At some point I have a numeric(28,10) and I cast it in money (I know its bad but for legacy reason I have to return money) in the same time I also have to set the sign (multiplying by +1/-1).
In a first attempt I had cast the +/-1 to match the numeric type.
For the value 133.3481497944 we encounter a strange behavior (I have simplified the actual code in order to keep only the elements needed to demonstrate the problem):
SELECT CAST(CAST(133.3481497944 AS numeric(28,10))*cast(1 AS numeric(28,10)) AS money)
133.3482
which is not correctly rounded...
Removing the cast solve the problem
SELECT CAST(CAST(133.3481497944 AS numeric(28,10)) * 1 AS money)
133.3481
Did someone know what is happening in SQL? How can a multiplication by 1 and cast(1 AS numeric(28,10)) affect the result of the rounding?
When multiplying numerics, SQL uses the following rules to determine the precision and scale of the output:
p = p1 + p2 + 1
s = s1 + s2
which makes sense - you wouldn't want 1.5 * 2.5 to be truncated to one digit past the decimal. Nor would you want 101 * 201 to be limited to 3 digits of precision, giving you 20300 instead of 20301.
In your case that would result in a precision of 57 and a scale of 20, which isn't possible - the maximum precision and scale is 38.
If the resulting type is too big, decimal digits are sacrificed in order to preserve the integral (most significant) part of the result.
From the SQL Programmability & API Development Team Blog:
In SQL Server 2005 RTM (and previous versions), we decided preserve a minimum scale of 6 in both multiplication and division.
So your answer depands on how big and precise you need the multiplier to be. In order to preserve 10 digits of decimal precision. If the multiplier needs a scale bigger than 9, then decimal digits may be truncated. If you use a smaller precision and scale, you should be fine:
SELECT CAST(CAST(133.3481497944 AS numeric(28,10))*cast(1 AS numeric(9,7)) AS money)
yields 133.3481.
I don't see any ROUNDing here. I only see casting. Don't assume that it will round, when you CAST. Historically, when we cast the environment truncates (SQL server or not) or behaves not as we expect - especially when we're talking about FLOATs.
SELECT
CAST(CAST(133.3481497944 AS numeric(28,10))*cast(1 AS numeric(28,10)) AS money) --Your original,
CAST(1 AS numeric(28,10)) --Just the 1 casted,
CAST(133.3481497944 AS numeric(28,10)) --Your expected calculation,
CAST(133.3481497944 AS numeric(28,10))*cast(1 AS numeric(28,10)) -- The actual calculation
SELECT
CAST(133.3481497944 AS numeric(28,10))*cast(1.5 AS numeric(28,10)),
CAST(133.3481497944 AS numeric(28,10))*1.5,
CAST((133.3481497944*1) AS money),
133.3481497944*1
Returns
133.3482
1.0000000000
133.3481497944
133.348150
200.022225
200.02222469160
133.3481
133.3481497944
So as mentioned above, there really isn't any true rounding, but a loss of precision during the cast. As to exactly why, I don't know. Most likely during the calculation(multiplication) while using the Numeric(28,10) it cuts off some precision.
I added the second lines to show that really you may not need your numeric casting.
Related
SQL Server decimal function not working as intended.
To test with sample data, I created a table and inserted values to it.
Then, I tried to run decimal function on these values.
CREATE TABLE TEST_VAL
(
VAL float
)
SELECT * FROM TEST_VAL
Output:
VAL
----------
16704.405
20382.135
2683.135
SELECT CAST(VAL AS DECIMAL(15, 2)) AS NEWVAL
FROM TEST_VAL;
Output:
NEWVAL
-------------
16704.40
20382.13
2683.14
I expected same formatting for all 3 values. But, for third value it returns ceiling round off value.
This is due to the nature of floating point numbers being inexact and being in binary. But I want to demonstrate how this is working.
The issue is that a decimal such as 0.135 cannot be represented exactly. As the floating point representation, it would typically be something like:
0.134999999234243423
(Note that these numbers as with all representations of values in this answer are made up. They are intended to be representative to make the point.)
