I have a dict that contains tuples as keys with a sympy expression and the number of arguments in each tuple and as value a customized translation. I want to check if the sympy expression is a trigonometric function i.e a function of those: https://omz-software.com/pythonista/sympy/modules/mpmath/functions/trigonometric.html
Do you guys know a nice command to check this? I could just think of a list in which I name [sin, cos, tan] etc and then check if key is contained in it. But I'd be really happy if there was a nicer solution.
Can one generally classify sympy expression and check to which class of expression they belong?
default_semantic_latex_table = {
(sympy.functions.elementary.trigonometric.sin, 1): FormatTemplate(r"\sin#{$0}"), (sympy.functions.special.polynomials.jacobi, 4): FormatTemplate(r"\JacobipolyP{{$0}{$1}{$2}#{$3}}"),
(sympy.functions.elementary.trigonometric.cos, 2): FormatTemplate(r"\cos#{$0,$2}")
}
for k, v in default_semantic_latex_table:
# check if key is instance of a sympy trigonometric function
# if so and the number of args is not equal to one, throw a warning
If the if condition evaluates to True, I want to check the second element of the tuple/the number of args. How would I do that?
I expected k to be a tuple like (sympy.functions.special.polynomials.jacobi, 4) yet if I test it k turns out to be just sympy.functions.special.polynomials.jacobi. How can I get the second tuple element?
I'd be really glad if someone could help!
You can import the TrigonometricFunction base class from sympy and use the issubclass method to perform the tests.
Note that you have to use dictionary_name.items() in order to loop over the keys and values of a dictionary:
from sympy import *
import sympy
from sympy.functions.elementary.trigonometric import TrigonometricFunction
import warnings
default_semantic_latex_table = {
(sympy.functions.elementary.trigonometric.sin, 1): "a",
(sympy.functions.elementary.trigonometric.cos, 2): "b"
}
for k, v in default_semantic_latex_table.items():
t, n = k
if issubclass(t, TrigonometricFunction) and (n != 1):
warnings.warn("Your warning: function=%s with n=%s" % (t, n))
Related
I have a question similar to this one: How to substitute multiple symbols in an expression in sympy? but in reverse.
I have a sympy expression with numerical values and symbols alike. I would like to substitute all numerical values with symbolic constants. I appreciate that such query is uncommon for sympy. What can I try next?
For example, I have:
-0.5967695*sin(0.15280747*x0 + 0.89256966) + 0.5967695*sin(sin(0.004289882*x0 - 1.5390939)) and would like to replace all numbers with a, b, c etc. ideally in a batch type of way.
The goal is to then apply trig identities to simplify the expression.
I'm not sure if there is already such a function. If there is not, it's quite easy to build one. For example:
import string
def num2symbols(expr):
# wild symbol to select all numbers
w = Wild("w", properties=[lambda t: isinstance(t, Number)])
# extract the numbers from the expression
n = expr.find(w)
# get a lowercase alphabet
alphabet = list(string.ascii_lowercase)
# create a symbol for each number
s = symbols(" ".join(alphabet[:len(n)]))
# create a dictionary mapping a number to a symbol
d = {k: v for k, v in zip(n, s)}
return d, expr.subs(d)
x0 = symbols("x0")
expr = -0.5967695*sin(0.15280747*x0 + 0.89256966) + 0.5967695*sin(sin(0.004289882*x0 - 1.5390939))
d, new_expr = num2symbols(expr)
print(new_expr)
# out: b*sin(c + d*x0) - b*sin(sin(a + f*x0))
print(d):
# {-1.53909390000000: a, -0.596769500000000: b, 0.892569660000000: c, 0.152807470000000: d, 0.596769500000000: e, 0.00428988200000000: f}
I feel like dict.setdefault was made for this purpose in Python :-)
>>> c = numbered_symbols('c',cls=Dummy)
>>> d = {}
>>> econ = expr.replace(lambda x:x.is_Float, lambda x: sign(x)*d.setdefault(abs(x),next(c)))
>>> undo = {v:k for k,v in d.items()}
Do what you want with econ and when done (after saving results to econ)
>>> econ.xreplace(undo) == expr
True
(But if you change econ the exact equivalence may no longer hold.) This uses abs to store symbols so if the expression has constants that differ by a sign they will appear in econ with +/-ci instead of ci and cj.
