I have a time variable that is expressed as a character in SAS. Example: 0:04 0:12 0:01 0:11 etc. I would like to convert it to a numeric variable 0.04 0.12 0.01 etc.
Using this code:
data work.set2; set work.set;
TIME2 = input(TIME, best4.);
;
run;
creates a new column with nothing but missing values. Can you advice on what to improve in my code?
SAS stores dates and times as numbers, time is the number of seconds. I think converting it to a SAS time is your best option. And there is a significant difference between 0.1 and 10 seconds because one is 6 seconds and one is 10 seconds. For example if you had 0.1 and 0.2 and took the difference that's 0.1 -> is that now a 10 or 6 second difference. You really need to think this through on how you want to interpret it and using your approach will be problematic.
The difference in times will not be reflected correctly.
Also, is 0:04 4 seconds or 4 minutes. The standard connotation would be 4 minutes, which is 240 seconds.
Here's how you can convert it:
data have;
x = '0:04';output;
x = '0:12';output;
x = '0:11'; output;
x = '1:00'; output;
x = '4:25'; output;
run;
data want;
set have;
sas_time = input(x, time.);
sas_time2 = sas_time;
format sas_time2 time4.;
/*if it's seconds*/
seconds = input(scan(x, 1, ':'), 8.)*60 + input(scan(x, 2, ':'), 8.);
run;
proc print data=want;run;
If your times are of type string:
WITHDOTS=translate(TIME2,'.',':');
Source:
https://communities.sas.com/t5/Base-SAS-Programming/Find-And-Replace-within-a-string/td-p/45104
Related
I have been asked to write some code in SAS that rounds a number up but only if the digit in the thousandth place is greater than one. For example, 78.858 would obviously round up to 78.86 but would also want to take 78.852 and round up to 78.86.
I would just do it in two operations. Use the normal ROUND() function. Then check how much it changed. And then based on that difference decide whether or not to add an extra hundredth.
Example:
data have;
input x ;
cards;
78.858
78.86
78.852
78.8515
;
data want;
set have;
round=round(x,0.01);
diff = x-round;
if diff > 0.001 then round=round+0.01 ;
run;
Results
OBS x round diff
1 78.8580 78.86 -.0020
2 78.8600 78.86 0.0000
3 78.8520 78.86 0.0020
4 78.8515 78.86 0.0015
Cheers!
Long story short I'm trying to deal with SAS rounding function and floating point precision.
Basically, I need to round half down to two decimals so, i.e. 1.235 should round to 1.23 and not 1.24
Here follow an example:
number
desired_output
1.230
1.23
1.235
1.23
1.2355
1.24
1.2305
1.23
1.231
1.23
1.236
1.24
I have tried many ways without success (i.e. several combinations of round(), ceil(), floor() and rounde() functions) but, to replicate the exercise, here below some tests:
data test;
input number desired_output;
datalines;
1.230 1.23
1.235 1.23
1.2355 1.24
1.2305 1.23
1.231 1.23
1.236 1.24
;
run;
data test;
set test;
round_01=round(number,.01);
round_ceil_01=ceil(round_01*100)/100;
round_floor_01=floor(round_01*100)/100;
round_even=rounde(number,.01);
less_half_rounding_factor=round(number-0.0005,.01);
run;
Thank you in advance!
How about:
round(number,0.01) - 0.01*(mod(number,0.01)=0.005)
Remove 0.01 when the next digit is exactly 5.
Test: Let's generate numbers to 5 decimal places and keep only those where the ROUND() function differs from the "round_down" logic above.
data test;
do integer=1 to 100000 ;
number=integer/100000 ;
round=round(number,0.01);
round_down = round(number,0.01) - 0.01*(mod(number,0.01)=0.005);
if round ne round_down then output;
end;
run;
Now let's check if any of them are those that are different are not those where the 3 least significant decimal places are 500, that is exactly X.XX500 .
data test2;
set test;
where 500 ne mod(integer,1000) ;
run;
So there were 100 cases where ROUND and ROUND_DOWN differed and they were all the cases where the value had a 5 in the thousands place and zeros after that.
