Add numbers without carrying

Question

Lets say I have two numbers,

Number 1:

Number 2:

You see adding them left to right without the carry on adds up to 3625 how would I do this in python? I'm not sure but I need it to add 9+6 which is 15 but not to carry on the 1 when it adds 8+4 is there any way to add like this in python? And the end result i'm looking for is

Since it doesn't carry on remaining numbers

you can try spliting the numbers by converting them to strings and adding them separately. — monkut, Oct 08 '13 at 01:33
Does it matter if there is a more efficient way of doing it? Surely that would be the most legible, straightforward, and easy-to-understand way of doing it. — Waleed Khan, Oct 08 '13 at 01:36
When you say efficient, do you mean *fast* or *fewest lines of code*? Because if you mean the first, is it really going to slow down your code? — SethMMorton, Oct 08 '13 at 01:44
Hm, a little of both but I won't want a huge code to do this, roipii has that but I'm not sure how the zip works or what it's exactly doing — ruler, Oct 08 '13 at 01:59

FMc · Accepted Answer · 2013-10-08T02:41:35.003

This will work with varying N of integers of varying lengths:

from itertools import izip_longest

nums = [1646,
        2089,
         345]

revs = [str(n)[-1::-1] for n in nums]        # nums as reversed strings
izl  = izip_longest(*revs, fillvalue = '0')  # zip the digits together
xsum = lambda ds: str(sum(map(int, ds)))[-1] # digits -> last digit of their sum
rres = ''.join(xsum(ds) for ds in izl)       # result, but as reversed string
res  = int(rres[-1::-1])                     # result: 3960

Similar idea, but using map rather than izip_longest. I like this one better.

revs = [str(n)[-1::-1] for n in nums]           # nums as reversed strings
dsum = lambda *ds: sum(int(d or 0) for d in ds) # str-digits -> their sum
sums = map(dsum, *revs)[-1::-1]                 # digit sums, in order
ones = [str(s % 10) for s in sums]              # last digits of the sums
res  = int(''.join(ones))                       # result: 3960

Clever to flip it around and use izip_longest. :) – roippi Oct 08 '13 at 02:21 — roippi, Oct 08 '13 at 02:21

score 4 · Answer 2 · answered Oct 08 '13 at 01:53

4

Because I like to take poetic license with open-ended requests for 'more efficient':

In [49]: a,b = 1646,2089

In [50]: ''.join(str(sum(map(int,x)))[-1] for x in zip(str(a),str(b)))
Out[50]: '3625'

answered Oct 08 '13 at 01:53

roippi

25,533
4
48
73

hm, that looks good but can you explain what the zip is doing? – ruler Oct 08 '13 at 01:55
1

@ruler you take 2 iterables and 'zip' them together (like a zipper) to make a number of 2-tuples equal to the length of the shortest iterable. But honestly this is more useful as code golf than anything else. Write your own function that does the above - albeit in more lines - that a human can actually read. Just because I managed to do it in one line does not make it 'pythonic'. – roippi Oct 08 '13 at 02:05
All respect to roppi, this is very clever ([golfy ;)](http://codegolf.stackexchange.com)) code, but @ruler its not code you should consider for production for the very reason that you aren't sure what its doing. It will fall down if the strings are different lengths. – Oct 08 '13 at 02:13
@roipi can enumerate be used for this? – ruler Oct 08 '13 at 02:42
1

I think it's a bit clearer without the map: `int(''.join(str(int(a)+int(b))[-1] for a, b in izip(str(n1), str(n2))) )` – monkut Oct 08 '13 at 02:46

score 4 · Answer 3 · answered Oct 08 '13 at 02:22

This will work even if the numbers of digits is not the same in both the numbers.

num1, num2, i, temp = 1646, 2089, 10, 0
total = num1 + num2
while num1 or num2:
    if (num1 % 10) + (num2 % 10) > 9: temp += i
    num1, num2, i = num1 // 10, num2 // 10, i * 10

print total - temp

Output

shengy · Answer 4 · 2013-10-08T02:34:33.707

3

The following code demonstrates what you want, but please just consider this as a hint. The var names and lack of exception handling will jeopardize your real code.

