Develop Reference

How will I convert decimals into accurate fractions or accurate decimals?

For some reason it shows an error message: TypeError: argument should be a string or a Rational instance
import cmath
from fractions import Fraction
#Function
# Quadratic equatrion solver
def solver(a_entry, b_entry, c_entry):
a = int(a_entry)
b = int(b_entry)
c = int(c_entry)
d = (b*b) - (4*a*c)
sol1 = (-b-cmath.sqrt(d)/(2*a))
sol2 = (-b+cmath.sqrt(d)/(2*a))
sol3 = Fraction(sol1)
sol4 = Fraction(sol2)
print(f"Value of x1 = {sol3} and value of x2 = {sol4}")
solver(1, 2, 3)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 8, in solver
File "/usr/lib/python3.10/fractions.py", line 139, in __new__
raise TypeError("argument should be a string "
TypeError: argument should be a string or a Rational instance
I am a new programmer and I saw that this code generates a weird number (example: 5.42043240824+0j {inaccurate values})
when i give random values. So I want it to give either an accurate decimal values or in fraction. The fraction method dosen't work for some reason. Can someone please help. Alot of thanks.

2 things wrong in your code :
Use math instead of cmath as cmath is used for complexed values (it will always returns a complexe value, even 1+0j) which is not compatible with Fraction.
Be careful you wrote : (-b-cmath.sqrt(d)/(2*a)) but is should be ((-b-cmath.sqrt(d))/(2*a))
Also, the solution might no exist. For example, resolving 1x^2 + 3x + 10 has no answer (your fonction does not cross x axe). It still has complexe answer(s).
To avoid this you can use a try except to catch errors. OR you can validate that d^2 is greater than 4ac because you can't sqrt negative values (except with complexe values ;) ) :
def solver():
a = int(entry.get())
b = int(entry1.get())
c = int(entry2.get())
d = (b*b) - (4*a*c)
if d < 0:
text = "no real answer ! The function doesn't cross X axe !"
label2.configure(text = text)
else:
sol1 = ((-b-math.sqrt(d))/(2*a))
sol2 = ((-b+math.sqrt(d))/(2*a))
sol3 = Fraction(sol1)
sol4 = Fraction(sol2)
label2.configure(text = f"Value of x1 = {sol3} and value of x2 = {sol4}")
Hope it helps

The issue with sqrt
It appears that you do not want to evaluate the square roots to numerical approximations. But that is exactly what cmath.sqrt and math.sqrt do: they calculate numerical approximations of square roots.
For instance:
import math
print( math.sqrt(2) )
# 1.4142135623730951
If you are not interested in numerical approximations, then I suggest using a library for symbolic calculus. The best-known library for symbolic calculus in python is called sympy. This module has a sympy.sqrt function that will simplify a square root as much as it can, but without returning a numerical approximation:
import sympy
print( sympy.sqrt(9) )
# 3
print( sympy.sqrt(2) )
# sqrt(2)
print( sympy.sqrt(18) )
# 3*sqrt(2)
More information about sympy: https://docs.sympy.org/latest/tutorials/intro-tutorial/intro.html
Other advice
When you write a program, it is most usually a good idea to cleanly separate the parts of the code that deal with algorithms, maths, and logic, from the parts of the code that deal with input and output. I suggest writing two functions, one that solves quadratic equations, and one that does input and output:
import sympy
# returns solutions of a x**2 + b x + c == 0
def solver(a, b, c):
Delta = b*b - 4*a*c
sol1 = (-b - sympy.sqrt(Delta)) / (2*a)
sol2 = (-b + sympy.sqrt(Delta)) / (2*a)
return (sol1, sol2)
# ask for user input and solve an equation
def input_equation_output_solution():
a = int(entry.get())
b = int(entry1.get())
c = int(entry2.get())
sol1, sol2 = solver(a, b, c)
label2.configure(text = f"Value of x1 = {sol1} and value of x2 = {sol2}")

