How to divide a variable with the number of inputs - python-3.x

I have a variable s and i want to divide that variable with the number of inputs there are. I am a beginner, can anyone tell me the code that can perform that action?
I haven't tried anything because i don't know the specific code that can perform that action.
x = int(input("Insert x: "))
y = int(input("Insert y: "))
s = x + y
print(s/number of inputs)
I expect the output of s/number of inputs.

Not sure why this is needed but you can define your own function that inputs and increments a global counter:
number_of_inputs = 0
def my_input(msg):
global number_of_inputs
x = input(msg)
number_of_inputs += 1
return x
x = int(my_input("Insert x: "))
y = int(my_input("Insert y: "))
s = x + y
print(s / number_of_inputs)
Input:
Insert x: 10
Insert y: 20
Output:
15.0

Related

How can I do number is bigger or smaller than a percentage of a veritable?

Lets say x = 10
y is an input number a user can input which could be any number.
I want to say if y is larger than two percents of x or less than two percents of x than its pass. anything else is print fail.
I think this is what you want?
if y > 0.02 * x or y < 0.02 * x:
// pass
else:
//fail
This will achieve the same goal :
if y != 0.02 * x:
// pass
else:
//fail
From your question, it can be depicted that you want the if statement to be executed only if y is less than or greater than 2% of x, in that case you do the following:
if y != 0.02 * x: //statement

How to solve Equation without any Modules in Python?

I want to solve this equation without any Modules(NumPy, Sympy... etc.)
Px + Qy = W
(ex. 5x + 6y = 55)
Thanks.
It is a very crude way to do this, but you can use brute-force technique, as I said in comment under your question. It can probably be optimized a lot, gives only int outputs, but overall shows the method:
import numpy as np
# Provide the equation:
print("Provide a, b and c to evaluate in equation of form {ax + by - c = 0}")
a = float(input("a: "))
b = float(input("b: "))
c = float(input("c: "))
x_range = int(input("x-searching range (-a, a): "))
y_range = int(input("y-searching range (-b, b): "))
error = float(input("maximum accepted error from the exact solution: "))
x_range = np.arange(-x_range, x_range, 1)
y_range = np.arange(-y_range, y_range, 1)
for x in x_range:
for y in y_range:
if -error <= a * x + b * y - c <= error:
print("Got an absolute error of {} or less with numbers x = {} and y = {}.".format(error, x, y))
Example output for a = 1, b = 2, c = 3, x_range = 10, y_range = 10, error = 0.001:
Got an error of 0.001 or less with numbers x = -9 and y = 6.
Got an error of 0.001 or less with numbers x = -7 and y = 5.
Got an error of 0.001 or less with numbers x = -5 and y = 4.
Got an error of 0.001 or less with numbers x = -3 and y = 3.
Got an error of 0.001 or less with numbers x = -1 and y = 2.
Got an error of 0.001 or less with numbers x = 1 and y = 1.
Got an error of 0.001 or less with numbers x = 3 and y = 0.
Got an error of 0.001 or less with numbers x = 5 and y = -1.
Got an error of 0.001 or less with numbers x = 7 and y = -2.
Got an error of 0.001 or less with numbers x = 9 and y = -3.
I am using numpy, but not a built-in function to solve the equation itself, just to create an array. This can be done without it, of course.
There are thousands of ways to solve an equation with python.
One of those is:
def myfunc (x=None, y=None):
return ((55-6*y)/5.0) if y else ((55-5*x)/6.0)
print(myfunc(x=10)) # OUTPUT: 0.833333333333, y value for x == 10
print(myfunc(y=42)) # OUTPUT: -39.4, x value for y == 42
You simply define inside a function the steps required to solve the equation.
In our example, if we have y value we subtract 6*y to 55 then we divide by 5.0 (we add .0 to have a float as result), otherwise (means we have x) we subtract 5*x from 55 and then we divide by 6.0
with the same principle, you can generalize:
def myfunc (x=None, y=None, P=None, Q=None, W=None):
if not W:
return P*x + Q*y
elif not x:
return (W-Q*y)/float(P)
elif not y:
return (W-P*x)/float(Q)
elif not P:
return (W-Q*y)/float(x)
elif not Q:
return (W-P*x)/float(y)
print(myfunc(x=10, P=5, Q=6, W=55)) # OUTPUT: 0.833333333333, y value for x == 10
print(myfunc(y=42, P=5, Q=6, W=55)) # OUTPUT: -39.4, x value for y == 42
check this QA for some other interesting ways to approach this problem

how to define a certain value to equal to constant just for the first iteration?

