How can I read inputs as numbers?

Question

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Why are x and y strings instead of ints in the below code?

(Note: in Python 2.x use raw_input(). In Python 3.x use input(). raw_input() was renamed to input() in Python 3.x)

play = True

while play:

    x = input("Enter a number: ")
    y = input("Enter a number: ")

    print(x + y)
    print(x - y)
    print(x * y)
    print(x / y)
    print(x % y)

    if input("Play again? ") == "no":
        play = False

Answer 1

n=int(input())
for i in range(n):
    n=input()
    n=int(n)
    arr1=list(map(int,input().split()))

the for loop shall run 'n' number of times . the second 'n' is the length of the array. the last statement maps the integers to a list and takes input in space separated form . you can also return the array at the end of for loop.

Answer 2

Convert to integers:

my_number = int(input("enter the number"))

Similarly for floating point numbers:

my_decimalnumber = float(input("enter the number"))

Answer 3

TLDR

• Python 3 doesn't evaluate the data received with input function, but Python 2's input function does (read the next section to understand the implication).
• Python 2's equivalent of Python 3's input is the raw_input function.

Python 2.x

There were two functions to get user input, called input and raw_input. The difference between them is, raw_input doesn't evaluate the data and returns as it is, in string form. But, input will evaluate whatever you entered and the result of evaluation will be returned. For example,

>>> import sys
>>> sys.version
'2.7.6 (default, Mar 22 2014, 22:59:56) \n[GCC 4.8.2]'
>>> data = input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
(22, <type 'int'>)

The data 5 + 17 is evaluated and the result is 22. When it evaluates the expression 5 + 17, it detects that you are adding two numbers and so the result will also be of the same int type. So, the type conversion is done for free and 22 is returned as the result of input and stored in data variable. You can think of input as the raw_input composed with an eval call.

>>> data = eval(raw_input("Enter a number: "))
Enter a number: 5 + 17
>>> data, type(data)
(22, <type 'int'>)

Note: you should be careful when you are using input in Python 2.x. I explained why one should be careful when using it, in this answer.

But, raw_input doesn't evaluate the input and returns as it is, as a string.

>>> import sys
>>> sys.version
'2.7.6 (default, Mar 22 2014, 22:59:56) \n[GCC 4.8.2]'
>>> data = raw_input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
('5 + 17', <type 'str'>)

Python 3.x

Python 3.x's input and Python 2.x's raw_input are similar and raw_input is not available in Python 3.x.

>>> import sys
>>> sys.version
'3.4.0 (default, Apr 11 2014, 13:05:11) \n[GCC 4.8.2]'
>>> data = input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
('5 + 17', <class 'str'>)

---

Solution

To answer your question, since Python 3.x doesn't evaluate and convert the data type, you have to explicitly convert to ints, with int, like this

x = int(input("Enter a number: "))
y = int(input("Enter a number: "))

You can accept numbers of any base and convert them directly to base-10 with the int function, like this

>>> data = int(input("Enter a number: "), 8)
Enter a number: 777
>>> data
511
>>> data = int(input("Enter a number: "), 16)
Enter a number: FFFF
>>> data
65535
>>> data = int(input("Enter a number: "), 2)
Enter a number: 10101010101
>>> data
1365

The second parameter tells what is the base of the numbers entered and then internally it understands and converts it. If the entered data is wrong it will throw a ValueError.

>>> data = int(input("Enter a number: "), 2)
Enter a number: 1234
Traceback (most recent call last):
  File "<input>", line 1, in <module>
ValueError: invalid literal for int() with base 2: '1234'

For values that can have a fractional component, the type would be float rather than int:

x = float(input("Enter a number:"))

---

Apart from that, your program can be changed a little bit, like this

while True:
    ...
    ...
    if input("Play again? ") == "no":
        break

You can get rid of the play variable by using break and while True.

Answer 4

For multiple integer in a single line, map might be better.

arr = map(int, raw_input().split())

If the number is already known, (like 2 integers), you can use

num1, num2 = map(int, raw_input().split())