Imprecise results of logarithm and power functions in Python

I am trying to complete the following exercise:
https://www.codewars.com/kata/whats-a-perfect-power-anyway/train/python
I tried multiple variations, but my code breaks down when big numbers are involved (I tried multiple variations with solutions involving log and power functions):
Exercise:
Your task is to check wheter a given integer is a perfect power. If it is a perfect power, return a pair m and k with m^k = n as a proof. Otherwise return Nothing, Nil, null, None or your language's equivalent.
Note: For a perfect power, there might be several pairs. For example 81 = 3^4 = 9^2, so (3,4) and (9,2) are valid solutions. However, the tests take care of this, so if a number is a perfect power, return any pair that proves it.
The exercise uses Python 3.4.3
My code:
import math
def isPP(n):
for i in range(2 +n%2,n,2):
a = math.log(n,i)
if int(a) == round(a, 1):
if pow(i, int(a)) == n:
return [i, int(a)]
return None
Question:
How is it possible that I keep getting incorrect answers for bigger numbers? I read that in Python 3, all ints are treated as "long" from Python 2, i.e. they can be very large and still represented accurately. Thus, since i and int(a) are both ints, shouldn't the pow(i, int(a)) == n be assessed correctly? I'm actually baffled.

(edit note: also added integer nth root bellow)
you are in the right track with logarithm but you are doing the math wrong, also you are skipping number you should not and only testing all the even number or all the odd number without considering that a number can be even with a odd power or vice-versa
check this
>>> math.log(170**3,3)
14.02441559235585
>>>
not even close, the correct method is described here Nth root
which is:
let x be the number to calculate the Nth root, n said root and r the result, then we get
rn = x
take the log in any base from both sides, and solve for r
logb( rn ) = logb( x )
n * logb( r ) = logb( x )
logb( r ) = logb( x ) / n
blogb( r ) = blogb( x ) / n
r = blogb( x ) / n
so for instance with log in base 10 we get
>>> pow(10, math.log10(170**3)/3 )
169.9999999999999
>>>
that is much more closer, and with just rounding it we get the answer
>>> round(169.9999999999999)
170
>>>
therefore the function should be something like this
import math
def isPP(x):
for n in range(2, 1+round(math.log2(x)) ):
root = pow( 10, math.log10(x)/n )
result = round(root)
if result**n == x:
return result,n
the upper limit in range is to avoid testing numbers that will certainly fail
test
>>> isPP(170**3)
(170, 3)
>>> isPP(6434856)
(186, 3)
>>> isPP(9**2)
(9, 2)
>>> isPP(23**8)
(279841, 2)
>>> isPP(279841)
(529, 2)
>>> isPP(529)
(23, 2)
>>>
EDIT
or as Tin Peters point out you can use pow(x,1./n) as the nth root of a number is also expressed as x1/n
for example
>>> pow(170**3, 1./3)
169.99999999999994
>>> round(_)
170
>>>
but keep in mind that that will fail for extremely large numbers like for example
>>> pow(8191**107,1./107)
Traceback (most recent call last):
File "<pyshell#90>", line 1, in <module>
pow(8191**107,1./107)
OverflowError: int too large to convert to float
>>>
while the logarithmic approach will success
>>> pow(10, math.log10(8191**107)/107)
8190.999999999999
>>>
the reason is that 8191107 is simple too big, it have 419 digits which is greater that the maximum float representable, but reducing it with a log produce a more reasonable number
EDIT 2
now if you want to work with numbers ridiculously big, or just plain don't want to use floating point arithmetic altogether and use only integer arithmetic, then the best course of action is to use the method of Newton, that the helpful link provided by Tin Peters for the particular case for cube root, show us the way to do it in general alongside the wikipedia article
def inthroot(A,n):
if A<0:
if n%2 == 0:
raise ValueError
return - inthroot(-A,n)
if A==0:
return 0
n1 = n-1
if A.bit_length() < 1024: # float(n) safe from overflow
xk = int( round( pow(A,1/n) ) )
xk = ( n1*xk + A//pow(xk,n1) )//n # Ensure xk >= floor(nthroot(A)).
else:
xk = 1 << -(-A.bit_length()//n) # power of 2 closer but greater than the nth root of A
while True:
sig = A // pow(xk,n1)
if xk <= sig:
return xk
xk = ( n1*xk + sig )//n
check the explanation by Mark Dickinson to understand the working of the algorithm for the case of cube root, which is basically the same for this
now lets compare this with the other one
>>> def nthroot(x,n):
return pow(10, math.log10(x)/n )
>>> n = 2**(2**12) + 1 # a ridiculously big number
>>> r = nthroot(n**2,2)
Traceback (most recent call last):
File "<pyshell#48>", line 1, in <module>
nthroot(n**2,2)
File "<pyshell#47>", line 2, in nthroot
return pow(10, math.log10(x)/n )
OverflowError: (34, 'Result too large')
>>> r = inthroot(n**2,2)
>>> r == n
True
>>>
then the function is now
import math
def isPPv2(x):
for n in range(2,1+round(math.log2(x))):
root = inthroot(x,n)
if root**n == x:
return root,n
test
>>> n = 2**(2**12) + 1 # a ridiculously big number
>>> r,p = isPPv2(n**23)
>>> p
23
>>> r == n
True
>>> isPPv2(170**3)
(170, 3)
>>> isPPv2(8191**107)
(8191, 107)
>>> isPPv2(6434856)
(186, 3)
>>>
now lets check isPP vs isPPv2
>>> x = (1 << 53) + 1
>>> x
9007199254740993
>>> isPP(x**2)
>>> isPPv2(x**2)
(9007199254740993, 2)
>>>
clearly, avoiding floating point is the best choice