I have a the following code (as an example):
answ = []
for i in range(1, 3):
x = y + 2
y = 3 + x
answ.append(y)
Where x and y are simultaneously determined. How can I determine them simultaneously? Or how can I assume that for the first loop y=0 (so x will equal to 2) and then starting from the second iteration 'y' = 3 + x.
Just set y to 0 before the for loop:
answ = []
y = 0
for i in range(1, 3):
x = y + 2
y = 3 + x
append.answ(y)
Like this:
x, y = y + 2, 3 + x
Right now, x and y are local variables, meaning they are deleted at the end of each iteration of the for loop. If you want them to carry over from the previous iteration, you need to define x and y outside of the for loop. This puts them above and the scope of the loop, so they don't get redefined.
Another thing, your for loop runs 2 times, because computers count from 0 and don't include the last number. Because i is not used in your loop at all, it would be better to do:
for i in range(2):
# code
But I am guessing that you want your loop to run three times, in which case you would write:
for i in range(3):
# code
I also noticed that you wrote append.answ(y) instead of answ.append(y). append is a member function of answ, so you would call it the second way.
Anyways, here is the final code for your program:
answ = []
x = 0
y = 0
for i in range(3):
x = y + 2
y = 3 + x
answ.append(y)

Better way to solve simultaneous linear equations programmatically in Python

I have the following code that solves simultaneous linear equations by starting with the first equation and finding y when x=0, then putting that y into the second equation and finding x, then putting that x back into the first equation etc...
Obviously, this has the potential to reach infinity, so if it reaches +-inf then it swaps the order of the equations so the spiral/ladder goes the other way.
This seems to work, tho I'm not such a good mathematician that I can prove it will always work beyond a hunch, and of course some lines never meet (I know how to use matrices and linear algebra to check straight off whether they will never meet, but I'm not so interested in that atm).
Is there a better way to 'spiral' in on the answer? I'm not interested in using math functions or numpy for the whole solution - I want to be able to code the solution. I don't mind using libraries to improve the performance, for instance using some sort of statistical method.
This may be a very naive question from either a coding or maths point of view, but if so I'd like to know why!
My code is as follows:
# A python program to solve 2d simultaneous equations
# by iterating over coefficients in spirals
import numpy as np
def Input(coeff_or_constant, var, lower, upper):
val = int(input("Let the {} {} be a number between {} and {}: ".format(coeff_or_constant, var, lower, upper)))
if val >= lower and val <= upper :
return val
else:
print("Invalid input")
exit(0)
def Equation(equation_array):
a = Input("coefficient", "a", 0, 10)
b = Input("coefficient", "b", 0, 10)
c = Input("constant", "c", 0, 10)
equation_list = [a, b, c]
equation_array.append(equation_list)
return equation_array
def Stringify_Equations(equation_array):
A = str(equation_array[0][0])
B = str(equation_array[0][1])
C = str(equation_array[0][2])
D = str(equation_array[1][0])
E = str(equation_array[1][1])
F = str(equation_array[1][2])
eq1 = str(A + "y = " + B + "x + " + C)
eq2 = str(D + "y = " + E + "x + " + F)
print(eq1)
print(eq2)
def Spiral(equation_array):
a = equation_array[0][0]
b = equation_array[0][1]
c = equation_array[0][2]
d = equation_array[1][0]
e = equation_array[1][1]
f = equation_array[1][2]
# start at y when x = 0
x = 0
infinity_flag = False
count = 0
coords = []
coords.append([0, 0])
coords.append([1, 1])
# solve equation 2 for x when y = START
while not (coords[0][0] == coords[1][0]):
try:
y = ( ( b * x ) + c ) / a
except:
y = 0
print(y)
try:
x = ( ( d * y ) - f ) / e
except:
x = 0
if x >= 100000 or x <= -100000:
count = count + 1
if count >= 100000:
print("It\'s looking like these linear equations don\'t intersect!")
break
print(x)
new_coords = [x, y]
coords.append(new_coords)
coords.pop(0)
if not ((x == float("inf") or x == float("-inf")) and (y == float("inf") or y == float("-inf"))):
pass
else:
infinity_flag if False else True
if infinity_flag == False:
# if the spiral is divergent this switches the equations around so it converges
# the infinity_flag is to check if both spirals returned infinity meaning the lines do not intersect
# I think this would mostly work for linear equations, but for other kinds of equations it might not
x = 0
a = equation_array[1][0]
b = equation_array[1][1]
c = equation_array[1][2]
d = equation_array[0][0]
e = equation_array[0][1]
f = equation_array[0][2]
infinity_flag = False
else:
print("These linear equations do not intersect")
break
y = round(y, 3)
x = round(x, 3)
print(x, y)
equation_array = []
print("Specify coefficients a and b, and a constant c for equation 1")
equations = Equation(equation_array)
print("Specify coefficients a and b, and a constant c for equation 1")
equations = Equation(equation_array)
print(equation_array)
Stringify_Equations(equation_array)
Spiral(equation_array)

