Round number to next highest greatest place
1
Im trying to create a function that will return the next highest (for lack of a better term) "round" number, based on the greatest place digit (left most digit).
For example:
17 > 20
328 > 400
18564 > 20000
//Already round numbers will stay the same:
500 > 500
I know I can just do something like this:
int customRound(int i)
{
string s = i.ToString();
if (int.Parse(s.Substring(1)) > 0)
{
string greatestDigit = s.Substring(0, 1);
string digit = (int.Parse(greatestDigit) + 1).ToString();
return int.Parse(digit + string.Empty.PadRight(s.Length - 1, '0'));
}
return i;
}
But that just feels really hacky and Im sure theres a more elegant and mathematical way to do it.
c# math rounding
asked Nov 21 '18 at 17:16
Brian TackerBrian Tacker
46211431
add a comment |
1
Im trying to create a function that will return the next highest (for lack of a better term) "round" number, based on the greatest place digit (left most digit).
For example:
17 > 20
328 > 400
18564 > 20000
//Already round numbers will stay the same:
500 > 500
I know I can just do something like this:
int customRound(int i)
{
string s = i.ToString();
if (int.Parse(s.Substring(1)) > 0)
{
string greatestDigit = s.Substring(0, 1);
string digit = (int.Parse(greatestDigit) + 1).ToString();
return int.Parse(digit + string.Empty.PadRight(s.Length - 1, '0'));
}
return i;
}
But that just feels really hacky and Im sure theres a more elegant and mathematical way to do it.
c# math rounding
asked Nov 21 '18 at 17:16
Brian TackerBrian Tacker
46211431
Does 1 "round" to 10, 91 "round" to 100, etc
– PaulF
Nov 21 '18 at 17:27
add a comment |
1
1
1
1
Im trying to create a function that will return the next highest (for lack of a better term) "round" number, based on the greatest place digit (left most digit).
For example:
17 > 20
328 > 400
18564 > 20000
//Already round numbers will stay the same:
500 > 500
I know I can just do something like this:
int customRound(int i)
{
string s = i.ToString();
if (int.Parse(s.Substring(1)) > 0)
{
string greatestDigit = s.Substring(0, 1);
string digit = (int.Parse(greatestDigit) + 1).ToString();
return int.Parse(digit + string.Empty.PadRight(s.Length - 1, '0'));
}
return i;
}
But that just feels really hacky and Im sure theres a more elegant and mathematical way to do it.
c# math rounding
asked Nov 21 '18 at 17:16
Brian TackerBrian Tacker
46211431
Im trying to create a function that will return the next highest (for lack of a better term) "round" number, based on the greatest place digit (left most digit).
For example:
17 > 20
328 > 400
18564 > 20000
//Already round numbers will stay the same:
500 > 500
I know I can just do something like this:
int customRound(int i)
{
string s = i.ToString();
if (int.Parse(s.Substring(1)) > 0)
{
string greatestDigit = s.Substring(0, 1);
string digit = (int.Parse(greatestDigit) + 1).ToString();
return int.Parse(digit + string.Empty.PadRight(s.Length - 1, '0'));
}
return i;
}
But that just feels really hacky and Im sure theres a more elegant and mathematical way to do it.
c# math rounding
c# math rounding
asked Nov 21 '18 at 17:16
Brian TackerBrian Tacker
46211431
asked Nov 21 '18 at 17:16
Brian TackerBrian Tacker
46211431
asked Nov 21 '18 at 17:16
Brian TackerBrian Tacker
46211431
asked Nov 21 '18 at 17:16
Brian TackerBrian Tacker
46211431
asked Nov 21 '18 at 17:16
Brian TackerBrian Tacker
46211431
46211431
Does 1 "round" to 10, 91 "round" to 100, etc
– PaulF
Nov 21 '18 at 17:27
add a comment |
Does 1 "round" to 10, 91 "round" to 100, etc
– PaulF
Nov 21 '18 at 17:27
Does 1 "round" to 10, 91 "round" to 100, etc
– PaulF
Nov 21 '18 at 17:27
Does 1 "round" to 10, 91 "round" to 100, etc
– PaulF
Nov 21 '18 at 17:27
add a comment |
2 Answers
2
active
oldest
votes
6
You can use Math.Log10 to determine the order-of-magnitude of the number (previous number that is a power of 10), then round up to the next multiple of it:
int customRound(int i)
{
var digits = (int)Math.Floor(Math.Log10(i));
var unit = (int)Math.Pow(10, digits);
return (int)(Math.Ceiling((double)i / unit) * unit);
}
answered Nov 21 '18 at 17:27
Klaus GütterKlaus Gütter
2,45311321
Perfect solution, exactly what I was looking for, thank you.
