I am working with sagemath which uses the python language. In the polynomial ring over the integer ring, I want to define a set whose elements have degree are less than a given number and the absolute value of coefficients are less than a given number.
How to achieve this? For a polynomial, I have defined the degree function
and the max_coefficient function already.
For example,
(x^3-3*x-5).degree(x) will return 3
max_coefficient(x^3-3*x-5) will return 5
The following are my codes.
R=Polynomialring(ZZ,x)
def A(deg_bound,coefficient_bound):
S=set()
for poly in R:
if poly.degree(x)<=deg_bound and max_coefficient(poly)<=coefficient_bound:
S=S.add(poly)
return S
But sagemath tells me I can't do for in the polynomial ring.
It is rather unclear what sort of objects are in R, but you seemto know how to manipulate them...
There is one error that may be causing you trouble:
S = S.add(poly)
is first adding poly to S, then assigning None to S, which is rather unfortunate.
try to replace it with:
S.add(poly)
which accumulates distinct objects into S
Related
I am trying to get integrated values from np.array, list of values. Not the surface under the function, but values. I have values of acceleration and want to get values of velocity.
So let's say I have an arry like:
a_x = np.array([111.2, 323.2, 123.3, 99.38, 65.23, -0.19, -34.67])
And I try to get integrated values from this array to get the values of velocity.
If I use lets say simps, quad, trapz, I get the one number (surface).
So how do you integrate np.array values and get integrated values that you can store in a list?
You can't do it by the way you want it, because you didn't understand the process behind it. If you are given acceleration, then using the following equation:
You are able only to find INDEFINITE integral, you know the acceleration, but you don't know starting conditions, thus your solution can't be empty.
As the solution to each of those questions is: "Find velocity given an acceleration", then the solution would be v(t)=integral of a(t)dt+c, where your acceleration is constant, so it doesn't rely on t and it can be written as v(t)=at+c, but still - we don't know anything about how long acceleration lasted and what is the starting condition.
But answering the question about getting values which can be stored in a list - you do it by indexing your values of np.array:
import numpy as np
a_x = np.array([111.2,323.2,123.3])
#Gets first value
print(a_x[0])
If I use lets say simps, quad, trapz, I get the one number (surface).
Because quad,simps,or trapz are methods used for given points, which return value of integral with those given points with corresponding method, for example:
numpy.trapz(y, x=None, dx=1.0, axis=- 1)
if x isn't specified (as in your case), it assumes that you want to use trapeze to estimate the field under the value y of given points with x equal distribution of x. It has to give one value.
My question is about matching this function: N = 0.13*(s^a), where s and a are variables, to a value. I am trying to find all values of s and a that satisfy N = 100 and N = 10,000,000. S is bounded from 0 to 101 and a is bounded from 3 to 8. And I would like to visualize the results possibly by graphing it with the axes being s and a, like a 2D plot. The algorithms I found that were similar to what I need seemed to all want to find the minimum or maximum of a function instead of matching it to a value. I have hit a wall and I don't know if my coding skills are high enough to write my own algorithm. Any help would be greatly appreciated! Thanks in advance!
This can easily be converted to a minimization problem. Simply minimize this function:
abs(0.13 * s ^ a - 100)
Replace the 100 with 10,000,000 for the second part. It will take some modification to find all values of s and a, rather than just one pair. This could be done by fixing an s value and minimizing over a, then repeating for different s values.
Why does the following code return a ValueError?
from scipy.optimize import fsolve
import numpy as np
def f(p,a=0):
x,y = p
return (np.dot(x,y)-a,np.outer(x,y)-np.ones((3,3)),x+y-np.array([1,2,3]))
x,y = fsolve(f,(np.ones(3),np.ones(3)),9)
ValueError: setting an array element with a sequence.
The basic problem here is that your function f does not satisfy the criteria required for fsolve to work. These criteria are described in the documentation - although arguably not very clearly.
The particular things that you need to be aware of are:
the input to the function that will be solved for must be an n-dimensional vector (referred to in the docs as ndarray), such that the value of x you want is the solution to f(x, *args) = 0.
the output of f must be the same shape as the x input to f.
