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

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.

CodePudding user response：

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

CodePudding user response：

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)

CodePudding user response：

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).