The number of 9s is actually larger. And the subsequent digits are just representative. In this representation, we wouldn't see a problem with truncating the value. After all 0.1349999 should round to the same value as 0.13499.
In binary, this looks different:
0.11101000010101 10011 10011 10011 10011 . . .
---------------- --------------
~0.135 "arbitrary" repeating pattern
(Note: The values are made up!)
That is, the "infinite" portion of binary representation is not a bunch of repeating 1s or repeating 0s; it has a pattern. This is analogous the inverse of most numbers in base 10 For instance, 1/7 has a repeating component of six digits, 142857. We tend to forget this, because common inverses are either exact (1/5 = 0.2) or have a single repeating digit (1/6 = 0.166666...). 1/7 is the first case that is not so simple -- and almost all decimals are like this. For rational numbers, there is always a repeating sequence regardless of base and it is never longer than dividend (number at the bottom) minus 1).
We can think of this as all decimal representations (regardless of base) always have some number of digits that are repeating. For an exact representation, the repeating portion is 0. For others it is rarely one digit. Usually, it is multiple digits. And it is a fun exercise in mathematics to characterize this. But all that is important is that the repeating portion has 1s and 0s.
Now, what is happening. A floating point number has three parts:
a magnitude. This is a number of bits that represent the exponent.
an integer portion, which is the number before the decimal point.
an integer portion, which is the number after the decimal point.
(Actually, the last two are really one integer, but I find it much easier to explain this by splitting them into two components.)
Only a fixed number of bits are available for the two integer portions. What does this look like? Once again the representative patterns are something like this:
0.135 0 11101000010101100111001110
1.135 1 11101000010101100111001110
2.135 10 1110100001010110011100111
4.135 100 111010000101011001110011
8.135 1000 11101000010101100111001
16.136 10000 1110100001010110011100
-----------^ part before the decimal
------------------^ part after the decimal
Note: This is leaving off the magnitude portion of the decimal representation.
As you can see, digits get chopped off from the end. But sometimes it is 0 that gets chopped off -- so there is no change in the value being represented. And sometimes it is a 1. And there is a change.
With this, you might be able to see how the values essentially fluctuate, say:
0.135 --> 0.135000000004
1.135 --> 0.135000000004
2.135 --> 0.135000000004
4.135 --> 0.135000000001
8.135 --> 0.135999999997
16.135 --> 0.135999999994
These are then rounded differently, which is what you are seeing.
I put together this little db<>fiddle, so you can see how the rounding changes around powers of two.
Perhaps this could be explained if we extend the precision of the three numbers in the first query:
16704.4050
20382.1349
2683.1351
Rounding each of the above to only two decimal places, which is what a cast to DECIMAL(10,2) would do, would yield:
16704.40
20382.13
2683.14
Would this be of use:
select CONVERT(DECIMAL(15,2), ROUND(VAL, 2, 1)) AS NEWVAL
from TEST_VAL;
Here is the DEMO for SQLServer 2012 : DEMO
first question : why they are not same value?
because their type is different , CAST(VAL as decimal(4,2)) will format like ##.## not ##.### so in your case it get ceiling round value.
Why not use the same type ?
CREATE TABLE T
(
[VAL] DECIMAL(8,3)
);
INSERT INTO T ([VAL])
VALUES (16704.405), (20382.135), (2683.135);
SELECT * FROM T
Output:
VAL
-----------
16704.405
20382.135
2683.135
db<>fiddle here
or you can cast AS DECIMAL(8, 3)
SELECT CAST(VAL AS DECIMAL(8,3)) AS NEWVAL
FROM T;
We don't have decimal data type in BigQuery now. So I have to use float
But
In Bigquery float division
0.029*50/100=0.014500000000000002
Although
0.021*50/100=0.0105
To round the value up
I have to use round(floatvalue*10000)/10000.
Is this the right way to deal with decimal data type now in BigQuery?
Depends on your coding preferences - for example you can just use simple ROUND(floatvalue, 4)
Depends on how exactly you need to round - up or down - you can respectively adjust expression
For example ROUND(floatvalue + 0.00005, 4)
See all rounding functions for BigQuery Standard SQL at below link
https://cloud.google.com/bigquery/docs/reference/standard-sql/functions-and-operators#rounding-functions
Note that this question deserves a different answer now.
