Как найти максимальное нечетное число питон

n = int(input("Введите количество элементов, которые будут в списке:"))
b = []
for i in range(n):
    a = int(input("Элемент:"))
    b.append(a)

c = []
d = []
for i in b:
    if i % 2 == 0:
        c.append(i)
    else:
        d.append(i)

c.sort()
d.sort()
print("Минимальное четное число:",c[0])
print("Максимальное нечетное число",d[-1])

3

Approach

Avoid using if-stmts to find maximum. Use python builtin max. Use either generator or filter to find only the odd numbers.

Using builtins like this is safer/more reliable because it is simpler to compose them, the code is well-tested, and the code executes mostly in C (rather than multiple byte code instructions).

Code

def find_largest_odd(*args):
    return max(arg for arg in args if arg & 1)

or:

def find_largest_odd(*args):
    return max(filter(lambda x: x & 1, args))

Test

>>> def find_largest_odd(*args):
...     return max(arg for arg in args if arg & 1)
... 
>>> print find_largest_odd(1, 3, 5, 7)
7
>>> print find_largest_odd(1, 2, 4, 6)
1

and:

>>> def find_largest_odd(*args):
...     return max(filter(lambda x: x & 1, args))
>>> print find_largest_odd(1, 3, 5, 7)
7
>>> print find_largest_odd(1, 2, 4, 6)
1

If you pass an empty sequence or provide only even numbers, you will get a ValueError:

>>> find_largest_odd(0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in find_largest_odd
ValueError: max() arg is an empty sequence

References

• filter
• max
• generators

answered Mar 31, 2013 at 19:19

try:
    largest = max(val for val in (x,y,z) if val % 2)
    print(largest)
except ValueError:
    print('Even')

Note that sorted is a O(n log n) operation, while max is O(n). The speed difference may not matter for sequences that are this short, but it is good practice to use the best tool for the job.

answered Mar 31, 2013 at 18:24

This is the finger exercise from Ch.2 of Introduction to Computation and Programming Using Python by John V. Guttag. It’s the recommended text for the MITx: 6.00x Introduction to Computer Science and Programming. The book is written to be used with Python 2.7.

‘Write a program that examines three variables x,y and z, and prints the largest odd number among them. If none of them are odd, it should print a message to that effect.’

At this point the book has only introduced variable assignment and conditional branching programs and the print function. I know as I’m working through it. To that end this is the code I wrote:

if x%2 == 0:
    if y%2 == 0:
        if z%2 == 0:
            print "None of them are odd"
        else:
            print z
    elif y > z or z%2 == 0:
        print y
    else:
        print z
elif y%2 == 0:
    if z%2 == 0 or x > z:
        print x
    else:
        print z
elif z%2 == 0:
    if x > y:
        print x
    else:
        print y
else:
    if x > y and x > z:
        print x
    elif y > z:
        print y
    else:
        print z

It seems to work for all the combinations I have tried, but it would be useful to know if it can be shortened bearing in mind my points from paragraph three. I appreciate it has been aswered, but by providing this answer, other users are more likely to come across it when searching for answers to the question in the book.

answered Nov 8, 2013 at 23:05

something like this:

def get_max_odd(*lis):
    try:
         return sorted(i for i in lis if i%2)[-1] #IndexError if no odd item found
    except IndexError:    
         return "even"


In [8]: get_max_odd(1,2,3)
Out[8]: 3

In [9]: get_max_odd(2,4,6)
Out[9]: 'even'

In [10]: get_max_odd(2,5,6)
Out[10]: 5

In [11]: get_max_odd(2,4,6,8,9,10,20)
Out[11]: 9

answered Mar 31, 2013 at 18:41

I am working my way through the same book by Guttag (and Python 2.7). Others reading it may take cognisance that lists (or slicing them) haven’t been introduced yet though I have used one in my solution (I think it works fine!?!). And you don’t really need to reverse the list order if you would rather slice from the end instead of the beginning of the list.

First I created an empty list (lst). But before I use it I check if all of x, y and z are not odd (Guttag asks if ‘none of them are odd’). Then, each variable in turn is checked to see if its odd and if it is, it gets added to the list. I sorted the list in a descending order (i.e. largest odd numbers to the front).. then check to make sure that it has at least one element (no point in printing an empty list) and then print the first element.

x,y,z = 111,45,67

lst = []
if x%2==0 and y%2==0 and z%2==0:
    print 'none of x,y or z is an odd number'
else:
    if x%2!=0:
        lst.append(x)
    if y%2!=0:
        lst.append(y)
    if z%2!=0:
        lst.append(z)
lst.sort(reverse = True)
if len(lst)!=0:
    print lst[:1]

answered Mar 26, 2014 at 12:56

As per in the comments, the post by AfDev and others will throw a TypeError if x is even, as it will attempt

if y > None:

Edit: The commentor’s solution (initialising largest = 0 instead of largest = None) will not work either if the highest odd is negative.

