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Comparing R and Python Vectorization and Optimization

In the R language, optimization can be achieved by using purrr::map() or furrr::future_map() functions. However, I am not sure how does optimization works for np.array() methods. Indeed, I would like to understand how does Python and R scales out to parallel processing [1, 2] in terms of complexity and performance.
Thus, the following questions arise:
How does the optimization of np.array() in Python works comparing to purrr::map() and furrr::future_map() functions in the R language?
By doing a simple tictoc test on purrr/furrr, I can observe that we have a big win from vectorization in both cases. Nonetheless, I can also notice that the results seem to show that the R language is just fundamentally faster.
Python
import time
a = np.random.rand(1000000)
b = np.random.rand(1000000)
tic = time.time()
c = np.dot(a,b)
toc = time.time()
print ("Vectorized version:" + str(1000*(toc-tic)) +"ms")
c = 0
tic = time.time()
for i in range(1000000):
c += a[i]*b[i]
toc = time.time()
print("For loop:" + str(1000*(toc-tic)) +"ms")
Output
Vectorized version: 54.151296615600586ms
For loop: 676.0082244873047ms
R
a <- runif(1000000,0,1)
b <- runif(1000000,0,1)
c = 0
tictoc::tic()
c = sum(a * b)
tictoc::toc()
c = 0
tictoc::tic()
for (i in 1:length(a)) {
c = a[i]*b[i] + c
}
tictoc::toc()
Output
Vectorized version: 0.013 sec elapsed
For loop: 0.065 sec elapsed
References
[1] Ross Ihaka & Robert Gentleman (1996) R: A Language for Data Analysis and Graphics, Journal of Computational and Graphical Statistics, 5:3, 299-314, DOI: 10.1080/10618600.1996.10474713
[2] S. van der Walt, S. C. Colbert and G. Varoquaux, "The NumPy Array: A Structure for Efficient Numerical Computation," in Computing in Science & Engineering, vol. 13, no. 2, pp. 22-30, March-April 2011, doi: 10.1109/MCSE.2011.37

I believe numpy wraps some of its "primitive" objects in wrapper classes which are, themselves, Python (eg. this one).
When looking at the R mirror source, I conversely find an array class that's basically native code (aka C).
That extra indirection layer alone could explain the difference in speed, I guess.

Sympy simplification of maximum

I don't understand why Sympy won't return to me the expression below simplified (not sure its a bug in my code or a feature of Sympy).
import sympy as sp
a = sp.Symbol('a',finite = True, real = True)
b = sp.Symbol('b',finite = True, real = True)
sp.assumptions.assume.global_assumptions.add(sp.Q.positive(b))
sp.assumptions.assume.global_assumptions.add(sp.Q.negative(a))
sp.simplify(sp.Max(a-b,a+b))
I would expect the output to be $a+b$, but Sympy still gives me $Max(a-b,a+b)$.
Thanks; as you can see I am a beginner in Sympy so any hints/help are appreciated.

Surely the result should be a + b...
You can do this by setting the assumptions on the symbol as in:
In [2]: a = Symbol('a', negative=True)
In [3]: b = Symbol('b', positive=True)
In [4]: Max(a - b, a + b)
Out[4]: a + b
You are trying to use the new assumptions system but that system is still experimental and is not widely used within sympy. The new assumptions are not used in core evaluation so e.g. the Max function has no idea that you have declared global assumptions on a and b unless those assumptions are declared on the symbols as I show above.

Error with quantifier in Z3Py

I would like Z3 to check whether it exists an integer t that satisfies my formula. I'm getting the following error:
Traceback (most recent call last):
File "D:/z3-4.6.0-x64-win/bin/python/Expl20180725.py", line 18, in <module>
g = ForAll(t, f1(t) == And(t>=0, t<10, user[t].rights == ["read"] ))
TypeError: list indices must be integers or slices, not ArithRef
Code:
from z3 import *
import random
from random import randrange
class Struct:
def __init__(self, **entries): self.__dict__.update(entries)
user = [Struct() for i in range(10)]
for i in range(10):
user[i].uid = i
user[i].rights = random.choice(["create","execute","read"])
s=Solver()
f1 = Function('f1', IntSort(), BoolSort())
t = Int('t')
f2 = Exists(t, f1(t))
g = ForAll(t, f1(t) == And(t>=0, t<10, user[t].rights == ["read"] ))
s.add(g)
s.add(f2)
print(s.check())
print(s.model())

