topic Re: Format $Z was not found or could not be loaded in SAS Programming

Format $Z was not found or could not be loaded

raivester — Tue, 27 Oct 2020 21:11:07 GMT

I have a variable that is character that I am trying to convert numeric and ensure leading zeros. I have attempted the following, but get the message: Format $Z was not found or could not be loaded.

length ocounty $2. oid $6.;
	if NCRPV1   ne . then ocounty 	= strip(put(NCRPV1,z2.));

Can someone help resolve this?

Re: Format $Z was not found or could not be loaded

novinosrin — Tue, 27 Oct 2020 21:18:04 GMT

Try to make you numeric and then convert back to char using Z format

strip(put(input(NCRPV1,32.),z2.));

You can't apply a Z format for a char value. You need your value to standard numeric data. So you need INPUT function first

Re: Format $Z was not found or could not be loaded

Kurt_Bremser — Tue, 27 Oct 2020 21:17:13 GMT

To convert character to numeric, you need the INPUT function.

ocounty  = input(NCRPV1,2.);
format ocounty z2.;

But if the data is some kind of a code for something, keep it as character.

Re: Format $Z was not found or could not be loaded

Tom — Tue, 27 Oct 2020 21:22:25 GMT

Character variables require character formats. So the data step compiler converted the attempt to use the numeric Z format with the character variable NCRPV1 into a request for the character format $Z. SAS does not delivery such a format and it couldn't find a user defined one either.

You first need to figure out whether NCRPV1 is supposed to be numeric or character in order to decide if you need to change your program or change your data.

Re: Format $Z was not found or could not be loaded

SASKiwi — Tue, 27 Oct 2020 21:17:26 GMT

The error suggests your variable NCRPV1 is already character. If you want to add leading zeros then you will need to do this:

if NCRPV1   ne '' then ocounty 	= strip(put(input(NCRPV1, 2.),z2.));