How to convert string to integer without using library functions in Python?

Question

I am trying to convert

a = "546"

to

a = 546

without using any library functions.

Presumably "library functions" includes builtin functions like int, right? Considering that this is homework, we'd like to see what you've thought about/found/tried so far, and what you got stuck with. — Izaak van Dongen
– Izaak van Dongen, Commented Sep 25, 2017 at 13:10
Maybe you should ask what google terms your teacher thought you'd find useful and share what you did search for so we know what you've looked at that isn't helpful? Apart from that - what's your understanding of what a "library functions" is? — Jon Clements
– Jon Clements, Commented Sep 25, 2017 at 13:10
Looking at the final version of technically correct answer, and that it took 2-3 iterations to root away all the builtin functions, I must say this is a pretty sadistic task. — luk32
– luk32, Commented Sep 25, 2017 at 13:33

Stefan Pochmann · Accepted Answer · 2017-09-25 13:49:56Z

21

The "purest" I can think of:

>>> a = "546"
>>> result = 0
>>> for digit in a:
        result *= 10
        for d in '0123456789':
            result += digit > d

>>> result
546

Or using @Ajax1234's dictionary idea if that's allowed:

>>> a = "546"
>>> value = {'0':0, '1':1, '2':2, '3':3, '4':4, '5':5, '6':6, '7':7, '8':8, '9':9}
>>> result = 0
>>> for digit in a:
        result = 10 * result + value[digit]

>>> result
546

edited Sep 25, 2017 at 13:49

answered Sep 25, 2017 at 13:34

Stefan Pochmann

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1 Comment

E-WAVE Over a year ago

Why multiplying by the result by 10 though ? because i don't see any effect given that the original value of the result variable is zero

Ajax1234 · Accepted Answer · 2017-09-25 13:28:55Z

4

You can keep a dictionary that stores the string and integer values of a numeric key, and then iterate over the string. While iterating over the string, you can use enumerate to keep track of the index and then raise 10 to that power minus 1 and then multiply by the corresponding key from the dictionary:

a = "546"
length = 0
for i in a:
   length += 1
d = {'1': 1, '0': 0, '3': 3, '2': 2, '5': 5, '4': 4, '7': 7, '6': 6, '9': 9, '8': 8}
count = 0
counter = 0
for i in a:
   count += (10**(length-counter-1)*d[i])
   counter += 1
print(count)

Output:

edited Sep 25, 2017 at 13:28

answered Sep 25, 2017 at 13:21

Ajax1234

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7 Comments

dabadaba Over a year ago

and enumerate :)

Ajax1234 Over a year ago

I completely forgot about the OP's requirement barring all builtin functions, not just int :) Please see my recent edit.

dabadaba Over a year ago

range and len are also built-in functions aren't they? you can just do for i in a

Izaak van Dongen Over a year ago

You might be able to avoid using range or len by first slicing with [::-1] and then incrementing the power manually. However, I think that the builtin functions thing is probably more of a problem with the question's phrasing than this answer - where does a function stop? Isn't a list indexing a call to list.__getitem__?

dabadaba Over a year ago

I mean, I think we're all probably overkilling it. I guess the teacher was trying to disuade students from using int(), but I'd think using any other common built-in function like range or len is fine.

|

Nate Jenson · Accepted Answer · 2017-09-25 13:24:44Z

3

The trick is that 546 = 500 + 40 + 6, or 5*10^2 + 4*10^1 + 6*10^0.

Note how the exponent is just the index (in reverse). Using that, you can generalize this approach into a function:

def strToInt(number):
    total = 0                             # this is where we accumulate the result
    pwr = len(number) - 1                 # start the exponent off as 2
    for digit in number:                  # digit is the str "5", "4", and "6"
        digitVal = ord(digit) - ord('0')  # using the ascii table, digitVal is the int value of 5,4, and 6.
        total += digitVal * (10 ** pwr)   # add 500, then 40, then 6
        pwr -= 1                          # make sure to drop the exponent down by one each time
    return total

And you can use it like so:

>>> strToInt("546")
546

answered Sep 25, 2017 at 13:24

Nate Jenson

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2 Comments

Izaak van Dongen Over a year ago

@Evan and this results in good, readable, educative code. I think this answer is more in the spirit of the task than trying to golf any builtin words out of your code.

Gino Mempin Over a year ago

ord is a library function.

Shyambeer Singh · Accepted Answer · 2020-03-08 07:30:37Z

1

def stringToInt(s):
    result = 0
    value = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
    for digit in s:
        result = 10 * result + value[digit]

    return result

answered Mar 8, 2020 at 7:30

Shyambeer Singh

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Comments

Mohideen bin Mohammed · Accepted Answer · 2021-07-06 05:10:27Z

0

def int(a):
ty = a.__class__.__name__
out = 0
di = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4,
      '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
if ty not in ("str", "int", "float", "bytes"):
    raise TypeError("unsupported format")
if a.__class__ == float:
    return a.__floor__()
elif a.__class__ == int:
    return a
else:
    ind = 0
    for val in a[::-1]:
        if val not in di:
            raise ValueError("invalid input")
        out += di[val]*(10**ind)
        ind += 1
        #print(out, di[val])
    return out

print(int("55"))
55

answered Jul 6, 2021 at 5:10

Mohideen bin Mohammed

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Comments

faizan baig · Accepted Answer · 2017-09-25 13:38:21Z

-1

You can loop through the string and perform the operation on each character using ord.

example:

a="546"
num=0
for i in a:
     num = num * 10 + ord(i) - ord('0')

edited Sep 25, 2017 at 13:38

answered Sep 25, 2017 at 13:37

faizan baig

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1 Comment

Stefan Pochmann Over a year ago

ord is clearly a library function and thus not allowed.

Deep · Accepted Answer · 2019-06-16 04:59:04Z

-1

astr = "1234"
num = 0
for index,val in enumerate(astr[::-1]):
    res = (ord(val) - ord('0')) * (10 ** index)
    num += (res)

answered Jun 16, 2019 at 4:59

Deep

1

1 Comment

mastisa Over a year ago

ord is a library function

AlexK · Accepted Answer · 2022-09-29 21:18:22Z

-1

a=input()
r=0
for i in a:
  r=r*10+(ord(i)-ord("0"))
print(r)
print(type(r))

edited Sep 29, 2022 at 21:18

AlexK

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answered Sep 26, 2022 at 9:36

Kirti

1

1 Comment

AlexK Over a year ago

Two previous answerers already used ord() and they were told that it's not allowed.

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