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Daily Challenge #276 - Unwanted Dollar Signs

dev.to staff on July 30, 2020

If you're faced with an input box, like this: +--------------+ Enter the price of the item, in dolla...

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Josh • Jul 30 '20 • Edited

This is my solution in Python3

def remove_dollar(string: str) -> float:
  remove_char = string.replace('$', '' ).replace(' ', '')
  return float(remove_char)

EDIT: Removed space, thanks for correcting me guys

jpittiglio • Jul 30 '20

This was great, but failed (Python 3.8) the first test due to the added space - added an additional replace and passed all tests.

def remove_dollar(string:str) -> float:
  remove_char = string.replace('$', '').replace(' ', '')
  return float(remove_char)

Rafael Acioly • Jul 30 '20 • Edited

You could also use the method strip

return float(string.strip("$"))

strip doc:

S.strip([chars]) -> str

Return a copy of the string S with leading and trailing
whitespace removed.
If chars is given and not None, remove characters in chars instead.

Jose Angel Munoz • Jul 31 '20 • Edited

This doesn't work with string = "-$ 0.1" isn't it? I'd rather prefer the replace way :P

Josh • Jul 31 '20

Thank you!

Ardi • Jul 30 '20

you alse need to remove space from the string to avoid the exception

Mouad K. • Jul 30 '20

This is my solution in javascript:

const money_value = (str) => parseFloat(str.replace('$','').replace(' ',''));

A more readable version:

const money_value = (str) => {
    let result = str.replace('$',''); 
    result = result.replace(' ','');
    return parseFloat(result);
};

Lucas • Jul 31 '20

My solution in Typescript

const remove_dollar = (input: string) => Number(input.replace(/\$|\s/g, ""));

Nik • Aug 5 '20

I like this but I will do it without /g as you will encounter this replacement at the most once. Also space after $ maybe present or may not be so /\$?\s?/ would do I guess. What do you think?

Sourabh Choure • Jul 30 '20

In C with O(1):

#include <stdio.h>
#include <stdlib.h>

float money_value(char const * s)
{
    if(*s == '$')
        return atof(s+1);
    else if(*s=='-' && *(s+1)=='$')
        return -1*atof(s+2);
    else    
        return atof(s);
}

int main(void)
{
    printf("%f\n",money_value("$-.38"));
    printf("%f\n",money_value("-$ 0.1"));
    printf("%f\n",money_value("12.34"));
    printf("%f\n",money_value("$-2.3456"));
    printf("%f\n",money_value("$.2"));
}

Craig McIlwrath • Jul 31 '20

Is atof O(1)?

Sourabh Choure • Jul 31 '20

Well, technically atof is O(n), but while specifing the big O notation, I just abstracted all the library calls to O(1). And I think that doing this is the most stupid thing but it just helps me to analyse what could be the worst case scenario of my implementation. Hope it doesn't mislead any wrong information.

Steve Crow • Jul 30 '20

Here's Kotlin with a String extension function:

fun String.toMoneyValue() =
    this.filter { it.isDigit() || it in listOf('-', '.') }
        .toBigDecimal()

// Examples
"12.34".toMoneyValue()
"-0.89".toMoneyValue()
".11".toMoneyValue()
"007".toMoneyValue()

Rafi • Aug 8 '20

Ruby

def money_value(money)
  money_str = money.tr("$", "").tr(" ", "")
  # This is to remove trailing zeros
  if money_str.to_f == money_str.to_i
    return money_str.to_i
  end
  return money_str.to_f
end

Bubbler • Aug 7 '20

APL (using Dyalog APL):

      MoneyValue←{⍎⍵~'$ '}  ⍝ ⍵: dollar string
      MoneyValue '12.34'
12.34
      MoneyValue '-0.89'
¯0.89
      MoneyValue '.11'
0.11
      MoneyValue '007'
7
      MoneyValue '-$ 0.1'
¯0.1
      MoneyValue '$-2.3456'
¯2.3456
      MoneyValue '$.2'
0.2

(Use the bookmarklet on this post to see the code with APL font and syntax highlighting.)

Removes the characters $ and space using "set difference" ~, and evaluates ⍎ the resulting string as APL expression. The preceding - gets evaluated as the "negate" function, and APL's numeric notation uses ¯ (read "high minus") for the negative sign.

ptouchton • Aug 3 '20 • Edited

This was my solution typescript/javascript

 const arr: Array<string> = ['12.34', '-0.89', '-$ 0.1','007','$-2.3456','$.2','$. 7'];

    console.log(`array: ${arr}`);

    const solve = xs => {

      xs.forEach(x => {

        const num = +x.replace(/[\$,' ']/g, '');
        console.log(`num: ${num}`);

      })
    }

    let me = solve(arr);

  }

peter279k • Jul 31 '20

Here is the simple solution with Python:

Using the while loop and if...else conditions to check specific string pattern and replace unwanted character.

Then return s with casting to float value finally.

def money_value(s):
    s = s.replace(' ', '')
    s = s.replace('$' , '')

    if s == '':
        return 0.0

    if s[0] == '.':
        s = '0' + s

    if s[0] == '0' and s[1] == '0':
        s = s.replace('0', '')

    if len(s) < 4:
        if '.' not in s:
            s += '.'
        start_length = 0
        zero_len = 4 - len(s)

        while start_length < zero_len:
            s += '0'
            start_length += 1    

    return float(s)

fatalchemist • Jul 31 '20

solution in js

function money_value(input){
  return parseFloat(input.replace(/\$/g,'').replace(/\s/g,''));
}

Ajay • Jul 31 '20

Javascript

const convert = string => parseFloat(string.replace(/[^0-9.]/g, ''))

fatalchemist • Jul 31 '20

fails on negative values, due to replacing the '-'