I am passing initial conditions as string, to be used to solving an ODE in sympy.
It is a first order ode, so for example, lets take initial conditions as y(0):3 for example. From help
ics is the set of initial/boundary conditions for the differential
equation. It should be given in the form of {f(x0): x1,
f(x).diff(x).subs(x, x2): x3}
I need to pass this to sympy.dsolve. But sympify(ic) gives an error for some reason.
What other tricks to use to make this work? Here is MWE. First one shows it works without initial conditions being string (normal mode of operation)
from sympy import *
x = Symbol('x')
y = Function('y')
ode = Eq(Derivative(y(x),x),1+2*x)
sol = dsolve(ode,y(x),ics={y(0):3})
gives sol Eq(y(x), x**2 + x + 3)
Now the case when ics is string
from sympy import *
ic = "y(0):3"
x = Symbol('x')
y = Function('y')
ode = Eq(Derivative(y(x),x),1+2*x)
sol = dsolve(ode,y(x),ics={ sympify(ic) })
gives
SympifyError: Sympify of expression 'could not parse 'y(0):3'' failed,
because of exception being raised: SyntaxError: invalid syntax
(, line 1)
So looking at sympify webpage
sympify(a, locals=None, convert_xor=True, strict=False, rational=False, evaluate=None)
And tried changing different options as shown above, still the syntax error shows up.
I also tried
sol = dsolve(ode,y(x),ics= { eval(ic) } )
But this gives syntax error as well
Is there a trick to use to convert this initial conditions string to something sympy is happy with?
Python 4.7 with sympy 1.5
As temporary work around, currently I do this
from sympy import *
ic = "y(0):3"
ic = ic.split(":")
x = Symbol('x')
y = Function('y')
ode = Eq(Derivative(y(x),x),1+2*x)
sol = dsolve(ode,y(x),ics= {S(ic[0]):S(ic[1])} )
Which works. So the problem is with : initially, sympify (or S) do not handle : it seems.
You can use sympify('{y(0):3}').
I don't know what your actual goal is but I don't recommend parsing strings like this in general. The format for ICs is actually slightly awkward so that for a second order ODE it looks like:
ics = '{y(0):3, y(x).diff(x).subs(x, 0):1}'
If you're parsing a string then you can come up with a better syntax than that like
ics = "y(0)=3, y'(0)=1"
Also you should use parse_expr rather than converting strings with sympify or S:
https://docs.sympy.org/latest/modules/parsing.html#sympy.parsing.sympy_parser.parse_expr
I have the following expression
A=Symbol('A')
x=Symbol('x')
B=Symbol('B')
C=Symbol('C')
D=Symbol('D')
expression=((A**x-B-C)/(D-1))*(D-1)
n,d=fraction(expression)
I am getting following result:
n=A**x-B-C
d=1
My expected result is
n=(A**x-B-C)*(D-1)
d=(D-1)
Is there way in sympy or need to write customize function to handle that
Use UnevaluatedExpr() to prevent the expression from being evaluated.
from sympy import symbols, fraction, UnevaluatedExpr
A,x,B,C,D = symbols('A x B C D')
expression = (A**x-B-C)/(D-1)*UnevaluatedExpr(D-1)
n,d = fraction(expression)
print(n)
print(d)
This returns
(A**x - B - C)*(D - 1)
D - 1
See the Sympy Advanced Expression Manipulation doc page for more details.
I have got a very simple function to apply to each row of my dataframe:
def distance_ot(fromwp,towp,pl,plee):
` if fromwp[0:3]==towp[0:3]:
sxcord=pl.loc[fromwp,"XCORD"]
sycord=pl.loc[fromwp,"YCORD"]
excord=pl.loc[towp,"XCORD"]
eycord=pl.loc[towp,"YCORD"]
x=np.abs(excord-sxcord); y=np.abs(eycord-sycord)
distance=x+y
return distance
else:
x1=np.abs(plee.loc[fromwp[0:3],"exitx"]-pl.loc[fromwp,"XCORD"])
y1=np.abs(plee.loc[fromwp[0:3],"exity"]-pl.loc[fromwp,"YCORD"])
x2=np.abs(plee.loc[fromwp[0:3],"exitx"]-plee.loc[towp[0:3],"entryx"])
y2=np.abs(plee.loc[fromwp[0:3],"exity"]-plee.loc[towp[0:3],"entryy"])
x3=np.abs(plee.loc[towp[0:3],"entryx"]-pl.loc[towp,"XCORD"])
y3=np.abs(plee.loc[towp[0:3],"entryy"]-pl.loc[towp,"YCORD"])
distance=x1+x2+x3+y1+y2+y3
return distance
With this line it is called:
pot["traveldistance"]=pot.apply(lambda row: distance_ot(fromwp=row["from_wpadr"],towp=row["to_wpadr"],pl=pl,plee=plee),axis=1)
Where: fromwp and towp are both strings and xcord and ycord are floats. I tried using numba but for some reasons it does not enhance this performance. Any suggestions?
