With my code, I want to get the last two digits of an integer. But when I make xa positive number, it will take the first x digits, if it is a negative number, it will remove the first x digits.
Code:
number_of_numbers = 1
num = 9
while number_of_numbers <= 100:
done = False
num = num*10
num = num+1
while done == False:
num_last = int(repr(num)[x])
if num_last%14 == 0:
number_of_numbers = number_of_numbers + 1
done = True
else:
num = num + 1
print(num)
Why don't you extract the absolute value of the number modulus 100? That is, use
abs(num) % 100
to extract the last two digits?
In terms of performance and clarity, this method is hard to beat.
To get the last 2 digits of num
I would use a 1 line simple hack:
str(num)[-2:]
This would give a string. To get an int, just wrap with int:
int(str(num)[-2:])
Simpler way to extract last two digits of the number (less efficient) is to convert the number to str
and slice the last two digits of the number. For example:
# sample function
def get_last_digits(num, last_digits_count=2):
return int(str(num)[-last_digits_count:])
# ^ convert the number back to `int`
OR, you may achieve it via using modulo %
operator (more efficient) , (to know more, check How does % work in Python? ) as:
def get_last_digits(num, last_digits_count=2):
return abs(num) % (10**last_digits_count)
# ^ perform `%` on absolute value to cover `-`ive numbers
Sample run:
>>> get_last_digits(95432)
32
>>> get_last_digits(2)
2
>>> get_last_digits(34644, last_digits_count=4)
4644
to get the last 2 digits of an integer.
a = int(input())
print(a % 100)
You can try this:
float(str(num)[-2:])
abs(num) % 100 would not work to get 00 from a number such as 900
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