The question from the book assumes no knowledge of lists or loops so here’s my solution:

x, y, z = 2, -5, -9
largest = None

if x % 2 != 0:
    largest = x
if y % 2 != 0:
    if largest == None or y > largest: # if None it will not attempt 2nd conditional
        largest = y
if z % 2 != 0:
    if largest == None or z > largest:
        largest = z

print(largest) # prints -5 

answered Jul 7, 2021 at 13:36

This is the first script I have written in my career after contemplating on it since yesterday evening.

Actually I was looking for answer to refer as I was constantly getting errors but could not find anything satisfactory, here is one from me 🙂

x = 333312
y = 221569
z = 163678

if x%2 ==0 and y%2==0 and z%2==0:
    print "Only even numbers provided."

elif x%2==0:
    if y%2!=0 and z%2!=0:
        if y > z:
            print y
        else:
            print z
    else:
        if y%2!=0:
            print y
        if z%2!=0:
            print z


elif y%2==0:
    if x%2!=0 and z%2!=0:
        if x > z:
            print x
        else:
            print z
    else:
        if x%2!=0:
            print x
        if z%2!=0:
            print z



elif z%2==0:
    if x%2!=0 and y%2!=0:
                if x > y:
                    print x
                else:
                    print y
    else:
        if x%2!=0:
                print x
        if y%2!=0:
                print y

else:
    if x>y and x>z:
        print x
    elif y>z:
        print y   
    else:
        print z

answered Jan 11, 2015 at 8:31

I am also currently going through the book and came across this problem. I used an approach similar to some of the solutions above. However, instead of just printing the largest value, the program prints the variable holding the largest value as well.

The difference here becomes that rather than checking if x > y, x > z, etc., the program also checks for equality (i.e. x >= y, x >= z, etc.). For multiple equal values, the program will print all the variables that share the highest odd value.

Finger Exercise 2.2

Write a program that examines three variables- x, y, and z- and prints the
largest odd number among them.
If none of them are odd, it should print a message to that effect.

x, y, z = 5,7,7

if x%2 == 1 or y%2 == 1 or z%2 == 1: #at least one of the variables is odd
    if x%2 == 1:
        if y%2 == 1:
            if z%2 == 1: #x, y, and z are all odd
                if x >= y and x >= z:
                    print "x =", x, "is the largest odd number!"
                if y >= x and y >=z:
                    print "y =",y, "is the largest odd number!"
                if z >= x and z >= y:
                    print "z =", z, "is the largest odd number!"
            else: #z is even, but x and y are still odd
                if x >= y:
                    print "x =", x, "is the largest odd number!"
                if y >= x:
                    print "y =",y, "is the largest odd number!"
        elif z%2 == 1: #y is even, but x and z are odd
            if x >= z:
                print "x =", x, "is the largest odd number!"
            if z >= x:
                print "z =", z, "is the largest odd number!"
        else: #x is the only odd number
            print "x = ", x, "is the largest odd number!"
    if x%2 != 1 and y %2 == 1: #x is not odd but y is odd
    #could have done an elif here. But this makes the code easier to follow
        if z%2 == 1:#z is also odd
            if y >=z:
                print "y =",y, "is the largest odd number!"
            if z >= y:
                print "z =", z, "is the largest odd number!"
        else: #z is even. Hence, y is the only odd number
            print "y =", y, "is the largest odd number!"
    if x%2 != 1 and y%2 != 1 and z%2 == 1:#z is the only odd number
        print "z =", z, "is the largest odd number!"             
else:
    print "No odd number was input!"

This answers the last finger exercise in the Ch.02 of the book by Prof. John Guttag mentioned in earlier posts. Using lists seemed helpful to me. The same method can be used for the previous finger exercise in the same chapter.

import math



a=int(input("Enter first integer: ")) # Manual entering of ten integers#
b=int(input("Enter second integer: "))
c=int(input("Enter third integer: "))
d=int(input("Enter fourth integer: "))
e=int(input("Enter fifth integer: "))
f=int(input("Enter sixth integer: "))
g=int(input("Enter seventh integer: "))
h=int(input("Enter eighth integer: "))
i=int(input("Enter ninth integer: "))
j=int(input("Enter tenth integer: "))

master=[a,b,c,d,e,f,g,h,i,j]



def ifalleven(a): # Finding whether every integer in the list if even or not.#

    i=0
    list=[]
    while i<len(a):
        if a[i]%2==0:
            list.append(a[i])
        i=i+1
    return len(list)==len(a)



