I'm a rookie SQL Programmer, but have searched here and many other SQL Forums and can't figure out why my simple division script is still ignoring the decimals. I've CAST EVERYTHING as Decimal, but still don't get any in the output . .
(CAST((ABS(CAST(CAST(SUM(h4qqnb) AS DECIMAL(12,4)) - CAST(SUM(h4hdnb) AS DECIMAL(12,4))AS DECIMAL(12,4)))/CAST(SUM(h4hdnb) AS DECIMAL(12,4))) AS DECIMAL(10,2))*100)||'%' AS Count_Variance_Rate,
What am I missing?
thanks!
That's a seriously ugly expression, with far too much CASTing, but essentially what you're doing (just looking at the outermost CAST statement) is saying
CAST(someNumber as DECIMAL(10,2)
which is going to give you a number with two decimal places of precision. You're then multiplying that by 100, which is going to give you an integer.
If you're trying to get a percentage value formatted to two decimal places, you can do it like this, assuming that h4qqnb and h4hdnb are decimal fields to begin with:
concat(cast(cast(abs(sum(h4qqnb) - sum(h4hdnb)) / sum(h4hdnb) as decimal(10, 4)) * 100 as decimal(10, 2)), '%') as Count_Variance_Rate2
Working example at http://sqlfiddle.com/#!2/279752/5/0
Related
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 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.
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.
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
I have the following piece of code that always results in 0.00. I know for sure that the weight_score is an actual number. Can someone help me format it so that it results in a decimal number. Using SQL server 2008 and this code is in a view. Thanks a lot!
CAST(SUM(weight_score * (45/100)) as decimal(10,2)) As avg_score
45/100 is implicitly treated as an INT divided by another INT, which results in an INT that gets rounded to 0. Add a decimal to each to work around that.
CAST(SUM(weight_score * (45.0/100.0)) as decimal(10,2)) As avg_score
45/100 is treated as INT.So change it to 45.0/100.0.
select CAST(SUM(weight_score * (45.0/100.0)) as decimal(10,2)) As avg_score
Even if you write the following code you will get 0
select (45/100)
If you want to avoid the error, you can use the NULLIF function.