How do I round to the nearest hundred?

lucmove · 05-25-2020, 12:29 AM

I have this code:

Code:

someFunction()   {
	if	(someVar > 2000)	{return 2;}
	if	(someVar > 1900)	{return 1.9;}
	if	(someVar > 1800)	{return 1.8;}
	if	(someVar > 1700)	{return 1.7;}
	if	(someVar > 1600)	{return 1.6;}
	if	(someVar > 1500)	{return 1.5;}
	if	(someVar > 1400)	{return 1.4;}
	if	(someVar > 1300)	{return 1.3;}
	if	(someVar > 1200)	{return 1.2;}
	if	(someVar > 1100)	{return 1.1;}
	if	(someVar > 1000)	{return 1;}
	if	(someVar > 900)		{return 0.9;}
	if	(someVar > 800)		{return 0.8;}
	if	(someVar > 700)		{return 0.7;}
	if	(someVar > 600)		{return 0.6;}
	if	(someVar > 500)		{return 0.5;}
	if	(someVar > 400)		{return 0.4;}
	if	(someVar > 300)		{return 0.3;}
	if	(someVar > 200)		{return 0.2;}
	if	(someVar > 100)		{return 0.1;}
}

Of course it's dumb.

It goes up to 2,000 but "someVar" could be a lot larger than that, maybe millions or billions. What is the smart way to achieve what I want?

Please assume some kind of shell scripting or pseudo code.

TIA

scasey · 05-25-2020, 12:34 AM

Divide SomeVar by 1000.

Code:

someVar/1000

I believe most languages can do that...which would you like to use?

lucmove · 05-25-2020, 12:43 AM

That doesn't work.

900/1000 = 0.9. OK
956/1000 = 0.956. Wrong. I need 0.9.
9560/1000 = 9.56. Wrong. I need 9.5.

Any shell language or pseudo code will do.

MadeInGermany · 05-25-2020, 12:46 AM

But in post #1 you want
901 => 0.9
900 => 0.8
?

lucmove · 05-25-2020, 12:55 AM

Quote:

Originally Posted by MadeInGermany

But in post #1 you want
901 => 0.9
900 => 0.8
?

Good point. What I really want is not > (greater than) but rather >= (equal to or greater than).

Sorry.

syg00 · 05-25-2020, 12:56 AM

The awk manual has some interesting info on rounding modes - and "interesting" results that are not always classified as errors. Well worth you spend the time reading it. Might even be some code in there methinks.

Turbocapitalist · 05-25-2020, 12:57 AM

In pseudo code, generic rounding is often done like this:

Code:

result = int(number * 1000 + 0.5) / 1000

If you do not need the multiplication, leave it out. Some shells deal only with integers, however, so you might need some other language instead.

If it is a question of adding trailing zeros, then use printf()

lucmove · 05-25-2020, 01:07 AM

Quote:

Originally Posted by Turbocapitalist

In pseudo code, generic rounding is often done like this:

Code:

result = int(number * 1000 + 0.5) / 1000

If you do not need the multiplication, leave it out. Some shells deal only with integers, however, so you might need some other language instead.

If it is a question of adding trailing zeros, then use printf()

That doesn't work either.

Testing 900 (in Tcl):

Code:

[900 * 1000 + 0.5] / 1000
900.0005
[[900 * 1000] + 0.5] / 1000
900.0005
[900 * [1000 + 0.5]] / 1000
900.45

The right result should be 0.9.

Note: please no printf(), just math.

pan64 · 05-25-2020, 01:19 AM

probably:

Code:

result = int(number / 100 ) / 10

lucmove · 05-25-2020, 01:30 AM

Quote:

Originally Posted by pan64

probably:

Code:

result = int(number / 100 ) / 10

That is the same as dividing by 1,000. It doesn't work.

900 / 100 / 10
0.9
OK.

945 / 100 / 10
0.945
Wrong. Should be 0.9.

1945 / 100 / 10
1.9449999999999998

Wrong. Should be 1.9. I could do with 1.49, but 1.9 would be better.

7228 / 100 / 10
7.228
Wrong. Should be 7.2. I could do with 7.22, but 7.2 would be better.

Also, note that many languages have a rounding function, but:
1.9449999999999998 = 1.94. OK.
1.9459999999999998 = 1.95. Not what I want.

pan64 · 05-25-2020, 01:41 AM

you do not understand the replies. int ( something ) is a function and has a special meaning. This function cuts all the fraction part, so the return value is always an integer, especially the one which is not greater than the input, but not less than the input - 1 . The other name of this function is truncate.
This function is implemented in almost all languages. [almost] all the rounding related functions are based on this one.
Rounding is something like: int ( something + 0.5 )

astrogeek · 05-25-2020, 02:04 AM

According to your expected results in the original post, I suspect this is the math you are looking for:

Code:

num = (x-(x % 100))/1000

Given x = 945
num = .9

I think the question is confused by presenting it as a rounding problem, it is not what many would think of as rounding.

lucmove · 05-25-2020, 02:19 AM

Quote:

Originally Posted by astrogeek

Code:

num = (x-(x % 100))/1000

Given x = 945
num = .9

What symbol is that, x divided by 100, or x remainder 100?
Neither seems to work for me. How did you test it?

astrogeek · 05-25-2020, 02:24 AM

Quote:

Originally Posted by lucmove

What symbol is that, x divided by 100, or x remainder 100?
Neither seems to work for me. How did you test it?

It is x modulo 100 (probably what you mean by remainder), represented by % operator in most programming languages.

I didn't test it, I just wrote the math expression. You should be able to test it in any programming language which handles floating point numbers.

How did you test it?

MadeInGermany · 05-25-2020, 03:30 AM

Quote:

Originally Posted by pan64

probably:

Code:

result = int(number / 100 ) / 10

That's the solution. The int function rounds down to the next integer.

int(945 / 100) / 10
int(9.45) == 9
=> 0.9

int(1945 / 100) / 10
int(19.45) == 19
=> 1.9