import operator

n1 = 1646
n2 = 2089

l1 = [int(x) for x in str(n1)]
l2 = [int(x) for x in str(n2)]

res1 = map(operator.add, l1, l2)
res2 = [str(x)[-1] for x in res1]

num = "".join(res2)

print int(num)

edited Oct 08 '13 at 02:34

answered Oct 08 '13 at 01:50

shengy

9,461
4
37
61

Might I recommend changing the last loop to `num = "".join(res2)`? – SethMMorton Oct 08 '13 at 02:12
@SethMMorton Ofcourse:) I'm new to python too, I'll change it now. – shengy Oct 08 '13 at 02:34

score 1 · Answer 5 · answered Oct 08 '13 at 01:49

This is kind of ugly because I do not know how to index a digit in an integer, only a string, but it works.

If the two strings are always the same size:

A = str(1646)
B = str(2089)
result = ""

for i in range(0,len(A)):
    result += str((int(A[i]) + int(B[i]))%10)

result = int(result)

If the two strings are not always the same size, find which one is bigger (length wise). Say the length of the biggest is X and the others length is Y where X > Y. Append the first X-Y indexes of the bigger string onto result, then repeat the above with the remaining digits.

You also want to only take the units digit after adding `int(A[i]) + int(B[i])`, otherwise if the result is > 9... — William Gaul, Oct 08 '13 at 01:52

score 1 · Answer 6 · edited May 23 '17 at 10:31

The numbers might be different length, so convert both to strings, reverse order (makes indexing easier), reverse using extended slice ([::-1], Reverse a string in Python), reverse back,

result=""
A=str(1646)[::-1]
B=str(2089)[::-1]
for ndx in range(0,max(len(A),len(B)):
    result += str(int(A[ndx])+int(B[ndx]))
resut = int(result[::-1])

You can get carry pretty easily, and handle unequal length strings (explicitly),

#!/bin/env python
a=1646
b=20893
A=str(a)[::-1]
B=str(b)[::-1]
lenA = len(A)
lenB = len(B)
length = max(lenA,lenB)
print "length: "+str(length)

#string add,no carry
total=""
for ndx in range(0,length):
    digit = 0
    if(ndx<lenA):
        digit += int(A[ndx])
    if(ndx<lenB):
        digit += int(B[ndx])
    digit = digit%10
    #print "digit: "+str(digit)+", carry: "+str(carry)
    total += str(digit)
print "a: " +str(a)+"+b: " +str(b)
result = total[::-1]
resut = int(result)
print "result: " +str(result)

#string add,with carry
total=""
carry=0
for ndx in range(0,length):
    digit = carry
    if(ndx<lenA):
        digit += int(A[ndx])
    if(ndx<lenB):
        digit += int(B[ndx])
    carry = digit/10
    digit = digit%10
    #print "digit: "+str(digit)+", carry: "+str(carry)
    total += str(digit)
if(carry>0):
    total += str(carry)
    #print "digit: "+str(digit)+", carry: "+str(carry)
print "a: " +str(a)+"+b: " +str(b)
result = total[::-1]
resut = int(result)
print "result: " +str(result)

HennyH · Answer 7 · 2013-10-08T05:06:09.987

First convert the numbers into strings so we can iterate over the digits:

>>> n1 = str(1646)
>>> n2 = str(2089)

We can then add the corresponding digits then do %10 to get the last digit. All numbers which result from adding two integers 0-9 will be <= 18, therefore %10 will always return the last digit.

>>> [(int(a)+int(b))%10 for a,b in zip(n1,n2)]
[3, 6, 2, 5]

We then convert each int into a string, join the digits and convert back to a int:

>>> int(''.join(map(str,((int(a)+int(b))%10 for a,b in zip(n1,n2)))))
3625

Or alternatively (converting the digits as you go):

>>> int(''.join(str((int(a)+int(b))%10) for a,b in zip(n1,n2)))
3625

Use izip_longest(...,fillvalue="0") to add numbers of different lengths.

Add numbers without carrying

7 Answers7