Good afternoon,
I am having issues with my code. The code I am going to show is a simplified version of the actual one, but the idea is the same.
Var1 = Symbol('Var1')
Var2 = Symbol('Var2')
A = 20
B = 30
Var1 = solve(12+A*B+Var1, Var1)
Var2 = solve(Var1+A+B+Var2, Var2)
print(Var1,Var2)
What is the issue is that for example print(Var1) gives me back the numerical solution of the equation which is -612, but when it comes to print(Var2) it displays -Var1 -50 instead of recognizing that Var1 became a number.
This is the library I import:
from sympy.solvers import solve
from sympy import Symbol
Any idea how to make it understand that Var1 became a number ?
I did try to assign a new variable and then use it in the Var2 equation but it gave me an error.
solve returns a list of (possibly multiple) solutions. sol[0] will give you the first (and in your case only) solution. you may then substitute that solution in your second equation:
from sympy.solvers import solve
from sympy import Symbol
Var1 = Symbol('Var1')
Var2 = Symbol('Var2')
A = 20
B = 30
sol1 = solve(12+A*B+Var1, Var1) # [-612]
eqn2 = (Var1+A+B+Var2).subs({'Var1': sol1[0]}) # Var2 - 562
sol2 = solve(eqn2, Var2) # [562]
print(sol1,sol2) # [-612] [562]
print(sol1[0],sol2[0]) # -612 562
Related
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.
I am trying to create a function which will simulate a poison process for a changeable dt and total time, and have the following:
def compound_poisson(lamda,mu,sigma,dt,T):
points = pd.Series(0)
out = pd.Series(0)
inds = simple_poisson(lamda,dt,T)
for ind in inds.index:
if inds[ind+dt] > inds[ind]:
points[ind+dt] = np.random.normal(mu,sigma)
else:
points[ind+dt] = 0
out = out.append(np.cumsum(points),ignore_index=True)
out.index = np.linspace(0,T,int(T/dt + 1))
return out
However, I receive a "KeyError: 0.010000000000000002", which should not be in the index at all. Is this a result of being lax with float objects?
In short, yes, it's a floating point error. It's quite hard to know how you got there, but probably something like this:
>>> 0.1 * 0.1
0.010000000000000002
Maybe use round?
I'm trying to use MATLAB engine to call a MATLAB function in Python, but I'm having some problems. After manage to deal with NumPy arrays as input in the function, now I have some error from MATLAB:
MatlabExecutionError: Undefined function 'simple_test' for input
arguments of type 'int64'.
My Python code is:
import numpy as np
import matlab
import matlab.engine
eng = matlab.engine.start_matlab()
eng.cd()
Nn = 30
x= 250*np.ones((1,Nn))
y= 100*np.ones((1,Nn))
z = 32
xx = matlab.double(x.tolist())
yy = matlab.double(y.tolist())
Output = eng.simple_test(xx,yy,z,nargout=4)
A = np.array(Output[0]).astype(float)
B = np.array(Output[1]).astype(float)
C = np.array(Output[2]).astype(float)
D = np.array(Output[3]).astype(float)
and the Matlab function is:
function [A,B,C,D] = simple_test(x,y,z)
A = 3*x+2*y;
B = x*ones(length(x),length(x));
C = ones(z);
D = x*y';
end
Is a very simple example but I'm not able to run it!
I know the problem is in the z variable, because when I define z=32 the error is the one I mentioned, and when I change for z=32. the error changes to
MatlabExecutionError: Undefined function 'simple_test' for input
arguments of type 'double'.
but I don't know how to define z.
Since a couple of hours, I am trying to print a simple time vector in a txt file using Python.
import numpy as np
Tp = 2000 * 10**(-9)
dt = Tp / (90000)
t = np.linspace(0,Tp,dt)
timing = open("time.txt","w")
for ii in range(len(t)) :
timing.write(str(t[ii]))
timing.write("\n")
timing.close()
But I still get an empty file and I don't understand at all why.
Maybe I have to be more specific in the function with the precision I want.
Since I have a lot of small numbers (4e-10 ..) to process I would like to understand a general method to write variable (not the entire vector at once) on a txt file with a exponential notation (In Matlab it's kind of automatic I think).