The premise of the question is "We don't have decimal data type in BigQuery now."
But now we do: You can use NUMERIC:
SELECT CAST('0.029' AS NUMERIC)*50/100
# 0.0145
Just make your column is NUMERIC instead of FLOAT64, and you'll get the desired results.
Rounding up in most SQL dialects is not a built-in function unless you're fortunate enough to be rounding up to an integer. In this case, CEIL is a quick and reliable solution.
In the case of rounding decimals up, we can also leverage CEIL, albeit with a couple of additional steps.
The procedure:
Multiply your value to move the last decimal to the tenths position. Ex. 18.234 becomes 1823.4 by multiplying by 100. (n * 100)
Use CEIL() to round up to the nearest integer. In our example, CEIL(n) = 1824.
Divide this result by the same figure used in step 1. In our example, n / 100 = 18.24.
Simplifying these steps leaves us with the below logic:
SELECT CEIL(value * 100) / 100 as rounded_up;
The same logic can be used to round down using the FLOOR function as such:
FLOOR(value * 100) / 100 AS rounded_down;
Thanks to #Mureinik for this answer.
We are doing some validation of data which has been migrated from one SQL Server to another SQL Server. One of the things that we are validating is that some numeric data has been transferred properly. The numeric data is stored as a float datatype in the new system.
We are aware that there are a number of issues with float datatypes, that exact numeric accuracy is not guaranteed, and that one cannot use exact equality comparisons with float data. We don't have control over the database schemas nor data typing and those are separate issues.
What we are trying to do in this specific case is verify that some ratio values were transferred properly. One of the specific data validation rules is that all ratios should be transferred with no more than 4 digits to the right of the decimal point.
So, for example, valid ratios would look like:
.7542
1.5423
Invalid ratios would be:
.12399794301
12.1209377
What we would like to do is count the number of digits to the right of the decimal point and find all cases where the float values have more than four digits to the right of it. We've been using the SUBSTRING, LEN, STR, and a couple of other functions to achieve this, and I am sure it would work if we had numeric fields typed as decimal which we were casting to char.
However, what we have found when attempting to convert a float to a char value is that SQL Server seems to always convert to decimal in between. For example, the field in question shows this value when queried in SQL Server Enterprise Manager:
1.4667
Attempting to convert to a string using the recommended function for SQL Server:
LTRIM(RTRIM(STR(field_name, 22, 17)))
Returns this value:
1.4666999999999999
The value which I would expect if SQL Server were directly converting from float to char (which we could then trim trailing zeroes from):
1.4667000000000000
Is there any way in SQL Server to convert directly from a float to a char without going through what appears to be an intermediate conversion to decimal along the way? We also tried the CAST and CONVERT functions and received similar results to the STR function.
SQL Server Version involved: SQL Server 2012 SP2
Thank you.
Your validation rule seems to be misguided.
An SQL Server FLOAT, or FLOAT(53), is stored internally as a 64-bit floating-point number according to the IEEE 754 standard, with 53 bits of mantissa ("value") plus an exponent. Those 53 binary digits correspond to approximately 15 decimal digits.
Floating-point numbers have limited precision, which does not mean that they are "fuzzy" or inexact in themselves, but that not all numbers can be exactly represented, and instead have to be represented using another number.
For example, there is no exact representation for your 1.4667, and it will instead be stored as a binary floating-point number that (exactly) corresponds to the decimal number 1.466699999999999892708046900224871933460235595703125. Correctly rounded to 16 decimal places, that is 1.4666999999999999, which is precisely what you got.
Since the "exact character representation of the float value that is in SQL Server" is 1.466699999999999892708046900224871933460235595703125, the validation rule of "no more than 4 digits to the right of the decimal point" is clearly flawed, at least if you apply it to the "exact character representation".