def findodd(a): # Finding the greatest odd integer on the list.#
    i=0
    list=[]
    while i<len(a):
        if a[i]%2!=0:
            list.append(a[i])
        i=i+1
    print (max(list))



def greatoddoreven(list): # Finding the greatest odd integer on the list or if none are odd, print a message to that effect.#
    if ifalleven(list)==True:
        print ('No odd numbers found!')
    else:
        findodd(list)



greatoddoreven(master)

answered Apr 17, 2016 at 6:35

x=input('x= ')
y=input('y= ')
z=input('z= ')
largest = None
if x%2 == 0 and y%2 == 0 and z%2 == 0:
    print('none of them is odd')
else:
    if x%2 != 0:
        largest = x
    if y%2 != 0:
        if y > largest:
            largest = y
    if z%2 != 0:
        if z > largest:
            largest = z
    print(largest)

answered Feb 25, 2017 at 5:33

just finished working the same problem. My answer seems different from the other ones but seems to be working fine.( Or am I missing something?!) so here is an alternative:

With only if/else:

    x = 4
    y = 7
    z = 7

    if x%2 and y%2 and z%2 == 1:
        if x > y and x > z:
            print x
        elif y > z:
            print y
        else:
            print z
    elif x%2 and y%2 == 1:
        if x > y:
            print x
        else:
            print y
    elif x%2 and z%2 == 1 :
        if x > z:
            print x
        else:
            print z
    elif y%2 and z%2 == 1:
        if y > z:
            print y
        else:
            print z
    elif x%2 == 1:
        print x
    elif y%2 == 1:
        print y
    elif z%2 == 1:
        print z
    else:
        print "there are no odd numbers"

as far as I know it works with negative, numbers, big numbers, … if not let me know!

answered Mar 8, 2017 at 5:46

Here is my interpretation, I am doing the same task which is part of the Book Introduction to Computation and Programming Using Python, 2nd ed, John Guttag

# edge cases 100, 2, 3 and 100,2,-3 and 2,2,2
x = 301
y = 2
z = 1

# what if the largest number is not odd? We need to discard any non-odd numbers

# if none of them are odd - go straight to else
if (x % 2 != 0) or (y % 2 != 0) or (z % 2 != 0):
    #codeblock if we have some odd numbers
    print("ok")
    
    # initialising the variable with one odd number, so we check each in turn
    if (x % 2 != 0):
        largest = x
    elif (y % 2 != 0):
        largest = y
    elif (z % 2 != 0):
        largest = z

    # here we check each against the largest
    # no need to check for x as we did already

    if (y % 2 != 0):
        if y > largest:
            largest = y
    
    if (z % 2 != 0):
        if z > largest:
            largest = z
    
    print("The largest odd number is:", largest)
    print("The numbers were", x, y, z)
        
else: 
    print("No odd number found")`

answered Nov 13, 2020 at 14:57

Here is my code for Guttag’s Finger exercise 2:

def is_odd(x):
    """returns True if x is odd else returns False"""
    if x % 2 != 0:
        return(True)
    else:
        return(False)

def is_even(x):
    """returns True if x is even else returns False"""
    if x % 2 == 0:
        return(True)
    else:
        return(False)

def largest_odd(x, y, z):
    """Returns the largest odd among the three given numbers"""
    if is_odd(x) and is_odd(y) and is_odd(z):
        return(max(x, y, z))
    elif is_odd(x) and is_odd(y) and is_even(z):
        return(max(x, y))
    elif is_odd(x) and is_even(y) and is_odd(z):
        return(max(x, z))
    elif is_even(x) and is_odd(y) and is_odd(z):
        return(max(y, z))
    elif is_odd(x) and is_even(y) and is_even(z):
        return(x)
    elif is_even(x) and is_odd(y) and is_even(z):
        return(y)
    elif is_even(x) and is_even(y) and is_odd(z):
        return(z)
    else:
        return("There is no odd number in the input")

answered Dec 25, 2020 at 6:54

Assuming we aren’t going to use something fancy 🙂 like the max and sort functions in python…I think it’s safe to say than an “assignment” statement is fair game….we do after all have to assign x,y and z.

As such: I created my own variable called “largest” initialized as zero.
I assign that variable the first odd number of x,y,z (interrogated in that order).
Then I simply check the remaining values and if each one is greater than “largest” and is also an odd number then I make it the value of “largest” — and check the next value.
Once all the values are checked, “largest” holds the value of the largest odd number or zero (which means I never did find an odd number and I print a message to that effect.

This is the same method you could use to read an arbitrary number (perhaps 10) of values from a user and print the largest odd one after all values are entered. Just check each value as it’s entered against the current “largest” one.

