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I have a double while loop, and it does not seem to be working because of some logic I'm doing incorrectly

I have a double while loop, and it does not seem to be working because of some logic I'm doing incorrectly. I'm not sure what is wrong exactly, but I feel the code may be too complicated and somewhere, there is an error.
enter code here
import math
print("How many numbers am I estimating John?")
count = int(input("COUNT> "))
print("Input each number to estimate.")
better_guess = 0
initial_guess = 10
i = 0
j = 0
t = 1
list = []
for j in range(count):
num = float(input("NUMBER> "))
list.append(num)
j = j + 1
if j == count:
print("The square roots are as follows:")
while i <= len(list):
while t != 0 :
initial_guess = 10
better_guess = (initial_guess + (list[i])/initial_guess) / 2
if initial_guess == better_guess:
print(f"OUTPUT After {t} iterations, {list[i]}^0.5 = {better_guess}")
i = i + 1
break
initial_guess = better_guess
i = i + 1

There are some errors in your code, #x pie has pointed out some of them but not all. The most important is that you are need to initalize t for every number in the list, so you can get the iterations for the numbers separately. Also, t needs to be incremented in the while loop, not inside the if block.
You can also clean up the code considerably, for example the j variable is not being used, list comprehension can be used to shorten the code (pythonic way), and iterating over lists can be done with for num in list.
Putting this altogether produces this:
count = int(input('How many numbers am I estimating John? \nCOUNT> '))
print("Input each number to estimate.")
list = [float(input(f'NUMBER {i+1}> ')) for i in range(count)]
print("The square roots are as follows:")
for num in list:
initial_guess = 10
t = 0
while True:
better_guess = (initial_guess + num/initial_guess) / 2
t += 1
if initial_guess == better_guess:
print(f"OUTPUT After {t} iterations, {num}^0.5 = {better_guess}")
break
initial_guess = better_guess
Sample run:
How many numbers am I estimating John?
COUNT> 4
Input each number to estimate.
NUMBER 1> 1
NUMBER 2> 9
NUMBER 3> 16
NUMBER 4> 4
The square roots are as follows:
OUTPUT After 9 iterations, 1.0^0.5 = 1.0
OUTPUT After 7 iterations, 9.0^0.5 = 3.0
OUTPUT After 7 iterations, 16.0^0.5 = 4.0
OUTPUT After 8 iterations, 4.0^0.5 = 2.0

#viggnah is right, and I just ignored the num of iterations. But I think #viggnah 's num of iterations are 1 bigger than the actual num of iterations. E.g., if input is 4 and initial guess is 2, the iteration should be 0 rather than 1. Also I add except in case of illegal input.
I suppose the following code works as you expect.
import math
print("How many numbers am I estimating John?")
count = int(input("COUNT> "))
print("Input each number to estimate.")
better_guess = 0
initial_guess = 10
i = 0
j = 0
t = 1
list = []
while True:
try:
num = float(input("NUMBER> "))
list.append(num)
j = j + 1
if j == count:
print("The square roots are as follows:")
break
except:
print("Invalid input! Try again.")
while i < len(list):
initial_guess = 10
t = 0
while True:
better_guess = (initial_guess + (list[i])/initial_guess) / 2
if initial_guess == better_guess:
print(f"OUTPUT After {t} iterations, {list[i]}^0.5 = {better_guess}")
break
t = t + 1
initial_guess = better_guess
i = i + 1
You need to understand:
We only need initialize guess once for each number. So do it in the first while loop;
tneeds to be updated when initial_guess==better_guess rather than i, I believe this is a clerical error;
initial_guessneeds to be updated in the second loop;

Is there a way I can compare the same variable and choose the higher one and print it?

The value of y is changing every time the while loops continues, I tried the watchpoints modules but I need a way to find out the highest y has ever been and print it, I'll print the code that I'm trying to get to work.
import random
from watchpoints import watch
def alg(I):
print(I)
x = 1
while i > 1:
if (i % 2) == 0:
i = int(i / 2)
x = x + 1
else:
i = int(3 * i + 1)
x = x + 1
print(I)
y = 1
if i in range(1, 9):
y = 1
if i in range(10, 99):
y = 2
if i in range(100, 999):
y = 3
if i in range(1000, 9999):
y = 4
watch(a)#I want to know when y reachest the highest
print(y, "is the max number of caracters")#then print it
print("numero passaggi = ", str(x))
print("1: choice")
print("2: random")
type = int(input(" 1 or 2: "))
if type == 1:
i = input("Enter a number: ")
alg(int(I))
elif type == 2:
i = random.randint(1, 100) # 10^9
alg(I)
else:
print("Enter 1 or 2")
I want to know when y reaches the highest and print it below.