What you might be able to do, however, is to round the stored number to fewer decimal places, so that the small error at the end of the decimals is hidden. Converting to a character representation rounded to 15 instead of 16 places (remember those "15 decimal digits" mentioned at the beginning?) will give you 1.466700000000000, and then you can check that all decimals after the first four are zeroes.
You can try using cast to varchar.
select case when
len(
substring(cast(col as varchar(100))
,charindex('.',cast(col as varchar(100)))+1
,len(cast(col as varchar(100)))
)
) = 4
then 'true' else 'false' end
from tablename
where charindex('.',cast(col as varchar(100))) > 0
For this particular number, don't use STR(), and use a convert or cast to varchar. But, in general, you will always have precision issues when storing in float... it's the nature of the storage of that datatype. The best you can do is normalize to a NUMERIC type and compare with threshold ranges (+/- .0001, for example). See the following for a breakdown of how the different conversions work:
declare #float float = 1.4667
select #float,
convert(numeric(18,4), #float),
convert(nvarchar(20), #float),
convert(nvarchar(20), convert(numeric(18,4), #float)),
str(#float, 22, 17),
str(convert(numeric(18,4), #float)),
convert(nvarchar(20), convert(numeric(18,4), #float))
Instead of casting to a VarChar you might try this: cast to a decimal with 4 fractional digits and check if it's the same value as before.
case when field_name <> convert(numeric(38,4), field_name)
then 1
else 0
end
The issue you have here is that float is an approximate number data type with an accuracy of about seven digits. That means it approaches the value while using less storage than a decimal / numeric. That's why you don't use float for values that require exact precision.
Check this example:
DECLARE #t TABLE (
col FLOAT
)
INSERT into #t (col)
VALUES (1.4666999999999999)
,(1.4667)
,(1.12399794301)
,(12.1209377);
SELECT col
, CONVERT(NVARCHAR(MAX),col) AS chr
, CAST(col as VARBINARY) AS bin
, LTRIM(RTRIM(STR(col, 22, 17))) AS rec
FROM #t
As you see the float 1.4666999999999999 binary equals 1.4667. For your stated needs I think this query would fit:
SELECT col
, RIGHT(CONVERT(NVARCHAR(MAX),col), LEN(CONVERT(NVARCHAR(MAX),col)) - CHARINDEX('.',CONVERT(NVARCHAR(MAX),col))) AS prec
from #t
I'm trying the following using SQL Server 2008 R2 and SQL Server 2012 and I get the same results in both.
If I write the following statement:
select round(4.005, 2)
I get the expected result: 4.01
However if I write the following statements:
declare #result float
select #result = 4.005
select round(#result, 2)
I get an unexpected result: 4
But if I replace float with real in the previous statements:
declare #result real
select #result = 4.005
select round(#result, 2)
I get the expected result.
Can anyone tell me why is this happening?
because you haven't specify the number of bits that are used to store the mantissa of the float number in scientific notation , try this,
declare #result float(5)
select #result = 4.005
select round(#result, 2)
SQLFiddle Demo
float and real (Transact-SQL)
This has to do with float being an approximate data type.
see: http://technet.microsoft.com/en-us/library/ms187912(v=sql.105).aspx
The float and real data types are known as approximate data types. The behavior of float and real follows the IEEE 754 specification on approximate numeric data types.
Approximate numeric data types do not store the exact values specified for many numbers; they store an extremely close approximation of the value. For many applications, the tiny difference between the specified value and the stored approximation is not noticeable. At times, though, the difference becomes noticeable. Because of the approximate nature of the float and real data types, do not use these data types when exact numeric behavior is required, such as in financial applications, in operations involving rounding, or in equality checks. Instead, use the integer, decimal, money, or smallmoney data types.
Avoid using float or real columns in WHERE clause search conditions, especially the = and <> operators. It is best to limit float and real columns to > or < comparisons.
The IEEE 754 specification provides four rounding modes: round to nearest, round up, round down, and round to zero. Microsoft SQL Server uses round up. All are accurate to the guaranteed precision but can result in slightly different floating-point values. Because the binary representation of a floating-point number may use one of many legal rounding schemes, it is impossible to reliably quantify a floating-point value.
yes I had the same issue. I tested out some solutions...