Smoothing values (neighbors between 1-9)

Instructions: Compute and store R=1000 random values from 0-1 as x. moving_window_average(x, n_neighbors) is pre-loaded into memory from 3a. Compute the moving window average for x for the range of n_neighbors 1-9. Store x as well as each of these averages as consecutive lists in a list called Y.
My solution:
R = 1000
n_neighbors = 9
x = [random.uniform(0,1) for i in range(R)]
Y = [moving_window_average(x, n_neighbors) for n_neighbors in range(1,n_neighbors)]
where moving_window_average(x, n_neighbors) is a function as follows:
def moving_window_average(x, n_neighbors=1):
n = len(x)
width = n_neighbors*2 + 1
x = [x[0]]*n_neighbors + x + [x[-1]]*n_neighbors
# To complete the function,
# return a list of the mean of values from i to i+width for all values i from 0 to n-1.
mean_values=[]
for i in range(1,n+1):
mean_values.append((x[i-1] + x[i] + x[i+1])/width)
return (mean_values)
This gives me an error, Check your usage of Y again. Even though I've tested for a few values, I did not get yet why there is a problem with this exercise. Did I just misunderstand something?
The instruction tells you to compute moving averages for all neighbors ranging from 1 to 9. So the below code should work:
import random
random.seed(1)
R = 1000
x = []
for i in range(R):
num = random.uniform(0,1)
x.append(num)
Y = []
Y.append(x)
for i in range(1,10):
mov_avg = moving_window_average(x, n_neighbors=i)
Y.append(mov_avg)
Actually your moving_window_average(list, n_neighbors) function is not going to work with a n_neighbors bigger than one, I mean, the interpreter won't say a thing, but you're not delivering correctness on what you have been asked.
I suggest you to use something like:
def moving_window_average(x, n_neighbors=1):
n = len(x)
width = n_neighbors*2 + 1
x = [x[0]]*n_neighbors + x + [x[-1]]*n_neighbors
mean_values = []
for i in range(n):
temp = x[i: i+width]
sum_= 0
for elm in temp:
sum_+= elm
mean_values.append(sum_ / width)
return mean_values
My solution for +100XP
import random
random.seed(1)
R=1000
Y = list()
x = [random.uniform(0, 1) for num in range(R)]
for n_neighbors in range(10):
Y.append(moving_window_average(x, n_neighbors))

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