– Brian Tacker
Nov 21 '18 at 17:32
This solution does leave values 1-9 unchanged - is that correct?
– PaulF
Nov 21 '18 at 17:40
@PaulF as it does 10, 20, ... That's how I understood the requirement.
– Klaus Gütter
Nov 21 '18 at 17:42
All other numbers round up to the next value ending with 0. If the requirement is to retain the same number of digits, then 91 should not round to 100.
– PaulF
Nov 21 '18 at 17:47
@PaulF but what else should 91 round to given the requirement "next highest"?
– Klaus Gütter
Nov 21 '18 at 17:49
|
show 1 more comment
1
int customRound(int i)
{
return (int)(Math.Ceiling(i / Math.Pow(10, i.ToString().Length-1)) * Math.Pow(10, i.ToString().Length - 1));
}
answered Nov 21 '18 at 17:32
Ehsan.SaradarEhsan.Saradar
45337
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
You can use Math.Log10 to determine the order-of-magnitude of the number (previous number that is a power of 10), then round up to the next multiple of it:
int customRound(int i)
{
var digits = (int)Math.Floor(Math.Log10(i));
var unit = (int)Math.Pow(10, digits);
return (int)(Math.Ceiling((double)i / unit) * unit);
}
answered Nov 21 '18 at 17:27
Klaus GütterKlaus Gütter
2,45311321
Perfect solution, exactly what I was looking for, thank you.
– Brian Tacker
Nov 21 '18 at 17:32
This solution does leave values 1-9 unchanged - is that correct?
– PaulF
Nov 21 '18 at 17:40
@PaulF as it does 10, 20, ... That's how I understood the requirement.
– Klaus Gütter
Nov 21 '18 at 17:42
All other numbers round up to the next value ending with 0. If the requirement is to retain the same number of digits, then 91 should not round to 100.
– PaulF
Nov 21 '18 at 17:47
@PaulF but what else should 91 round to given the requirement "next highest"?
– Klaus Gütter
Nov 21 '18 at 17:49
|
show 1 more comment
6
You can use Math.Log10 to determine the order-of-magnitude of the number (previous number that is a power of 10), then round up to the next multiple of it:
int customRound(int i)
{
var digits = (int)Math.Floor(Math.Log10(i));
var unit = (int)Math.Pow(10, digits);
return (int)(Math.Ceiling((double)i / unit) * unit);
}
answered Nov 21 '18 at 17:27
Klaus GütterKlaus Gütter
2,45311321
Perfect solution, exactly what I was looking for, thank you.
– Brian Tacker
Nov 21 '18 at 17:32
This solution does leave values 1-9 unchanged - is that correct?
– PaulF
Nov 21 '18 at 17:40
@PaulF as it does 10, 20, ... That's how I understood the requirement.
– Klaus Gütter
Nov 21 '18 at 17:42
All other numbers round up to the next value ending with 0. If the requirement is to retain the same number of digits, then 91 should not round to 100.
– PaulF
Nov 21 '18 at 17:47
@PaulF but what else should 91 round to given the requirement "next highest"?
– Klaus Gütter
Nov 21 '18 at 17:49
|
show 1 more comment
6
6
6
You can use Math.Log10 to determine the order-of-magnitude of the number (previous number that is a power of 10), then round up to the next multiple of it:
int customRound(int i)
{
var digits = (int)Math.Floor(Math.Log10(i));
var unit = (int)Math.Pow(10, digits);
return (int)(Math.Ceiling((double)i / unit) * unit);
}
answered Nov 21 '18 at 17:27
Klaus GütterKlaus Gütter
2,45311321
You can use Math.Log10 to determine the order-of-magnitude of the number (previous number that is a power of 10), then round up to the next multiple of it:
int customRound(int i)
{
var digits = (int)Math.Floor(Math.Log10(i));
var unit = (int)Math.Pow(10, digits);
return (int)(Math.Ceiling((double)i / unit) * unit);
}
answered Nov 21 '18 at 17:27
Klaus GütterKlaus Gütter
2,45311321
answered Nov 21 '18 at 17:27
Klaus GütterKlaus Gütter
2,45311321
answered Nov 21 '18 at 17:27
Klaus GütterKlaus Gütter
2,45311321
answered Nov 21 '18 at 17:27
Klaus GütterKlaus Gütter
2,45311321
2,45311321
Perfect solution, exactly what I was looking for, thank you.