SELECT round(cast(3.175 as float), 2) as roundingBAD,
CAST(round(TRY_CONVERT(DECIMAL(28,2), cast(3.175 as float)), 2) as
nvarchar(max)) as roundconvertBAD, cast(CAST(round(CAST(3.175 as
DECIMAL(18,10)), 4) as decimal(18,2)) as nvarchar(max)) as
roundconvert3, cast(FORMAT(round(CAST(3.175 as DECIMAL(18,10)), 2),
'0.######') as nvarchar(max)) as roundconvert4, cast(CAST(CAST(3.175
as DECIMAL(18,10)) as decimal(18,2)) as nvarchar(max)) as
roundconvert5P, cast(CAST(CAST(3.175 as DECIMAL(18,10)) as
decimal(18,1)) as nvarchar(max)) as roundconvert1DP
output: 3.17 3.17 3.18 3.18 3.18 3.2
I went with the last two options in my production Microsoft SQL Server code for rounding floats correctly.
I have a column X which is full of floats with decimals places ranging from 0 (no decimals) to 6 (maximum). I can count on the fact that there are no floats with greater than 6 decimal places. Given that, how do I make a new column such that it tells me how many digits come after the decimal?
I have seen some threads suggesting that I use CAST to convert the float to a string, then parse the string to count the length of the string that comes after the decimal. Is this the best way to go?
You can use something like this:
declare #v sql_variant
set #v=0.1242311
select SQL_VARIANT_PROPERTY(#v, 'Scale') as Scale
This will return 7.
I tried to make the above query work with a float column but couldn't get it working as expected. It only works with a sql_variant column as you can see here: http://sqlfiddle.com/#!6/5c62c/2
So, I proceeded to find another way and building upon this answer, I got this:
SELECT value,
LEN(
CAST(
CAST(
REVERSE(
CONVERT(VARCHAR(50), value, 128)
) AS float
) AS bigint
)
) as Decimals
FROM Numbers
Here's a SQL Fiddle to test this out: http://sqlfiddle.com/#!6/23d4f/29
To account for that little quirk, here's a modified version that will handle the case when the float value has no decimal part:
SELECT value,
Decimals = CASE Charindex('.', value)
WHEN 0 THEN 0
ELSE
Len (
Cast(
Cast(
Reverse(CONVERT(VARCHAR(50), value, 128)) AS FLOAT
) AS BIGINT
)
)
END
FROM numbers
Here's the accompanying SQL Fiddle: http://sqlfiddle.com/#!6/10d54/11
This thread is also using CAST, but I found the answer interesting:
http://www.sqlservercentral.com/Forums/Topic314390-8-1.aspx
DECLARE #Places INT
SELECT TOP 1000000 #Places = FLOOR(LOG10(REVERSE(ABS(SomeNumber)+1)))+1
FROM dbo.BigTest
and in ORACLE:
SELECT FLOOR(LOG(10,REVERSE(CAST(ABS(.56544)+1 as varchar(50))))) + 1 from DUAL
A float is just representing a real number. There is no meaning to the number of decimal places of a real number. In particular the real number 3 can have six decimal places, 3.000000, it's just that all the decimal places are zero.
You may have a display conversion which is not showing the right most zero values in the decimal.
Note also that the reason there is a maximum of 6 decimal places is that the seventh is imprecise, so the display conversion will not commit to a seventh decimal place value.
Also note that floats are stored in binary, and they actually have binary places to the right of a binary point. The decimal display is an approximation of the binary rational in the float storage which is in turn an approximation of a real number.
So the point is, there really is no sense of how many decimal places a float value has. If you do the conversion to a string (say using the CAST) you could count the decimal places. That really would be the best approach for what you are trying to do.
I answered this before, but I can tell from the comments that it's a little unclear. Over time I found a better way to express this.
