Develop Reference

sympy .subs() built-in function doesn't work

I'm trying to implement the multivariate chain rule using python, when I try to substitute one of the elements using the built-in sympy function expr.subs() I won't get an answer.
The exercise goes as follows:
Compute df/dx of the following function using the chain rule:
a = x^2
b = exp(a)
c = a + b
d = log(c)
e = sin(c)
f = d + e
⁡
And this is the code:
a, b, c, d, e, f = sym.symbols('a b c d e f')
f = d + e
dfd = sym.diff(f, d)
dfe = sym.diff(f, e)
df = sym.Matrix([[dfd, dfe]])
d = sym.log(c)
e = sym.sin(c)
ddc = sym.diff(d)
dde = sym.diff(e)
dd = sym.Matrix([[ddc],[dde]])
dfdd = df#dd
c = a + b
dca = sym.diff(c, a)
dcb = sym.diff(c, b)
dc = sym. Matrix([[dca, dcb]])
dfdc = dfdd#dc
a = x**2
b = sym.exp(a)
result = dfdc.subs(c, (a + b))
result
The result the function .subs() doesn't substitute anything and I don't know why, I tried substituting it using other ways to write the function, like:
dfdc.subs({c : (a + b)})
dfdc.subs(c, a + b)
And even tried to just substitute it for an integer to see it that would work and neither does it.
What am I doing wrong?

The c in your dfdc expression was created before you set c=a+b. So it still appears as c in dfdc. However, by the time you want to do your subs, the c symbol is declared as c=a+b, and so your substitute really reads
dfdc.subs(a+b, a+b)
And that does nothing.
In order to really use c there, do
dfdc.subs(sym.Symbol('c'), a+b)

Default Arguments in Python for Import

I have a small module written as:
Contents of psychometrics.py
def prob3pl(theta, D = 1.7, a, b, c):
result = c + (1 - c) / (1 + np.exp(-D * a * (theta - b)))
return(result)
def gpcm(theta, d, score, a, D=1.7):
Da = D * a
result = np.exp(np.sum(Da * (theta - d[0:score]))) / np.sum(np.exp(np.cumsum(Da * (theta - d))))
return(result)
if __name__ == '__main__':
gpcm(theta, d, score, a, D=1.7)
prob3pl(theta, D = 1.7, a, b, c)
Now using the python interpret I do the following:
import psychometrics as py
import numpy as np
py.prob3pl(0, a = 1, b= 0, c=0)
However, when running this yields
>>> py.prob3pl(0,a=1,b=0,c=0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: prob3pl() missing 1 required positional argument: 'D'
When I copy and paste the function into the interpreter it runs using the default value of D = 1 but when importing that isn't happening.
What error am I making such that the default value for D is not recognized when I import the module?
Thank you

Your code has syntax error -
SyntaxError: non-default argument follows default argument
Therefore, kindly change function prob3pl() as
def prob3pl(theta, a, b, c, D = 1.7):
result = c + (1 - c) / (1 + np.exp(-D * a * (theta - b)))
return(result)
Reason - In python function declaration, there should not be any non-default argument after any default argument. as D=1.7 here

Linking multiple python functions together

Simple question on how to link (or string together) multiple functions that depend on each other. I have the following example function (in Jupyter):
### First function
def function_one():
## process one
a = "one" + "two"
print(a)
## process two
b = "red" + "blue"
print(b)
## process three
c = "this" + "sucks"
print(c)
return a, b, c
### Second function
def function_two(a, b, c):
## process four
d = a + b
print(d)
## process five
e = b + c
print(e)
## process six
f = a + c
print(f)
return d, e, f
### Third function
def function_three():
g = a + b + c + d + e + f
print(g)
return g
### Calling functions
initial = function_one()
second = function_two(initial)
third = ... #I can't get past the following error to even link this function in
The first function works when called, but when I try to send that data downstream to the second function, I get this error:
onetwo
redblue
thissucks
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-10-7c5562b97c86> in <module>
1 initial = function_one()
----> 2 second = function_two(initial)
TypeError: function_two() missing 2 required positional arguments: 'b' and 'c'
How do I remedy this?