getting ZeroDivisionError: integer division or modulo by zero

I had written a simple pascal triangle code in python but I am getting a error
def factorial(n):
c=1
re=1
for c in range(n):
re = re * c;
return(re)
print "Enter how many rows of pascal triangle u want to show \n"
n=input();
i=1
c=1
for i in range(n):
for c in range(n-i-1):
print ""
for c in range(i):
a = factorial(i);
b = factorial(c);
d = factorial(i-c);
z = (a/(b*d));
print "%d" % z
print "\n"
ERROR:
Traceback (most recent call last):
File "/home/tanmaya/workspace/abc/a.py", line 19, in <module>
z = (a/(b*d));
ZeroDivisionError: integer division or modulo by zero

Your factorial() function returns 0 for any input because of how you defined your range.
The range builtin starts at 0 unless otherwise defined so:
for c in range(n):
re = re * c # no semicolons in Python
is doing:
re = re * 0
on the first iteration so for all subsequent iterations:
re = 0 * c
will always be 0
Start your range at 1 like so
for c in range(1, n):
re *= c # The *= operator is short hand for a = a * b
you can see this more explicityly:
>>> print(list(range(5)))
[0, 1, 2, 3, 4]
>>> print(list(range(1,5)))
[1, 2, 3, 4]
>>>
or instead of rolling your own function use the one that comes with Python:
>>> from math import factorial
>>> factorial(3)
6
Upon closer reading of your code it seems you tried to circumvent this by setting c = 1 outside your for loop. This is not going to work because the variables you declared outside the loop are being reassigned inside it.

ZeroDivisionError means that you were trying to divide or modulo, a number n with 0.
in your case, z = (a/(b*d)) resulted in z = (a/0)
Also, as #theB pointed out, your factorial function is incorrect.
Try fixing those.
Also, you don't really need ; in your code. It's usually the case we put ; when we want to make the code one liner.

