, and loop over it there, ie. outside your loop, have:
xfile = open("q1x.dat", "r")
yfile = open("q1y.dat", "r")
data = zip([line.split(" ")[1:3] for line in xfile], map(float, yfile))
And inside just:
for (x1,x2), y in data:
x[0] = x1
x[1] = x2
...

x = matrix([[0.],[0],[1]])
theta = matrix(zeros([3,1]))
for i in range(5):
grad = matrix(zeros([3,1]))
hess = matrix(zeros([3,3]))
[xfile, yfile] = [open('q1'+a+'.dat', 'r') for a in 'xy']
for xline, yline in zip(xfile, yfile):
x.transpose()[0,:2] = [map(float, xline.split(" ")[1:3])]
y = float(yline)
hypoth = 1 / (1 + math.exp(theta.transpose() * x))
grad += (y - hypoth) * x
hess -= hypoth * (1 - hypoth) * x * x.transpose()
theta += inv(hess) * grad
print "done"
print theta

the matrixes kept rounding to integers until I initialized one value
to 0.0. Is there a better way?
At the top of your code:
from __future__ import division
In Python 2.6 and earlier, integer division always returns an integer unless there is at least one floating point number within. In Python 3.0 (and in future division in 2.6), division works more how we humans might expect it to.
If you want integer division to return an integer, and you've imported from future, use a double //. That is
from __future__ import division
print 1//2 # prints 0
print 5//2 # prints 2
print 1/2 # prints 0.5
print 5/2 # prints 2.5

You could make use of the with statement.

the code that reads the files into lists could be drastically simpler
for line in open("q1x.dat", "r"):
x = map(float,line.split(" ")[1:])
y = map(float, open("q1y.dat", "r").readlines())