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Daily Challenge #163 - Significant Figures

dev.to staff on January 15, 2020

Setup Write a function that takes a number and returns the number of significant figures in a given number. Significant Figures are the...

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Idan Arye • Jan 15 '20

Trailing zeroes in a number with a decimal point are significant
78.200 has 5 significant figures
20.0 has 3 significant figures

Trailing zeroes in a number without a decimal point are not significant
1200 has 2 significant figures
345000 has 3 significant figures

Shouldn't it be the other way around?

schwepmo • Jan 15 '20

I just checked on Wikipedia and it is right the way that it is stated in the task. This is because if you add zeros after the decimal point, you imply that your number was "measured" with a higher precision.

Example: a normal ruler vs. a precision measuring devices

If you use the normal ruler you can only be certain your measured distance is 1.5cm

With a precision ruler you can be certain, that your measured distance is something like 1.500cm

SavagePixie • Jan 15 '20

Oh, so it's a measurement thing.

Interestingly enough, technically, according to Wikipedia we have no way of knowing whether any zero to the right of the last non-zero digit is significant or not, regardless of the presence of decimals. So the exercise is still not quite right.

schwepmo • Jan 15 '20

I feel like the exercise is fine. It defines a definite rule for deciding on which trailing zeros are significant and which aren't.

SavagePixie • Jan 15 '20

Yes, but I suppose it's changed to avoid the obvious string -> number -> string conversion that would take care of non-significant zeros.

schwepmo • Jan 15 '20

Well actually, no. See my response :D

Idan Arye • Jan 15 '20

Still doesn't explain why "Trailing zeroes in a number without a decimal point" are not significant, considering how each such zero increases the number by an order of magnitude and how appending .0 suddenly makes these zeroes significant.

schwepmo • Jan 15 '20

Check the Wikipedia article it explains that aswell. Short answer is, that trailing zeros are somewhat ambiguous and that you have to specify it by some rule, which is done by the task description.

Ryan Burmeister-Morrison • Jan 16 '20

Nim submission.

import strutils

proc significantFigures(n: float): int = formatFloat(n)
  .replace(".")
  .strip(chars={'0'})
  .len()

when isMainModule:
  assert significantFigures(1) == 1
  assert significantFigures(003) == 1
  assert significantFigures(3000) == 1
  assert significantFigures(404) == 3
  assert significantFigures(050030210) == 7
  assert significantFigures(0.1) == 1
  assert significantFigures(0.0) == 0

lusen / they / them 🏳️‍🌈🥑 • Feb 1 '20

in python

note: added the 11 and 1.1 test cases because the subtracting-from-both-ends solution can have edge cases around there

def sigfig(num):
  # ignore 0's on the far left
  ldx = 0
  while num[ldx] == '0' and ldx < len(num):
    ldx += 1

  # ignore 0's on the far right if no decimal place
  rdx = len(num) - 1
  if num.find('.') > 0:
    return (rdx - ldx) # the +1 from below is cancelled out by needing to subtract the '.'
  else:
    while num[rdx] == '0' and rdx > -1:
      rdx -= 1
    return (rdx - ldx) + 1

print(sigfig('11'), 2)
print(sigfig('1.1'), 2)
print(sigfig('1'), 1)
print(sigfig('003'), 1)
print(sigfig('3000'), 1)
print(sigfig('404'), 3)
print(sigfig('050030210'), 7)
print(sigfig('0.1'), 1)
print(sigfig('0.0'), 1)

SavagePixie • Jan 15 '20 • Edited

I think this should do the trick, I'll test it later.

JavaScript

const significant = str => str.includes('.')
   ? str.replace(/^0+/, '').length - 1
   : str.replace(/^0+|0+$/g, '').length

Andrei McMillan • Jan 16 '20

js solution :)

function get_significant_num(str) {
  var total_significant_nums = 0;

  if (typeof str === 'string') {

    str = str.trim().replace(/^(0)+/, '');

    if (/\./.test(str)) {
      total_significant_nums = str.match(/\d+/g).join('').length;
    } else {
      str = str.replace(/0+$/, '');
      total_significant_nums = str.length;
    }

  }

  return total_significant_nums;
}


console.log(get_significant_num('20.0'));//3
console.log(get_significant_num('78.200'));//5
console.log(get_significant_num('0.6'));//1
console.log(get_significant_num('0215'));//3
console.log(get_significant_num('4.357'));//4
console.log(get_significant_num('345000'));//3
console.log(get_significant_num('050030210'));//7

Ryan Burmeister-Morrison • Jan 16 '20

Good point! I didn’t account for trailing 0’s. I’ll update it later.