There are many ways to do this, but since you seem interested in "watching" the values of y as they change, one good option might be to make your alg function a generator that yields the values of y. The caller can then do whatever it wants with those values, including taking the max of them.
Note that instead of doing this kind of thing to figure out how many digits a number has:
if i in range(1, 9):
y = 1
if i in range(10, 99):
y = 2
if i in range(100, 999):
y = 3
if i in range(1000, 9999):
y = 4
you can just do:
y = len(str(i))
i.e. turn it into a string and then count the characters.
def alg(i: int):
x = 1
while i > 1:
if i % 2 == 0:
i = i // 2
x += 1
else:
i = 3 * i + 1
x = x + 1
print(f"i: {i}")
yield len(str(i))
print(f"numero passaggi = {x}")
print(f"Max number of digits: {max(alg(50))}")
i: 25
i: 76
i: 38
i: 19
i: 58
i: 29
i: 88
i: 44
i: 22
i: 11
i: 34
i: 17
i: 52
i: 26
i: 13
i: 40
i: 20
i: 10
i: 5
i: 16
i: 8
i: 4
i: 2
i: 1
numero passaggi = 25
Max number of digits: 2

The simplest method would be to create another variable to store the highest value that y has achieved and run a check each loop to see if the new y value is larger than the previous max.
Here is an example:
def exampleFunc():
i = 0
y = yMax = 0
while i < 20:
y = random.randint(1, 100)
if y > yMax:
yMax = y
i += 1
print(yMax)

The easiest way to approach something like this is to transform your function into one that returns all intermediate values, and then aggregate those (in this case, using the builtin max()).
from math import log10
def collatz_seq(n):
yield n
while n > 1:
if n % 2:
n = 3 * n + 1
else:
n //= 2
yield n
def print_stats(n):
seq = collatz_seq(n)
idx, val = max(enumerate(seq), key=lambda x: x[1])
digits = int(log10(val)) + 1
print(f"{digits} is the max number digits")
print(f"{idx} is the iteration number")
Here I use enumerate() to give me the index of each value, and I use math.log10() to obtain the number of digits in the number (minus one).

Factorization: What went wrong with d? [duplicate]

This question already has answers here:
What is a debugger and how can it help me diagnose problems?
(2 answers)
Closed 2 years ago.
Consider:
Enter image description here
Input: 20
17
999997
Output: 2^2 * 5
17
757 * 1321
My code:
a = int(input())
# Find the factors first
for i in range(2, a+1):
s = 0
b = a
d = 0
# See if it is a prime number
if a%i == 0:
for x in range(1, i+1):
if a%x == 0:
d = d + x
if (d-1)/i == 1:
d = 0
print(i)
else:
s = 0
b = a
d = 0
continue
d = 0
# I will see how many prime numbers
while(b>0):
if (b/i)%1 == 0:
s = s + 1
b = b/i
else:
b = 0
if b == 1:
b = 0
print(s)
I will find the factors first, and then see if it is a prime number. If so, I will see how many prime numbers it is
if i input 12, it outputs 2 2
Enter link description here

I believe you need the output of the following.
import math
a = int(input())
while (a % 2 == 0):
print(2)
a = int(a/2)
while (a % 3 == 0):
print(3)
a = int(a/3)
for i in range(5, math.ceil(math.sqrt(a)), 6):
while (a % i == 0):
print(i)
a = int(a / i)
while (a % (i + 2) == 0):
print(i + 2)
a = int(a / (i + 2))
if (a > 3):
print(a)
This will give you the prime factors for a given number. As I can understand, it is what you are looking for.