Thx
You have an error using linspace. Please check https://docs.scipy.org/doc/numpy/reference/generated/numpy.linspace.html
Try this:
import numpy as np
Tp = 2000 * 10**(-9)
# dt = Tp / 90000.0
dt = 90000
t = np.linspace(0,Tp,dt)
timing = open("time.txt","w")
for ii in range(len(t)) :
timing.write(str(t[ii]))
timing.write("\n")
timing.close()
I am doing an ordinal logistic regression, and following the guide here for the analysis: R Data Analysis Examples: Ordinal Logistic Regression
My dataframe (consult) looks like:
n raingarden es_score consult_case
garden_id
27436 7 0 3 0
27437 1 0 0 1
27439 1 1 1 1
37253 1 0 3 0
37256 3 0 0 0
I am at the part where I need to to create graph to test the proportional odds assumption, with the command in R as follows:
(s <- with(dat, summary(es_score ~ n + raingarden + consult_case, fun=sf)))
(es_score is an ordinal ranked score with values between 0 - 4; n is an integer; raingarden and consult_case, binary values of 0 or 1)
I have the sf function:
sf <- function(y) {
c('Y>=1' = qlogis(mean(y >= 1)),
'Y>=2' = qlogis(mean(y >= 2)),
'Y>=3' = qlogis(mean(y >= 3)))
}
in a utils.r file that I access as follows:
from rpy2.robjects.packages import STAP
with open('/file_path/utils.r', 'r') as f:
string = f.read()
sf = STAP(string, "sf")
And want to do something along the lines of:
R = ro.r
R.with(work_case_control, R.summary(formula, fun=sf))
The major problem is that the R withoperator is seen as a python keyword, so that even if I access it with ro.r.with it is still recognized as a python keyword. (As a side note: I tried using R's apply method instead, but got an error that TypeError: 'SignatureTranslatedAnonymousPackage' object is not callable ... I assume this is referring to my function sf?)
I also tried using the R assignment methods in rpy2 as follows:
R('sf = function(y) { c(\'Y>=1\' = qlogis(mean(y >= 1)), \'Y>=2\' = qlogis(mean(y >= 2)), \'Y>=3\' = qlogis(mean(y >= 3)))}')
R('s <- with({0}, summary(es_score~raingarden + consult_case, fun=sf)'.format(consult))
but ran into issues where the dataframe column names were somehow causing the error: RRuntimeError: Error in (function (file = "", n = NULL, text = NULL, prompt = "?", keep.source = getOption("keep.source"), :
<text>:1:19: unexpected symbol
1: s <- with( n raingarden
I could of course do this all in R, but I have a very involved ETL script in python, and would thus prefer to keep everything in python using rpy2 (I did try this using mord for scipy-learn to run my regreession, but it is pretty primitive).
Any suggestions would be most welcome right now.
EDIT
I tried various combinations #Parfait's suggestions, and qualifying the fun argument is syntactically incorrect, as per PyCharm interpreter (see image with red highlighting at end): ... it doesn't matter what the qualifier is, either, I always get an error
that SyntaxError: keyword can't be an expression.
On the other hand, with no qualifier, there is no syntax error: , but I do get the error TypeError: 'SignatureTranslatedAnonymousPackage' object is not callable when using the function sf as obtained:
from rpy2.robjects.packages import STAP
with open('/Users/gregsilverman/development/python/rest_api/rest_api/scripts/utils.r', 'r') as f:
string = f.read()
sf = STAP(string, "sf")
With that in mind, I created a package in R with the function sf, imported it, and tried various combos with the only one producing no error, being: print(base._with(consult_case_control, R.summary(formula, fun=gms.sf))) (gms is a reference to the package in R I made).
The output though makes no sense:
Length Class Mode
3 formula call
I am expecting a table ala the one on the UCLA site. Interesting. I am going to try recreating my analysis in R, just for the heck of it. I still would like to complete it in python though.
Consider bracketing the with call and be sure to qualify all arguments including fun:
ro.r['with'](work_case_control, ro.r.summary(formula, ro.r.summary.fun=sf))
Alternatively, import R's base package. And to avoid conflict with Python's named method with() translate the R name:
from rpy2.robjects.packages import importr
base = importr('base', robject_translations={'with': '_with'})
base._with(work_case_control, ro.r.summary(formula, ro.r.summary.fun=sf))
And be sure to properly create your formula. Consider using R's stats packages' as.formula to build from string. Notice too another translation is made due to naming conflict:
stats = importr('stats', robject_translations={'format_perc': '_format_perc'})
formula = stats.as_formula('es_score ~ n + raingarden + consult_case')