Hope this is what you are looking for
data test;
set test;
value = input(put(number,4.2)best.);
run;
Good day,
I had this issue where I was writing some numbers to database, which should have had value 0.1 in SAS, but for some bizarre reason appeared as 0.09 in SQL database. When I manually checked the dataset it showed 0.10 in format 12.2.
So what I do is check if the values are actually 0.1 or somewhat below this:
data _checking;
set publish_data;
if value < 0.1;
dummy = value*10000000;
run;
It appeared that number of observations fulfill the first condition. Ok... That explains why the values come out as 0.09. Rounding issue.
However, all dummy values come out as integers. I tried 10, 100, 1k, 10k all appear to come out as integers. (1, 10, 100 ...)
Next step I try:
data _checking2;
set _checking;
if dummy<10; /*Depending on the factorial*/
run;
This is consistent. Dummy retains the value 'a little below the value shown'.
I solved the issue by round(value,.1);
Questions:
How to observe the actual value stored in dataset? (Especially in case 'a little below')
If first condition if is true, then how can the checking with dummy still show integer values. (Because in computers epsilon has to have actual value)
2.b Or is this just a display issue? Or does SAS has flag for 'value minus epsilon'?
Answer 1:
The most precise and least human way to see the actual value is to observe the underlying IEEE bytes using HEX format.
Answer 2:
The default format for those new dummy variables is BEST12., so you won't see any small offsets if they are smaller than what best12. will show, or more precisely epsilon < 1e-(12-log10(x)). The SAS format could be considered a display issue in this case.
If your use case is that of a 'shown' value must be the actual value sent to a remote database then you will want to use ROUND prior to populating the remote tables.
data x;
x = 1/3; output;
x = 0.1 - 1e-13; output;
format x 12.2;
run;
data y;
set x;
put x= x= HEX16.;
xhex = x;
format xhex hex16.;
array dummy dummy1-dummy13;
do _n_ = 1 to 13;
dummy(_n_) = x * 10**_n_;
end;
run;
proc print data=y;
run;
data z;
do p = 0 to 10;
do q = 1 to 15;
array z z1-z15;
z(q) = 10**p + 10**-q;
end; output;
end;
drop p q;
run;
==== LOG ====
x=0.33 x=3FD5555555555555
x=0.10 x=3FB9999999997D74
==== PRINT ====
Obs x xhex dummy1 dummy2 dummy3 dummy4 dummy5 dummy6 dummy7
1 0.33 3FD5555555555555 3.33333 33.3333 333.333 3333.33 33333.33 333333.33 3333333.33
2 0.10 3FB9999999997D74 1.00000 10.0000 100.000 1000.00 10000.00 100000.00 1000000.00
Obs dummy8 dummy9 dummy10 dummy11 dummy12 dummy13
1 33333333.33 333333333.33 3333333333.3 33333333333 333333333333 3.3333333E12
2 10000000.00 100000000.00 1000000000.0 10000000000 100000000000 999999999999
You can try a different format. try 32.31 or best32.
Subtract 0.1-value and look at the result. Again, use a format with a lot of decimal places.
You are probably not seeing the value in the dummy variables because the epsilon is very small and the dummy is still getting rounded for display.
Try dummy=value*1e16 or higher.
Numbers in SAS are C doubles, fwiw.
I'm trying to create a program that takes the difference of two military times and get its time difference.
Example:
**AM to PM**
Time 1: 0900
Time 2: 1730
Time Difference: 8 hours 30 minutes
**PM to AM**
Time 1: 1200
Time 2: 1100
Time Difference: 23 hours 0 minutes
Using a couple of if than else statements, I was able to figure out how to convert from military hours into standard hours but I'm stuck with how to go about subtracting time. I was trying to come up with a way to do with on paper just with addition and subtraction but I haven't developed a method that works in all cases. Any help?
One option might be to split the time into two elements, the hours and the minutes.
First, take the time, ie. 1730 and divide by 100. If its an integer or similar, it should result in 17 hours (it will automatically round down).