You are mixing and matching Python and Z3 expressions, and while that is the whole point of Z3py, it definitely does not mean that you can mix/match them arbitrarily. In general, you should keep all the "concrete" parts in Python, and relegate the symbolic parts to "z3"; carefully coordinating the interaction in between. In your particular case, you are accessing a Python list (your user) with a symbolic z3 integer (t), and that is certainly not something that is allowed. You have to use a Z3 symbolic Array to access with a symbolic index.
The other issue is the use of strings ("create"/"read" etc.) and expecting them to have meanings in the symbolic world. That is also not how z3py is intended to be used. If you want them to mean something in the symbolic world, you'll have to model them explicitly.
I'd strongly recommend reading through http://ericpony.github.io/z3py-tutorial/guide-examples.htm which is a great introduction to z3py including many of the advanced features.
Having said all that, I'd be inclined to code your example as follows:
from z3 import *
import random
Right, (create, execute, read) = EnumSort('Right', ('create', 'execute', 'read'))
users = Array('Users', IntSort(), Right)
for i in range(10):
users = Store(users, i, random.choice([create, execute, read]))
s = Solver()
t = Int('t')
s.add(t >= 0)
s.add(t < 10)
s.add(users[t] == read)
r = s.check()
if r == sat:
print s.model()[t]
else:
print r
Note how the enumerated type Right in the symbolic land is used to model your "permissions."
When I run this program multiple times, I get:
$ python a.py
5
$ python a.py
9
$ python a.py
unsat
$ python a.py
6
Note how unsat is produced, if it happens that the "random" initialization didn't put any users with a read permission.

Python Sympy Latex Fraction won't print without factoring first

I'm working with Python Sympy, solving a quadratic, and wanting to print the result using LaTex. For example, if the result is x = (1 + sqrt(3))/2, I would like it to print via LaTex as \frac{1 + \sqrt{3}}{2}. However, Python Sympy either splits this into two fractions, as in \frac{1}{2} + \frac{\sqrt{3}}{2}, OR factors out the half, as in \frac{1}{2}(1 + \sqrt{3}). I have tried to return the numerator via sympy.fraction(expr) and have viewed other articles (Sympy - fraction manipulation and others), but none were able to produce the result.

Check out how to override the default printers.
import sympy
from sympy.printing.latex import LatexPrinter # necessary because latex is both a function and a module
class CustomLatexPrinter(LatexPrinter):
def _print_Add(self, expr):
n, d = expr.as_numer_denom()
if d == sympy.S.One:
# defer to the default printing mechanism
super()._print_Add(expr)
return
return r'\frac{%s}{%s}' % (sympy.latex(n), sympy.latex(d))
# doing this should override the default latex printer globally
# adopted from "Examples of overloading StrPrinter" in the sympy documentation
sympy.printing.latex = lambda self: CustomLatexPrinter().doprint(self)
print(sympy.printing.latex((1 + sympy.sqrt(3)) / 2)) # \frac{1 + \sqrt{3}}{2}

Formatting Complex Numbers

For a project in one of my classes we have to output numbers up to five decimal places.It is possible that the output will be a complex number and I am unable to figure out how to output a complex number with five decimal places. For floats I know it is just:
print "%0.5f"%variable_name
Is there something similar for complex numbers?

