Char to numeric with decimal points

[David_Billa]

I've a character variable called calc_rslt and it has values like  -259328.943733,-35860.382829,0. Now I want to convert it to numeric without rounding off the values.

When I tried with the Format 22.6 in the Input function, I could see it is rounding off the value. e.g. -259328.943733 is rounded to -259328.9437.

calc_rslt_num = input(calc_rslt,22.6); 

Also if it is 0 (character) in the Input value then I want to convert it to 0.000000 (numeric). It is also not working when I tried with if then clause as shown below.

If value=0 then value=0.000000

Desired results for the provided values should be -259328.943733,-35860.382829,0.000000

Any help?

[PaigeMiller]

When you do this creation of a new numeric variable, it has a default format of best12. However, the full decimal places are there, which can be seen by changing the format.

data a;
    calc_rslt='-259328.943733';
    calc_rslt_num = input(calc_rslt,22.6); 
    format calc_rslt_num best24.;
run;

Of course, this begs the question about why you have non-integer numbers in a character variable, which doesn't seem like a good thing to do.

[David_Billa]

@PaigeMiller how about my other question?

Also if it is 0 after converting from character to numeric then I want to convert it to 0.000000 (numeric). It is also not working when I tried with if then clause as shown in the Initial post

[PaigeMiller]

Internally, there is no difference between 0 and 0.000000, these are the same number. The only difference is the appearance, so you need to find a format (which has the effect of changing the appearance) that works on both -259328.943733 and 0.

[Tom · Accepted solution]

There is a limit to the number of decimal digits that can be exactly represented (without gaps) using the binary floating point representation SAS (and most software) use to store numbers.  But your example strings are no where near that limit.

 -259328.943733
-35860.382829
0 

So to convert those strings into numbers use the normal numeric informat. 

calc_rslt_num = input(calc_rslt,32.); 

Note do NOT include a number of decimal places in the informat, unless you know that the periods have been removed to save space. Otherwise if you use 22.6 informat and the string does not contain a period the number is divided by 10**6 to place the implied decimal point.

If you want the value to display with 6 digits to the right of the decimal place the use a format when displaying the value.

format calc_rslt_num 22.6 ;