wrong output Python program

Question

writing code to find out highest number in list

import sys

print ("Enter number of elements \n")

i = int(sys.stdin.read())

print ("Enter numbers \n")

input = sys.stdin.read()

numbers = input.split()

print ("Number of elements in list",len(numbers))

if (len(numbers)<=i):

    print ("Valid Inputs \n", len(numbers))

    max_index1=-1

    for x in range(len(numbers)):

        if ((max_index1==-1)or(numbers[x] >= numbers[max_index1])):

            max_index1=x

            print ("max index is",max_index1)

print ("Highest input number is", numbers[max_index1])

while executing this code I am getting correct output as below:

Enter number of elements

5
Enter numbers

10 45 32 23 21
Number of elements in list 5
Valid Inputs

 5
max index is 0
max index is 1
Highest input number is 45

but sometimes the output is totally wrong:

Enter number of elements

5
Enter numbers

10 45 32 6 21
Number of elements in list 5
Valid Inputs
 5
max index is 0
max index is 1
max index is 3
Highest input number is 6

can someone please help me to fix this code?

Are you aware that input is a reserved word, that your code blocks? — boardrider
– boardrider, Commented Nov 7, 2016 at 18:07

Jim Copley · Accepted Answer · 2016-11-07 00:27:45Z

1

What you want to do is use the max function. You can call max on an array of numbers and it will return the largest number.

num_array = [5, 3, 6, 12, 3, 5]
largest = max(num_array)
print("Max num: {}".format(largest))

answered Nov 7, 2016 at 0:27

Jim Copley

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Comments

Bakyuns · Accepted Answer · 2016-11-07 00:36:48Z

0

you should use int function in your code.

int(numbers[x]) >= int(numbers[max_index1]))

If you shouldn't, Python compare numbers[x] string and numbers[max_index1]

In dictionary order, "45" is behind "6"

edited Nov 7, 2016 at 0:36

answered Nov 7, 2016 at 0:33

Bakyuns

587 bronze badges

2 Comments

sudip Over a year ago

sure will give this a try, but how typecasting will help here?

Bakyuns Over a year ago

int typecasting makes a string integer. For instance, int("45") became integer 45. If not use int typecasting, you just compare "45" and "6". Therefore, as "4" is smaller than "6" in ascii code, your answer will be wrong. I think you want to compare the integers.

Moses Koledoye · Accepted Answer · 2016-11-07 00:48:28Z

The items in your list are strings, so they are being ordered lexicographically:

>>> '6' > '45'
True
>>> 6 > 45
False

You should convert the items in the list to integer type to do a numerical ordering by replacing:

input = sys.stdin.read()
numbers = input.split()

With:

numbers = [int(i) for i in input().strip().split()] # use 'raw_input' for Python 2

That reads the items in using the builtin input function, strips leading and trailing whitespaces, splits the string into a list and converts each item in the list to integer using a list comprehension.

Note that using input as a name is not a good idea as it shadows the builtin input function which you should probably be using in place of sys.stdin.read.

Mark Hannel · Accepted Answer · 2016-11-07 00:50:43Z

Currently your list numbers is a list of strings. To show this, you can print numbers.

>>> print(numbers)
['10', '42', '32', '6', '21']

In python 2.7, a string can be compared with another string with (possibly) surprising results

>>>  '6' > '42'
True

To avoid this problem, you can change numbers to be a list of integers

>>> numbers = map(int, numbers)
>>> print(numbers)
[10, 42, 32, 6, 21]

where map allows you to apply a function to each of the elements of numbers and return the result.

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