You could do it as is shown below using the str.format() method:
>>> n = 3.4+2.3j
>>> n
(3.4+2.3j)
>>> '({0.real:.2f} + {0.imag:.2f}i)'.format(n)
'(3.40 + 2.30i)'
>>> '({c.real:.2f} + {c.imag:.2f}i)'.format(c=n)
'(3.40 + 2.30i)'
To make it handle both positive and negative imaginary portions properly, you would need a (even more) complicated formatting operation:
>>> n = 3.4-2.3j
>>> n
(3.4-2.3j)
>>> '({0:.2f} {1} {2:.2f}i)'.format(n.real, '+-'[n.imag < 0], abs(n.imag))
'(3.40 - 2.30i)'
Update - Easier Way
Although you cannot use f as a presentation type for complex numbers using the string formatting operator %:
n1 = 3.4+2.3j
n2 = 3.4-2.3j
try:
print('test: %.2f' % n1)
except Exception as exc:
print('{}: {}'.format(type(exc).__name__, exc))
Output:
TypeError: float argument required, not complex
You can however use it with complex numbers via the str.format() method. This isn't explicitly documented, but is implied by the Format Specification Mini-Language documentation which just says:
'f'  Fixed point. Displays the number as a fixed-point number. The default precision is 6.
. . .so it's easy to overlook.
In concrete terms, the following works in both Python 2.7.14 and 3.4.6:
print('n1: {:.2f}'.format(n1))
print('n2: {:.2f}'.format(n2))
Output:
n1: 3.10+4.20j
n2: 3.10-4.20j
This doesn't give you quite the control the code in my original answer does, but it's certainly much more concise (and handles both positive and negative imaginary parts automatically).
Update 2 - f-strings
Formatted string literals (aka f-strings) were added in Python 3.6, which means it could also be done like this in that version or later:
print(f'n1: {n1:.2f}') # -> n1: 3.40+2.30j
print(f'n2: {n2:.3f}') # -> n2: 3.400-2.300j
In Python 3.8.0, support for an = specifier was added to f-strings, allowing you to write:
print(f'{n1=:.2f}') # -> n1=3.40+2.30j
print(f'{n2=:.3f}') # -> n2=3.400-2.300j

Neither String Formatting Operations - i.e. the modulo (%) operator) -
nor the newer str.format() Format String Syntax support complex types.
However it is possible to call the __format__ method of all built in numeric types directly.
Here is an example:
>>> i = -3 # int
>>> l = -33L # long (only Python 2.X)
>>> f = -10./3 # float
>>> c = - 1./9 - 2.j/9 # complex
>>> [ x.__format__('.3f') for x in (i, l, f, c)]
['-3.000', '-33.000', '-3.333', '-0.111-0.222j']
Note, that this works well with negative imaginary parts too.

For questions like this, the Python documentation should be your first stop. Specifically, have a look at the section on string formatting. It lists all the string format codes; there isn't one for complex numbers.
What you can do is format the real and imaginary parts of the number separately, using x.real and x.imag, and print it out in a + bi form.

>>> n = 3.4 + 2.3j
>>> print '%05f %05fi' % (n.real, n.imag)
3.400000 2.300000i

As of Python 2.6 you can define how objects of your own classes respond to format strings. So, you can define a subclass of complex that can be formatted. Here's an example:
>>> class Complex_formatted(complex):
... def __format__(self, fmt):
... cfmt = "({:" + fmt + "}{:+" + fmt + "}j)"
... return cfmt.format(self.real, self.imag)
...
>>> z1 = Complex_formatted(.123456789 + 123.456789j)
>>> z2 = Complex_formatted(.123456789 - 123.456789j)
>>> "My complex numbers are {:0.5f} and {:0.5f}.".format(z1, z2)
'My complex numbers are (0.12346+123.45679j) and (0.12346-123.45679j).'
>>> "My complex numbers are {:0.6f} and {:0.6f}.".format(z1, z2)
'My complex numbers are (0.123457+123.456789j) and (0.123457-123.456789j).'
Objects of this class behave exactly like complex numbers except they take more space and operate more slowly; reader beware.

Check this out:
np.set_printoptions(precision=2) # Rounds up to 2 decimals all float expressions
I've successfully printed my complexfloat's expressions:
# Show poles and zeros
print( "zeros = ", zeros_H , "\n")
print( "poles = ", poles_H )
out before:
zeros = [-0.8 +0.6j -0.8 -0.6j -0.66666667+0.j ]
poles = [-0.81542318+0.60991027j -0.81542318-0.60991027j -0.8358203 +0.j ]
out after:
zeros = [-0.8 +0.6j -0.8 -0.6j -0.67+0.j ]
poles = [-0.82+0.61j -0.82-0.61j -0.84+0.j ]