– Brian Tacker
Nov 21 '18 at 17:32
This solution does leave values 1-9 unchanged - is that correct?
– PaulF
Nov 21 '18 at 17:40
@PaulF as it does 10, 20, ... That's how I understood the requirement.
– Klaus Gütter
Nov 21 '18 at 17:42
All other numbers round up to the next value ending with 0. If the requirement is to retain the same number of digits, then 91 should not round to 100.
– PaulF
Nov 21 '18 at 17:47
@PaulF but what else should 91 round to given the requirement "next highest"?
– Klaus Gütter
Nov 21 '18 at 17:49
|
show 1 more comment
Perfect solution, exactly what I was looking for, thank you.
– Brian Tacker
Nov 21 '18 at 17:32
This solution does leave values 1-9 unchanged - is that correct?
– PaulF
Nov 21 '18 at 17:40
@PaulF as it does 10, 20, ... That's how I understood the requirement.
– Klaus Gütter
Nov 21 '18 at 17:42
All other numbers round up to the next value ending with 0. If the requirement is to retain the same number of digits, then 91 should not round to 100.
– PaulF
Nov 21 '18 at 17:47
@PaulF but what else should 91 round to given the requirement "next highest"?
– Klaus Gütter
Nov 21 '18 at 17:49
Perfect solution, exactly what I was looking for, thank you.
– Brian Tacker
Nov 21 '18 at 17:32
Perfect solution, exactly what I was looking for, thank you.
– Brian Tacker
Nov 21 '18 at 17:32
This solution does leave values 1-9 unchanged - is that correct?
– PaulF
Nov 21 '18 at 17:40
This solution does leave values 1-9 unchanged - is that correct?
– PaulF
Nov 21 '18 at 17:40
@PaulF as it does 10, 20, ... That's how I understood the requirement.
– Klaus Gütter
Nov 21 '18 at 17:42
@PaulF as it does 10, 20, ... That's how I understood the requirement.
– Klaus Gütter
Nov 21 '18 at 17:42
All other numbers round up to the next value ending with 0. If the requirement is to retain the same number of digits, then 91 should not round to 100.
– PaulF
Nov 21 '18 at 17:47
All other numbers round up to the next value ending with 0. If the requirement is to retain the same number of digits, then 91 should not round to 100.
– PaulF
Nov 21 '18 at 17:47
@PaulF but what else should 91 round to given the requirement "next highest"?
– Klaus Gütter
Nov 21 '18 at 17:49
@PaulF but what else should 91 round to given the requirement "next highest"?
– Klaus Gütter
Nov 21 '18 at 17:49
|
show 1 more comment
1
int customRound(int i)
{
return (int)(Math.Ceiling(i / Math.Pow(10, i.ToString().Length-1)) * Math.Pow(10, i.ToString().Length - 1));
}
answered Nov 21 '18 at 17:32
Ehsan.SaradarEhsan.Saradar
45337
add a comment |
1
int customRound(int i)
{
return (int)(Math.Ceiling(i / Math.Pow(10, i.ToString().Length-1)) * Math.Pow(10, i.ToString().Length - 1));
}
answered Nov 21 '18 at 17:32
Ehsan.SaradarEhsan.Saradar
45337
add a comment |
1
1
1
int customRound(int i)
{
return (int)(Math.Ceiling(i / Math.Pow(10, i.ToString().Length-1)) * Math.Pow(10, i.ToString().Length - 1));
}
answered Nov 21 '18 at 17:32
Ehsan.SaradarEhsan.Saradar
45337
int customRound(int i)
{
return (int)(Math.Ceiling(i / Math.Pow(10, i.ToString().Length-1)) * Math.Pow(10, i.ToString().Length - 1));
}
answered Nov 21 '18 at 17:32
Ehsan.SaradarEhsan.Saradar
45337
answered Nov 21 '18 at 17:32
Ehsan.SaradarEhsan.Saradar
45337
answered Nov 21 '18 at 17:32
Ehsan.SaradarEhsan.Saradar
45337
answered Nov 21 '18 at 17:32
Ehsan.SaradarEhsan.Saradar
45337
45337
add a comment |
add a comment |
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Does 1 "round" to 10, 91 "round" to 100, etc
– PaulF
Nov 21 '18 at 17:27