How to check last digit of number
Question:
Is there a way to get the last digit of a number. I am trying to find variables that end with “1” like 1,11,21,31,41,etc..
If I use a text variable I can simply put
print number[:-1]
but it works for variables with text(like “hello) but not with numbers. With numbers I get this error:
TypeError: 'int' object is not subscriptable
I am trying to see if there’s a better way to deal with numbers this way. I know a solution is to convert to a string and then do the above command but I’m trying to see if there’s another way I have missed.
Thanks so much in advance…
Answers:
Remainder when dividing by 10, as in
numericVariable % 10
This only works for positive numbers. -12%10 yields 8
Use the modulus operator with 10:
num = 11
if num % 10 == 1:
print 'Whee!'
This gives the remainder when dividing by 10, which will always be the last digit (when the number is positive).
So you want to access the digits in a integer like elements in a list; easiest way I can think of is:
n = 56789
lastdigit = int(repr(n)[-1])
# > 9
Convert n into a string, accessing last element then use int constructor to convert back into integer.
For a Floating point number:
n = 179.123
fstr = repr(n)
signif_digits, fract_digits = fstr.split('.')
# > ['179', '123']
signif_lastdigit = int(signif_digits[-1])
# > 9
I can’t add a comment yet, but I wanted to iterate and expand on what Jim Garrison said
Remainder when dividing by 10, as in
numericVariable % 10
This only works for positive numbers. -12%10 yields 8
While modulus (%) is working as intended, throw on an absolute value to use the modulus in this situation.
abs(numericVariable) % 10
By using iteration and the in built function of ‘digit’ the number is treated as binary and so it goes from backwards to forwards. Here is an example of a bit of code for you.
for digit in binary:
denary= denary*2 + int(digit)
Lostsoul, this should work:
number = int(10)
#The variable number can also be a float or double, and I think it should still work.
lastDigit = int(repr(number)[-1])
#This gives the last digit of the variable "number."
if lastDigit == 1 :
print("The number ends in 1!")
Instead of the print statement at the end, you can add code to do what you need to with numbers ending in 1.
Hope it helped!
Convert to string first:
oldint = 10101
newint = int(str(oldint)[-1:])
The simplest and most efficient way is to use the reminder :
last_digit = orginal_number % 10
This is a simple yet effective way to do it
if number < 0:
remainder = number % -10
else:
remainder = number % 10
Is there a way to get the last digit of a number. I am trying to find variables that end with “1” like 1,11,21,31,41,etc..
If I use a text variable I can simply put
print number[:-1]
but it works for variables with text(like “hello) but not with numbers. With numbers I get this error:
TypeError: 'int' object is not subscriptable
I am trying to see if there’s a better way to deal with numbers this way. I know a solution is to convert to a string and then do the above command but I’m trying to see if there’s another way I have missed.
Thanks so much in advance…
Remainder when dividing by 10, as in
numericVariable % 10
This only works for positive numbers. -12%10 yields 8
Use the modulus operator with 10:
num = 11
if num % 10 == 1:
print 'Whee!'
This gives the remainder when dividing by 10, which will always be the last digit (when the number is positive).
So you want to access the digits in a integer like elements in a list; easiest way I can think of is:
n = 56789
lastdigit = int(repr(n)[-1])
# > 9
Convert n into a string, accessing last element then use int constructor to convert back into integer.
For a Floating point number:
n = 179.123
fstr = repr(n)
signif_digits, fract_digits = fstr.split('.')
# > ['179', '123']
signif_lastdigit = int(signif_digits[-1])
# > 9
I can’t add a comment yet, but I wanted to iterate and expand on what Jim Garrison said
Remainder when dividing by 10, as in
numericVariable % 10
This only works for positive numbers. -12%10 yields 8
While modulus (%) is working as intended, throw on an absolute value to use the modulus in this situation.
abs(numericVariable) % 10
By using iteration and the in built function of ‘digit’ the number is treated as binary and so it goes from backwards to forwards. Here is an example of a bit of code for you.
for digit in binary:
denary= denary*2 + int(digit)
Lostsoul, this should work:
number = int(10)
#The variable number can also be a float or double, and I think it should still work.
lastDigit = int(repr(number)[-1])
#This gives the last digit of the variable "number."
if lastDigit == 1 :
print("The number ends in 1!")
Instead of the print statement at the end, you can add code to do what you need to with numbers ending in 1.
Hope it helped!
Convert to string first:
oldint = 10101
newint = int(str(oldint)[-1:])
The simplest and most efficient way is to use the reminder :
last_digit = orginal_number % 10
This is a simple yet effective way to do it
if number < 0:
remainder = number % -10
else:
remainder = number % 10