Reading pixels in python: "ValueError too many values to unpack"

I want to read a .tif file and count the number of pixels in an image and determine the density of objects, but when I attempt this y, x = np.indices(image.shape) it gives me then
Value Error (ValueError: too many values to unpack, File "<stdin>", line 1, in <module>).
My code is as follows:
import sys
import os
import numpy as np
from pylab import *
import scipy
import matplotlib.pyplot as plt
import math
#Function
def radial_plot(image):
y, x = np.indices(image.shape) # <----- Problem here???
center = np.array([(x.max()-x.min())/2.0, (x.max()-x.min())/2.0])
r = np.hypot(x - center[0], y - center[1])
ind = np.argsort(r.flat)- center[1])
r_sorted = r.flat[ind]
i_sorted = image.flat[ind]
r_int = r_sorted.astype(int)
deltar = r_int[1:] - r_int[:-1]
rind = np.where(deltar)[0]
nr = rind[1:] - rind[:-1]
csim = np.cumsum(i_sorted, dtype=float)
tbin = csim[rind[1:]] - csim[rind[:-1]]
radial_prof = tbin / nr
return rad
#Main
img = plt.imread('dat.tif')
radial_plot(img)

The issue is that you are attempting to assign more than two values to only two varibles:
>>> a, b = range(2) #Assign two values to two variables
>>> a
0
>>> b
1
>>> a, b = range(3) #Attempt to assign three values to two variables
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: too many values to unpack
In Python 2.x you can do the following:
>>> a, b = range(3)[0], range(3)[1:]
>>> a
0
>>> b
[1, 2]
Just for completeness, if you had Python 3.x, you could do Extended Iterable Unpacking:
>>> a, *b, c = range(5)
>>> a
0
>>> c
4
>>> b
[1, 2, 3]
Hope this helps

np.indices returns an array representing the indices of the grid. The error basically indicates that there are more than 2 values obtained by calling the indices method. Since it's returning a grid, you can assign it to a variable such as grid and then access the indices accordingly.
The crux of the error is the function call returns more than just 2 values, and in your code you are trying to 'squeeze' them into just 2 variables.
For eg.
s = "this is a random string"
x, y = s.split()
The above code gives you a value error since there are 5 strings obtained by calling split(), while I am trying to accommodate them into just 2 variables.

Math domain error in fermat

from math import sqrt
def fermatBook (n):
x=int(sqrt(n))
c=x**2-n
while (sqrt(c)!=int(sqrt(c))):
x=x+1
y=sqrt(c)
a=x+y
b=x-y
if a==1 or b==1:
print "The number is prime"
return a, b
error:
Traceback (most recent call last):
File "<pyshell#0>", line 1, in <module>
fermatBook (23867)
File "C:/Python27/fermatLivro.py", line 6, in fermatBook
while (sqrt(c)!=int(sqrt(c))):
ValueError: math domain error
I don't know what is going wrong with the program... Could someone help me ?

most likely your variable c is going negative:
Example
if you call:
n = 2
fermatBook(n)
it will assign the following values to the following variables:
x = int(sqrt(n)) = int(1.47...) = 1
c = x**2 - n = 1**2 - 2 = 1 - 2 = -1
This will likely happen alot on values of n whose square root is not an integer.
sqrt(n) >= int(sqrt(n)), n >= 0
Then when you call sqrt(c) it is out of the domain because it cannot handle negative values.
>>> from math import sqrt
>>> sqrt(-1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: math domain error
You should rather use something that can handle imaginary numbers, i.e. cmath
Or perform checks to assure this does not happen:
As an example...
if c < 0:
c = 0
As soon as you fix this however you are going to run into another problem:
This is an infinite loop:
while (sqrt(c)!=int(sqrt(c))):
x=x+1
you need to update c otherwise the condidtion will never change no matter how many times you increment x. You probably meant this?
while (sqrt(c)!=int(sqrt(c))):
x=x+1
c = x**2+n # <--- UPDATE c