Currently, your function takes a 2 member tuple of 1x3-arrays (in p) and a fixed scalar offset (in a). It returns a 3 member tuple of types (scalar,3x3 array, 1x3 array)
As you can see, neither condition 1 nor 2 is met.
It is hard to advise you on exactly how to fix this without being exactly sure of the equation you are trying to solve. It seems you are trying to solve some particular equation f(x,y,a) = 0 for x and y with x0 = (1,1,1) and y0 = (1,1,1) and a = 9 as a fixed value. You might be able to do this by passing in x and y concatenated (e.g. pass in p0 = (1,1,1,1,1,1) and in the function use x=p[:3] and y = p[3:] but then you must modify your function to output x and y concatenated into a 6-dimensional vector similarly. This depends on the exact function your are solving for and I can't work this out from the output of your existing f (i.e based on a dot product, outer product and sum based tuple).
Note that arguments that you don't pass in the vector (e.g. a in your case) will be treated as fixed values and won't be varied as part of the optimisation or returned as part of any solution.
Note for those who like the full story...
As the docs say:
fsolve is a wrapper around MINPACK’s hybrd and hybrj algorithms.
If we look at the MINPACK hybrd documentation, the conditions for the input and output vectors are more clearly stated. See the relevant bits below (I've cut some stuff out for clarity - indicated with ... - and added the comment to show that the input and output must be the same shape - indicated with <--)
1 Purpose.
The purpose of HYBRD is to find a zero of a system of N non-
linear functions in N variables by a modification of the Powell
hybrid method. The user must provide a subroutine which calcu-
lates the functions. The Jacobian is then calculated by a for-
ward-difference approximation.
2 Subroutine and type statements.
SUBROUTINE HYBRD(FCN,N,X, ...
...
FCN is the name of the user-supplied subroutine which calculates
the functions. FCN must be declared in an EXTERNAL statement
in the user calling program, and should be written as follows.
SUBROUTINE FCN(N,X,FVEC,IFLAG)
INTEGER N,IFLAG
DOUBLE PRECISION X(N),FVEC(N) <-- input X is an array length N, so is output FVEC
----------
CALCULATE THE FUNCTIONS AT X AND
RETURN THIS VECTOR IN FVEC.
----------
RETURN
END
N is a positive integer input variable set to the number of
functions and variables.
X is an array of length N. On input X must contain an initial
estimate of the solution vector. On output X contains the
final estimate of the solution vector.
I have a function which receives an integer as an input and depending on what range this input lies in, assigns to it a difficulty value. I know that this can be done using if else loops. I was wondering whether there is a more efficient/cleaner way to do it.
I tried to do something like this
TIME_RATING_KEY ={
range(0,46):1,
range(46,91):2,
range(91,136):3,
range(136,201):4,
range(201,10800):5,
}
But found out that we can use range as a key in dict(right?). So is there a better way to do this?
You can implement an interval tree. This kind of data structures are able to return all the intervals that intersect a given input point.
In your case intervals don't overlap, so they would always return 1 interval.
Centered interval trees run in O(log n + m) time, where m is the number of intervals returned (1 in your case). So this would reduce the complexity from O(n) to O(log n).
The idea of these interval trees is the following:
You consider the interval that encloses all the intervals you have
Take the center of that interval and partition the given intervals into those that end before that point, those that contain that point and those that start after it.
Recursively construct the same kind of tree for the intervals ending before the center and those starting after it
Keep the intervals that contain the center point in two sorted sequences. One sorted by starting point, and the other sorted by ending point
When searching go left or right depending on the center point. When you find an overlap you use binary search on the sorted sequence you want to check (this allows for looking up not only intervals that contain a given point but intervals that intersect or contain a given interval).
It's trivial to modify the data structure to return a specific value instead of the found interval.
This said, from the context I don't think you actually need to reduce the efficiency of this lookup and you should probably use the simpler and more readable solution since it would be more maintainable and there are less chances to make mistakes.
However reading about the mroe efficient data structure can turn out useful in the future.