Thanks to caiohamamura hint hereby the solution:
distance_ot(pl=pl,plee=plee)
pot.ix[pot.from_wpadr.str[0:3]==pot.to_wpadr.str[0:3],"traveldistance"]=pot["distance1"]
pot.ix[pot.from_wpadr.str[0:3]!=pot.to_wpadr.str[0:3],"traveldistance"]=pot["distance2"]
def distance_ot(pl,plee):
from_df = pl.loc[pot["from_wpadr"]]
to_df = pl.loc[pot["to_wpadr"]]
sxcord=from_df["XCORD"].values
sycord=from_df["YCORD"].values
excord=to_df["XCORD"].values
eycord=to_df["YCORD"].values
x=np.abs(excord-sxcord); y=np.abs(eycord-sycord)
pot["distance1"]=x+y
from_df2=plee.loc[pot["from_wpadr"].str[0:3]]
to_df2=plee.loc[pot["to_wpadr"].str[0:3]]
x1=np.abs(from_df2["exitx"].values-from_df["XCORD"].values)
y1=np.abs(from_df2["exity"].values-from_df["YCORD"].values)
x2=np.abs(from_df2["exitx"].values-to_df2["entryx"].values)
y2=np.abs(from_df2["exity"].values-to_df2["entryy"].values)
x3=np.abs(to_df2["entryx"].values-to_df["XCORD"].values)
y3=np.abs(to_df2["entryy"].values-to_df["YCORD"].values)
pot["distance2"]=x1+x2+x3+y1+y2+y3
Vectorize the distance_ot function to calculate all distances at once. I would begin populating a from_df and a to_df like the following:
import numpy as np
from_df = pl.loc[np.in1d(pl.loc.index, pot["from_wpadr"])
to_df = pl.loc[np.in1d(pl.loc.index, pot["to_wpadr"])
Then you can continue as in your function:
sxcord=from_df["XCORD"]
sycord=from_df["YCORD"]
excord=to_df["XCORD"]
eycord=to_df["YCORD"]
x=np.abs(excord-sxcord); y=np.abs(eycord-sycord)
distances=x+y
This will calculate all the distances at once. Your if clause can also be vectorized, your results will be calculated into different arrays, those which match the if clause and those that doesn't, you just have to keep track of the boolean array, so you can put them together in the dataframe afterwards:
first_three_equals = np.char.ljust(pot["from_wpadr"].values.astype(str), 3) \
== np.char.ljust(pot["to_wpadr"].values.astype(str), 3)
One may select elements in numpy arrays as follows
a = np.random.rand(100)
sel = a > 0.5 #select elements that are greater than 0.5
a[sel] = 0 #do something with the selection
b = np.array(list('abc abc abc'))
b[b==a] = 'A' #convert all the a's to A's
This property is used by the np.where function to retrive indices:
indices = np.where(a>0.9)
What I would like to do is to be able to use regular expressions in such element selection. For example, if I want to select elements from b above that match the [Aab] regexp, I need to write the following code:
regexp = '[Ab]'
selection = np.array([bool(re.search(regexp, element)) for element in b])
This looks too verbouse for me. Is there any shorter and more elegant way to do this?
There's some setup involved here, but unless numpy has some kind of direct support for regular expressions that I don't know about, then this is the most "numpytonic" solution. It tries to make iteration over the array more efficient than standard python iteration.
import numpy as np
import re
r = re.compile('[Ab]')
vmatch = np.vectorize(lambda x:bool(r.match(x)))
A = np.array(list('abc abc abc'))
sel = vmatch(A)