Consider pi as
(a) 3.141592653590
This shows pi as 11 decimal places. However this was rounded to 12 decimal places, as pi, to 14 digits is
(b) 3.1415926535897932
A computer or database stores values in binary. For a single precision float, pi would be stored as
(c) 3.141592739105224609375
This is actually rounded up to the closest value that a single precision can store, just as we rounded in (a). The next lowest number a single precision can store is
(d) 3.141592502593994140625
So, when you are trying to count the number of decimal places, you are trying to find how many decimal places, after which all remaining decimals would be zero. However, since the number may need to be rounded to store it, it does not represent the correct value.
Numbers also introduce rounding error as mathematical operations are done, including converting from decimal to binary when inputting the number, and converting from binary to decimal when displaying the value.
You cannot reliably find the number of decimal places a number in a database has, because it is approximated to round it to store in a limited amount of storage. The difference between the real value, or even the exact binary value in the database will be rounded to represent it in decimal. There could always be more decimal digits which are missing from rounding, so you don't know when the zeros would have no more non-zero digits following it.
Solution for Oracle but you got the idea. trunc() removes decimal part in Oracle.
select *
from your_table
where (your_field*1000000 - trunc(your_field*1000000)) <> 0;
The idea of the query: Will there be any decimals left after you multiply by 1 000 000.
Another way I found is
SELECT 1.110000 , LEN(PARSENAME(Cast(1.110000 as float),1)) AS Count_AFTER_DECIMAL
I've noticed that Kshitij Manvelikar's answer has a bug. If there are no decimal places, instead of returning 0, it returns the total number of characters in the number.
So improving upon it:
Case When (SomeNumber = Cast(SomeNumber As Integer)) Then 0 Else LEN(PARSENAME(Cast(SomeNumber as float),1)) End
Here's another Oracle example. As I always warn non-Oracle users before they start screaming at me and downvoting etc... the SUBSTRING and INSTRING are ANSI SQL standard functions and can be used in any SQL. The Dual table can be replaced with any other table or created. Here's the link to SQL SERVER blog whre i copied dual table code from: http://blog.sqlauthority.com/2010/07/20/sql-server-select-from-dual-dual-equivalent/
CREATE TABLE DUAL
(
DUMMY VARCHAR(1)
)
GO
INSERT INTO DUAL (DUMMY)
VALUES ('X')
GO
The length after dot or decimal place is returned by this query.
The str can be converted to_number(str) if required. You can also get the length of the string before dot-decimal place - change code to LENGTH(SUBSTR(str, 1, dot_pos))-1 and remove +1 in INSTR part:
SELECT str, LENGTH(SUBSTR(str, dot_pos)) str_length_after_dot FROM
(
SELECT '000.000789' as str
, INSTR('000.000789', '.')+1 dot_pos
FROM dual
)
/
SQL>
STR STR_LENGTH_AFTER_DOT
----------------------------------
000.000789 6
You already have answers and examples about casting etc...
This question asks of regular SQL, but I needed a solution for SQLite. SQLite has neither a log10 function, nor a reverse string function builtin, so most of the answers here don't work. My solution is similar to Art's answer, and as a matter of fact, similar to what phan describes in the question body. It works by converting the floating point value (in SQLite, a "REAL" value) to text, and then counting the caracters after a decimal point.
For a column named "Column" from a table named "Table", the following query will produce a the count of each row's decimal places:
select
length(
substr(
cast(Column as text),
instr(cast(Column as text), '.')+1
)
) as "Column-precision" from "Table";
The code will cast the column as text, then get the index of a period (.) in the text, and fetch the substring from that point on to the end of the text. Then, it calculates the length of the result.
Remember to limit 100 if you don't want it to run for the entire table!
It's not a perfect solution; for example, it considers "10.0" as having 1 decimal place, even if it's only a 0. However, this is actually what I needed, so it wasn't a concern to me.
Hopefully this is useful to someone :)
Probably doesn't work well for floats, but I used this approach as a quick and dirty way to find number of significant decimal places in a decimal type in SQL Server. Last parameter of round function if not 0 indicates to truncate rather than round.
CASE
WHEN col = round(col, 1, 1) THEN 1
WHEN col = round(col, 2, 1) THEN 2
WHEN col = round(col, 3, 1) THEN 3
...
ELSE null END