Then take 1730 and mod it by 100 to get 30 minutes.
int time1 = 900;
int time2 = 1730;
int diffHours = time2 / 100 - time1 / 100;
int diffMinutes = time2 % 100 - time1 % 100;
If you're unfamiliar with modulus (%), it just returns the remainder after dividing the two numbers, so 7 % 3 would be 1.
Assuming your military times are stored as int values in the range 0000..2400, then the following function will return the difference between two such values.
#include <cassert>
extern int timediff_minutes(int t1, int t2);
int timediff_minutes(int t1, int t2)
{
assert(t1 >= 0 && t1 <= 2400);
assert(t2 >= 0 && t2 <= 2400);
assert(t1 % 100 < 60 && t2 % 100 < 60);
int t1_mins = (t1 / 100) * 60 + (t1 % 100);
int t2_mins = (t2 / 100) * 60 + (t2 % 100);
return(t2_mins - t1_mins);
}
Note that 2400 is useful for representing the end of the day — and using 2400 is sanctioned by ISO 8601:2004 Data elements and interchange formats — Information interchange — Representation of dates and times.
4.2.3 Midnight
The complete representations in basic and extended format for midnight, in accordance with 4.2.2, shall be
expressed in either of the two following ways:
Basic format Extended format
a) 000000 00:00:00 (the beginning of a calendar day)
b) 240000 24:00:00 (the end of a calendar day)
The representations may have reduced accuracy in accordance with 4.2.2.3 or may be designated as a time
expression in accordance with 4.2.2.5. To represent midnight the representations may be expanded with a
decimal fraction containing only zeros in accordance with 4.2.2.4.
NOTE 1 Midnight will normally be represented as [00:00] or [24:00].
NOTE 2 The end of one calendar day [24:00] coincides with [00:00] at the start of the next calendar day, e.g. [24:00] on
12 April 1985 is the same as [00:00] on 13 April 1985. If there is no association with a date or a time interval both a) and b)
represent the same local time in the 24-hour timekeeping system.
NOTE 3 The choice of representation a) or b) will depend upon any association with a date, or a time interval.
Representations where [hh] has the value [24] are only preferred to represent the end of a time interval in accordance with
4.4 or recurring time interval in accordance with 4.5.
It is easy to transform a number into a alphanumeric value based on radix 16 in SAS by using the $HEX format. Now i'm looking for an easy way to do this with radix 36 (10 numerals & 26 letters).
Examples:
100 -> '2s'
2000 -> '1jk'
30000 -> 'n5c'
400000 -> '8kn4'
In Java you can do this by Integer.toString(mynumber, 36). Any ideas how to do this in SAS Base?
Unfortunately there is easy way to do it usings formats, but the following data step should solve the problem. It works for positive integers only.
data _null_;
infile cards;
input innumber;
number = innumber;
format base $32.;
alphabet = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ';
if number = 0 then base = '0';
else
do while (number ne 0);
mod = mod(number, length(alphabet));
div = floor(number / (length(alphabet)));
base = cats(substr(alphabet,mod+1,1),base);
number = div;
end;
put innumber= base=;
cards;
0
100
2000
30000
400000
;
run;
There is no built-in functionality for this. You can iteratively modulo 36 your number, then divide by 36 until what remains is zero. To convert the sequence of modulos you get you have to add 48decimal or 30hex to get the ascii-character in case of the digits 0-9 and 101decimal or 65hex to get the ascii-character in the case of the digits A-Z.
I suggest using PROC FCMP to create your own function that does the formatting. Then you can reuse the code whenever you want:
proc fcmp outlib=sasuser.funcs.Radix36;
function Radix36(innumber) $ 32; /* returns character string, length 32 */
number = innumber;
format base $32.;
alphabet = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ';
if number = 0 then base = '0';
else do while (number ne 0);
mod = mod(number, length(alphabet));
div = floor(number / (length(alphabet)));
base = cats(substr(alphabet,mod+1,1),base);
number = div;
end;
return (base);
endsub;
run;
/*****************************************************/
options cmplib=sasuser.funcs; /* add to search path */
data _null_;
input innumber;
base = Radix36(innumber); /* call function */
put innumber= base=;
datalines;
0
100
2000
30000
400000
;
run;