If the price charged for a crayon is p cents, then x thousand crayons
will be sold in a certain school store, where p(x)= 122-x/34 .
Using Python, calculate how many crayons must be sold to maximize
revenue.
I can solve this by hand much easily, the only problem is how can I do it using plain Python? I am using IDLE (Python GUI). I am new to Python and haven't downloaded any external libraries. Any help will be greatly appreciated.
What I've done up to this point is
import math
def f(x):
return (122-(x/34.0))
def g(x):
return x*f(x)
def h(x):
return (122-(2*x/34.0))
Use SymPy. It's simple, beautiful and powerful.
You can write down your equations with simpify(), like that:
p = simpify('122 - x/34')
And define symbols for symbolic evaluation with Symbol() and symbols().
With that you can do things like simply use solve() function for any given equation. i.e. x + 4 = 2x:
res = solve('x + 4 - 2*x')
It's pretty much the tool I use for any math work with python.
So, you should go and download an external library for this, as it's not functionality that python makes easy to implement natively. Also, if you're serious about doing mathematical computation in python I would suggest switching operating systems to something like OSX or linux, simply because compiling old FORTRAN libraries (required for much performant mathematical computing) is a huge pain on Windows.
You have to make use of the scipy library here, which has an optimize module. Specifically I would suggest using the optimize.minimize_scalar function. Docs can be found here.
>>> from scipy.optimize import minimize_scalar
>>> def g(x):
... return -(x*(122 - (x/34))) # inverse because you're minimizing.
>>> minimize_scalar(g, bounds=(1, 10000), method='bounded')
status: 0
nfev: 6
success: True
fun: -126514.0
x: 2074.0
message: 'Solution found.'
Related
I was reading the wiki page for Numba, and it says Numba is a "compiler". But then later on, it says that to use Numba, you import it like a package. I later looked up how to use Numba, and indeed, you just pip install it.
So now I am confused. I thought Numba was a compiler? But it seems to be used just like any other package, like numpy or pandas? What's the difference?
A compiler is a program that inputs something in human-readable form (usually a program in a specified language) and outputs a functionally equivalent stream in another, more machine-digestible form. Just as with any other transformation, it's equally viable as a command-line invocation or a function call.
As long as it's wrapped properly in a package for general use, it's perfect reasonable to deliver a compiler as a Python package.
Does that clear up the difficulty?
From what I have read at Numba documentation it's a package that you import to you project and then use the Numba decorator do indicate parts of your code that you would like to have compiled in JIT (Just in Time) in order to optimize them. Like in the following example:
from numba import jit
import random
#jit(nopython=True)
def monte_carlo_pi(nsamples):
acc = 0
for i in range(nsamples):
x = random.random()
y = random.random()
if (x ** 2 + y ** 2) < 1.0:
acc += 1
return 4.0 * acc / nsamples
When the monte_carlo_pi function is called Numba will have it compiled in order to optimize it, so there isn't a compilation step that you can take.
I'm interested in math but I don't know a lot about coding on python I want to write a code in python that calculate:
4*(1-(1/3)+(1/5)-(1/7)+...+(1/2n-1))
that convergence to pi. I want a python code so that I import n for example 1,2,3,1000,...and see the answer.
Here.
def pi_approx(n):
return 4*sum([((-1)**i)/(2*i+1) for i in range(n)])
I am trying to use the basic sin, cos, arctan, etc function from numpy, but I want to use gradians. I have search the doc without success, and search for other python modules without luck. Any suggestion on a python module i could use?
Or a function that will work. I have tried different methods to convert grad to rad and back to grad again, but no-one is working.
This one should be fine !
def gradFromRad(rad):
return 200*rad/math.pi
def radFromGrad(grad):
return math.pi*grad/200
Given 2 large arrays of 3D points (I'll call the first "source", and the second "destination"), I needed a function that would return indices from "destination" which matched elements of "source" as its closest, with this limitation: I can only use numpy... So no scipy, pandas, numexpr, cython...
To do this i wrote a function based on the "brute force" answer to this question. I iterate over elements of source, find the closest element from destination and return its index. Due to performance concerns, and again because i can only use numpy, I tried multithreading to speed it up. Here are both threaded and unthreaded functions and how they compare in speed on an 8 core machine.
import timeit
import numpy as np
from numpy.core.umath_tests import inner1d
from multiprocessing.pool import ThreadPool
def threaded(sources, destinations):
# Define worker function
def worker(point):
dlt = (destinations-point) # delta between destinations and given point
d = inner1d(dlt,dlt) # get distances
return np.argmin(d) # return closest index
# Multithread!
p = ThreadPool()
return p.map(worker, sources)
def unthreaded(sources, destinations):
results = []
#for p in sources:
for i in range(len(sources)):
dlt = (destinations-sources[i]) # difference between destinations and given point
d = inner1d(dlt,dlt) # get distances
results.append(np.argmin(d)) # append closest index
return results
# Setup the data
n_destinations = 10000 # 10k random destinations
n_sources = 10000 # 10k random sources
destinations= np.random.rand(n_destinations,3) * 100
sources = np.random.rand(n_sources,3) * 100
#Compare!
print 'threaded: %s'%timeit.Timer(lambda: threaded(sources,destinations)).repeat(1,1)[0]
print 'unthreaded: %s'%timeit.Timer(lambda: unthreaded(sources,destinations)).repeat(1,1)[0]
Retults:
threaded: 0.894030461056
unthreaded: 1.97295164054
Multithreading seems beneficial but I was hoping for more than 2X increase given the real life dataset i deal with are much larger.
All recommendations to improve performance (within the limitations described above) will be greatly appreciated!
Ok, I've been reading Maya documentation on python and I came to these conclusions/guesses:
They're probably using CPython inside (several references to that documentation and not any other).
They're not fond of threads (lots of non-thread safe methods)
Since the above, I'd say it's better to avoid threads. Because of the GIL problem, this is a common problem and there are several ways to do the earlier.
Try to build a tool C/C++ extension. Once that is done, use threads in C/C++. Personally, I'd only try SIP to work, and then move on.
Use multiprocessing. Even if your custom python distribution doesn't include it, you can get to a working version since it's all pure python code. multiprocessing is not affected by the GIL since it spawns separate processes.
The above should've worked out for you. If not, try another parallel tool (after some serious praying).
On a side note, if you're using outside modules, be most mindful of trying to match maya's version. This may have been the reason because you couldn't build scipy. Of course, scipy has a huge codebase and the windows platform is not the most resilient to build stuff.
When I run this program, I get no solution at the end, but there should be a solution ( I believe). Any idea what I am doing wrong? If you take away the Q from e2 equation it seems to work correctly.
#!/usr/bin/python
from sympy import *
a,b,w,r = symbols('a b w r',real=True,positive=True)
L,K,Q = symbols('L K Q',real=True,positive=True)
e1=K
e2=(K*Q/2)**(a)
print solve(e1-e2,K)
It works if we do the following:
Set Q=1 or,
Change e2 to e2=(K*a)(Q/2)**(a)
I would still like it to work in the original way though, as my equations are more complicated than this.
This is just a deficiency of solve. solve is based mostly on heuristics, so sometimes it isn't able to figure out how to solve an equation when it's given in a particular form. The workaround here is to just call expand_power_base on the expression, since SymPy is able to solve K - K**a*(Q/2)**a:
In [8]: print(solve(expand_power_base(e1-e2),K))
[(2/Q)**(a/(a - 1))]
It's also worth pointing out that the result of [] from solve does not in any way mean that there are no solutions, only that solve was unable to find any. See the first note at http://docs.sympy.org/latest/tutorial/solvers.html.