The simplest way is probably just to write a short function:
def convert(n, difficulties=[0, 46, 91, 136, 201]):
if n < difficulties[0]:
raise ValueError
for difficulty, end in enumerate(difficulties):
if n < end:
return difficulty
else:
return len(difficulties)
Examples:
>>> convert(32)
1
>>> convert(68)
2
>>> convert(150)
4
>>> convert(250)
5
As a side note: You can use a range as a dictionary key in Python 3.x, but not directly in 2.x (because range returns a list). You could do:
TIME_RATING_KEY = {tuple(range(0, 46)): 1, ...}
However that won't be much help!
Write the function sinusoid(a, w, n) that will return a list of ordered pairs representing n cycles of a sinusoid with amplitude a and frequency w. Each cycle should contain 180 ordered pairs.
So far I have:
def sinusoid(a,w,n):
return [a*sin(x) for x in range 180]
Please consider the actual functional form of a sinusoidal wave and how the frequency comes into the equation. (Hint: http://en.wikipedia.org/wiki/Sine_wave).
Not sure what is meant exactly by 'ordered pairs', but I would assume it means the x,y pairs. Currently you're only returning a list of single values. Also you might want to take a look at the documentation for Python's sin function.
Okay, we know this is a homework assignment and we're not going to do it for you. However, I'll give you a couple hints.
The instructions:
Write the function sinusoid(a, w, n) that will return a list of ordered pairs representing n cycles of a sinusoid with amplitude a and frequency w. Each cycle should contain 180 ordered pairs.
... translated into a bullet list of requirements:
Write a function
... named sinusoid()
... taking three arguments: a, w, and n
returning a list
... of n cycles(?)
... (each consisting of?) 180 "ordered pairs"
The example you've given does define a function, by the correct name, and taking the correct number of arguments. That's a start (not much of one, frankly, but it's something).
The obvious failings are that it doesn't use two of the arguments that are required and it doesn't return pairs of anything. It seems that it would return 180 numbers which are based on the argument supplied to its first parameter.
Surely you can do a bit better than that.
Let's start with a stub:
def sinusoid(a, w, n):
'''Return n cycles of the sinusoid for a given amplitude and frequence
where each cycle consists of 180 ordered pairs
'''
results = list()
# do stuff here
return results
That's a function, takes three arguments and returns a list. Now for that list to contain anything before we return it we'll have to append some things to it ... and the instructions tell us how many things it should return (n times 180) and what sorts of things they should be (ordered pairs).
That sounds quite a bit like we'll need a loop (for n) and another (for 180). Hmmm ...
That might look like:
for each_cycle in range(n):
for each_pair in range(180):
# do something here
results.append(something) # where something is a tuple ... an "ordered pair"
... or it might look like:
for each_cycle in range(n):
this_cycle = list()
for each_pair in range(180):
this_cycle.append(something)
results.extend(this_cycle)
... or it might even look like:
for each_pair in range(n*180):
results.append(something)
... though, frankly, that seems unlikely. (If you try flattening the inner loop to the outer loop in this way you might find that you're having to use modulo arithmetic to get n back out for some other intermediate computational purposes).
I have no idea what the instructor is actually asking for. It seems likely that the math.sin() function will be involved and I guess "ordered pairs" might be co-ordinates mapped to some sort of graphics subsystem and suitable for plotting a graph. I guess 180 of these to show the sinusoid wave through a full range of its values. Maybe you're supposed to multiply something by the amplitude and/or divide something else by the frequency and maybe you're supposed to even add something for each cycle ... some sort of offset to keep the plot moving towards the right or something.
But it seems like you might start with that stub of a function definition and try pasting in one or another of these loop bodies and then figuring out how to actually return meaningful values in the parts where I've used "something" as a placeholder.
Going with the assumption that these "ordered pairs" are co-ordinates, for plotting, then it seems likely that each of the things you append to your results should be of the form (x,y) where x is monotonically increasing (fancy way of saying it keeps going up, never goes down) and might even always be the range(0,n*180) and y is probably math.sin() of something involved a and w ... but that's just speculation on my part.