When returning multiple objects, you are actually returning a tuple. Ex:
>>> def foo():
>>> return 1, 2, 3
>>> type(foo())
<class 'tuple'>
So, right now the whole tuple is treated as argument a and hence b and c are missing. In order to pass that on as 3 separate arguments, you need to unpack that tuple:
initial = function_one()
second = function_two(*initial)
third = function_three(*initial, *second)

Assign the return values to variables then pass it to the second function:
a, b, c = function_one()
function_two(a, b, c)
For your third function you need it to accept parameters as well
def function_three(a, b, c, d, e, f):
g = a + b + c + d + e + f
print(g)
return g
Then putting it all together:
a, b, c = function_one()
d, e, f = function_two(a, b, c)
g = function_three(a, b, c, d, e, f)

Unable to crack Python code for GCD of 3 numbers?

I am trying to find the GCD of 3 numbers but have been unable to get around much success till now. I am using the following code
def gcd(a, b,c):
if sum(np.array(list([a,b,c]))%min(a,b,c))==0:
return min(a,b,c)
else:
x = np.array(list([a,b,c]))%min(a,b,c)
if sum((list([x]))%min(x))==0:
return min(x)
When I run this on an example say gcd(21,45,60), it gives me the below error
C:\Users\mmt8091\Anaconda3\lib\site-packages\ipykernel_launcher.py:6: RuntimeWarning: divide by zero encountered in remainder
What am i doing wrong here? Have been unable to find any other solutions on net as well. Please help

You do not need np.
Try this:
def gcd(*args):
a, b, *c = args
a, b = min(a, b), max(a, b)
gcd1 = b if not a else gcd(b % a, a)
for i in c:
gcd1 = gcd(gcd1, i)
return gcd1
print(gcd(21, 45, 60))

Error with quadratic formula script

I am making a script where you can input a, b and c into the quadratic formula and it will give you the answer.
It says that b is not defined when I run it.
from cmath import sqrt
qf = lambda a, b, c: (-b-cmath.sqrt((b**2) - (4*a*c)))/(2*a), (-b+cmath.sqrt((b**2) - (4*a*c)))/(2*a)
a, b, c = input('Enter a, b and c with spaces in between the values \n').split()
a = float(a) ; b = float(b) ; c = float(c)
Print(qf(a, b,c)
Traceback (most recent call last):
File "/storage/emulated/0/Download/.last_tmp.py", line 2, in
qf = lambda a, b, c: (-b-cmath.sqrt((b2) - (4*a*c)))/(2*a), (-b+cmath.sqrt((b2) - (4*a*c)))/(2*a)
NameError: name 'b' is not defined

Check this out:
from math import sqrt
def get_roots(a, b, c):
if a == 0:
raise ZeroDivisionError('Not a quadratic equation')
val = b ** 2 - 4 * a * c
if val < 0:
raise Exception('Imaginary Roots')
val = sqrt(val)
root1 = (-b - val) / (2 * a)
root2 = (-b + val) / (2 * a)
return root1, root2
a, b, c = input('Enter a, b and c with spaces in between the values \n').split()
a, b, c = float(a), float(b), float(c)
print(get_roots(a,b,c))

Lambda functions can only return one thing so you need to group your outputs into either a tuple or a list by adding an enclosing () or []. The python parser is getting to:
qf = lambda a, b, c: (-b-sqrt((b**2) - (4*a*c)))/(2*a)
and assuming the lambda function is over. It then begins reading:
, (-b+sqrt((b**2) - (4*a*c)))/(2*a)
and tries to interpret -b in the global scope (where it doesn't exist) this gives you your name error. If you were to get rid of all the variables and for a moment pretend the second result of the quadratic formula was always 0, you'd get a tuple with the first element being a lambda function and the second being the integer 0
>>>qf = lambda a, b, c: [(-b-sqrt((b**2) - (4*a*c)))/(2*a),0
>>>qf
(<function <lambda> at 0x000002B44EF72F28>, 0)
This doesn't quite get what you're after, because you want a single function rather than a tuple of two separate functions to compute each root separately.
Here's what it'd look like if you want the lambda to return a list:
qf = lambda a, b, c: [(-b-sqrt((b**2) - (4*a*c)))/(2*a), (-b+sqrt((b**2) - (4*a*c)))/(2*a)]