Yes, you do need to be careful to support negative odd numbers — you need to handle the assignment of “largest” to the first odd number whether or not it’s greater than zero.

answered Oct 6, 2013 at 4:29

It’s actually easier for me to to think about this problem with more values rather than only 3 (x,y,z). Suppose I show you a list of words Apple, dog, Banana, cat, monkey, Zebra, elephant… etc. Let’s say there are 26 words denoted by variables a,b,c,d,e,f…x,y, and z.

If you need to find the “largest word” (in dictionary order) that begins with a lowercase letter, you don’t need to compare every word with every other word. Our brains don’t typically sort the entire list into order and then pick from the end either…well…mine doesn’t anyway.

I just start at the front of the list and work my way through it. The first time I find a word that starts with a lowercase letter…I memorize it and then read until I get another lowercase word — then I check which is larger and memorize just that word…. I never have to go back and check any of the others. One pass on the list and I’m done. In the list above…my brain immediately discards “Zebra” because the case is wrong.

answered Oct 6, 2013 at 5:01

The question asks to print out the largest odd number from 3 variables -x, y, and z, and since it’s chapter 2 which only goes as far as describing conditionals I’m guessing the ans should be provided by using only conditionals.

The program can be completed by exhaustively writing all the possible combinations of odd or even. Since there is 3 variables and at any one time these variables can be either odd or even there is 2^3 = 8 combinations/ conditionals.
This can be shown by using a Truth Table but instead of using True and False we use ‘O’ for odd and ‘E’ for even.
The table will look like this Odd-Even Table

Each row is a conditional in the code which checks ‘oddness/evenness’. Inside each of these conditionals you will then check which variables is the largest odd number using nested conditionals for this.

x = int(input('Enter an integer: '))
y = int(input('Enter another integer: '))
z = int(input('Enter a final integer: '))

print('x is:', x)
print('y is:', y)
print('z is:', z)
print('n')

if x % 2 == 1 and y % 2 == 1 and z % 2 == 1:
    print('First conditional triggered.')
    if x > y and x > z:
        print(x, 'is the largest odd number.')
    elif y > x and y > z:
        print(y, 'is the largest odd number.')
    elif z > x and z > y:
        print(z, 'is the largest odd number.')
    else:
        print(x, 'is the largest odd number.')
    # This else clause covers the case in which all variables are equal.
    # As such, it doesn't matter which variable you use to output the 
    # largest as all are the largest.
elif x % 2 == 1 and y % 2 == 1 and z % 2 == 0:
    print('Second conditional is triggered.')
    if x > y:
        print(x, 'is the largest odd number.')
    elif y > x:
        print(y, 'is the largest odd number.')
    else:
        print(x, 'is the largest odd number.')
elif x % 2 == 1 and y % 2 == 0 and z % 2 == 1:
    print('Third conditional is triggered.')
    if x > z:
        print(x, 'is the largest odd number.')
    elif z > x:
        print(z, 'is the largest odd number.')
    else:
        print(x, 'is the largest odd number.')
elif x % 2 == 1 and y % 2 == 0 and z % 2 == 0:
    print('Fourth conditional is triggered.')
    print(x, 'is the largest odd number.')
elif x % 2 == 0 and y % 2 == 1 and z % 2 == 1:
    print('Fifth conditional is triggered.')
    if y > z:
        print(y, 'is the largest odd number.')
    elif z > y:
        print(z, 'is the largest odd number.')
    else:
        print(y, 'is the largest odd number.')
elif x % 2 == 0 and y % 2 == 1 and z % 2 == 0:
    print('Sixth conditional is triggered.')
    print(y, 'is the largest odd number.')
elif x % 2 == 0 and y % 2 == 0 and z % 2 == 1:
    print('Seventh conditional is triggered.')
    print(z, 'is the largest odd number.')
else:
    print('Eight conditional is triggered.')
    print('There is no largest odd number as all numbers are even.')

This method works fine for 3 variables but the complexity and amount of conditionals needed can rise sharply with more variables being added.

answered Sep 13, 2016 at 3:36

Just done this excercise and this seems like an efficient way of doing it although it doesn’t seem to handle negative numbers (I don’t think it’s supposed to at this level though):

x, y, z = 7, 6, 12
xo = 0
yo = 0
zo = 0

if x%2 == 0 and y%2 == 0 and z%2 == 0:
    print("None of the numbers are odd.")

if x%2 == 1:
    xo = x

if y%2 == 1:
    yo = y

if z%2 == 1:
    zo = z

if xo > yo and xo > zo:
    print(xo)

if yo > xo and yo > zo:
    print(yo)

if zo > xo and zo > yo:
    print(zo)

answered Dec 7, 2019 at 15:01