a = int(input("Enter a number:"))
for i in range(2, a + 1):
if a % i != 0:
continue
# SETTING THE DEFAULT VALUES AT THE BEGINNING OF EVERY ITERATION OF THE LOOP
s = 0
b = a
d = 0
for x in range(1, i + 1):
if b % x == 0:
d = d + x
if (d - 1) / i == 1:
d = 0
print(i)
else:
# s = 0 # NO LONGER NEEDED, AS WE RESET THEM AT THE BEGINNING OF THE LOOP
# b = a
# d = 0
continue
while b > 0:
if (b / i) % 1 == 0:
s = s + 1
b = b / i
else:
b = 0
if b == 1:
b = 0
print(s)
a /= i**s # THIS LINE IS IMPORTANT
You were close. You forgot to set the default values at the beginning of every iteration of the loop, so they sometimes didn't have the right values ; and you should set a to a different value by dividing it by the factor you found (i**s, so i to the power of s).
As has been mentioned, your code also follows an odd coding style. I suggest you stop putting newlines between each statement, and start separating operators with spaces (example: range(3+5) is bad, range(3 + 5) is more readable)

You are using too many loops here and that's why you are getting too much confused. Here is the code which serve the same purpose (if I understand your problem correctly)
a = int(input("Enter a number: "))
i = 2
factors = []
while i <= a:
if (a%i) == 0:
factors.append(i)
a = a/i
else:
i = i + 1
print(factors)
here I am returning a list, if you want you can change the type accordingly.
Here are the inputs/outputs:
Enter a number: 17
[17]
Enter a number: 100
[2, 2, 5, 5]
Enter a number: 12
[2, 2, 3]

Finding even numbers with while as

I'm doing this assignment:
Write a program that prints all even numbers less than the input
number using the while loop.
The input format:
The maximum number N that varies from 1 to 200.
The output format:
All even numbers less than N in ascending order. Each number must be
on a separate line.
N = int(input())
i = 0
while 200 >= N >= 1:
i += 1
if i % 2 == 0 and N > i:
print(i)
and its output like:
10 # this is my input
2
4
6
8
but there is an error about time exceed.

The simple code would be:
import math
N = int(input(""))
print("1. " + str(N))
num = 1
while num < math.ceil(N/2):
print (str(num) + ". " + str(num * 2))
num += 1

The problem is that the while loop never stops
while 200 >= N >= 1 In this case because you never change the value of N the condition will always be true. Maybe you can do something more like this:
N = int(input())
if N > 0 and N <= 200:
i = 0
while i < N:
i += 2
print(i)
else
print("The input can only be a number from 1 to 200")

How can you sum these outputs in Python?

I made this code about a number and it's power. It will ask a number and it's power and show the output like a horizontal list.. Like
Number = 2
Power = 3.... then output will be like=
1
2
4
Number and power can be +/-.
But I want to sum those numbers like Sum = 7 after it shows
1
2
4
I have no idea how to do it after the output. I am new to programming maybe that's why can't figure out this problem.
Here is the code in Python :
A =float(input("Number:"))
B =float(input("Power:"))
print("Result of Powers:")
i = 0
while i < B:
print(A**i)
i = i + 1
while i >= B:
print(A**i)
i = i - 1

You could simplify this with numpy as follows
import numpy as np
A =float(input("Number:"))
B =int(input("Power:"))
print("Result of Powers:")
power = np.arange(B)
power_result = A ** power
sum_result = np.sum(power_result)
print(power_result)
print(sum_result)
I made B into an int, since I guess it makes sense. Have a look into the numpy documentation to see, what individual functions do.

You can create another variable to store the sum
and to print values on the same line use end=" " argument in the print function
a = float(input("Number:"))
b = int(input("Power:"))
sum = 0.0
i = 0
while b < 0:
ans = a**i
i = i - 1
print(ans, end=" ")
sum = sum + ans
b += 1
while b >= 0:
ans = a**i
i = i + 1
print(ans, end=" ")
sum = sum + ans
b -= 1
print("\nSum = " + str(sum))

I'm not sure what you want to achieve with the second loop. This works:
A =float(input("Number:"))
B =float(input("Power:"))
print("Result of Powers:")
i = 0
n_sum = 0
while i < B:
n_sum += A**i
print(A**i)
i = i + 1
while i >= B:
n_sum += A**i
print(A**i)
i = i - 1
print(n_sum)