I'm using SPSS and hoping to create a new variable that houses the sum of a group of previous variables. The catch is: if any of the variables are recorded as "9" then I need to leave them out.
Is there a command that makes this relatively easy? I had thought of doing DO IF, but it seems like that would get extremely tedious with 6 variables as I'd have to code each possible instance unless I'm mistaken. If there is an easier way that I just haven't been able to find, I would be forever grateful!
This is what I'm thinking would need to be done, but could absolutely be wrong:
DO IF (var1<=4 AND var2<=4 AND var3<=4 AND var4<=4 AND var5<=4 AND var6<=4).
COMPUTE newvar=SUM(var1,var2,var3,var4,var5,var6).
ELSE IF (var1<=4 AND var2<=4 AND var3<=4 AND var4<=4 AND var5<=4 AND var6=9).
COMPUTE newvar=SUM(var1,var2,var3,var4,var5).
/* And so on and so forth for each possible variation of the sum.
It's not that this didn't work, I'm just checking to see if there's an easier way before committing the time to it.
Here are some ways to make it shorter:
missing values var1 to var6(9).
compute newvar=sum(var1 to var6).
Now if you don't want to just mark value 9 as missing, but to get rid of it all together:
recode var1 to var6(9=sysmis).
compute newvar=sum(var1 to var6).
If you don't want to change anything in the data, you can go:
missing values var1 to var6(9).
compute newvar=sum(var1 to var6).
missing values var1 to var6().
If all those sulutions are problematic for you, here's the classicle loop solution:
compute newvar=0.
do repeat vr=var1 to var6.
if vr<>9 newvar=newvar+vr.
end repeat.
exe.
Here's another solution just for fun:
count nines=var1 to var6(9).
compute newvar=sum(var1 to var6) - nines*9.
Related
I have SPSSmodeler stream which is now used and updated every week constantly to generate a certain dataset. A raw data for this stream is also renewed on a weekly basis.
In part of this stream, there is a chunk of nodes that were necessary to modify and update manually every week, and the sequence of this part is below: Type Node => Restructure Node => Aggregate Node
To simplify the explanation of those nodes' role, I drew an image of them as bellow.
Because the original raw data is changed weekly basis, the range of Unit value above is always varied, sometimes more than 6 (maybe 100) others less than 6 (maybe 3). That is why somebody has to modify there and update those chunk of nodes on a weekly basis until now. *Unit value has a certain limitation (300 for now)
However, now we are aiming to run this stream automatically without touching any human operations on it that we need to customize there to work perfectly, automatically. Please help and will appreciate your efforts, thanks!
In order to automatize, I suggest to try to use global nodes combined with clem scripts inside the execution (default script). I have a stream that calculates the first date and the last date and those variables are used to rename files at the end of execution. I think you could use something similar as explained here:
1) Create derive nodes to bring the unit values used in the weekly stream
2) Save this information in a table named 'count_variable'
3) Use a Global node named Global with a query similar to this:
#GLOBAL_MAX(variable created in (2)) (only to record the number of variables. The step 2 created a table with only 1 values, so the GLOBAL_MAX will only bring the number of variables).
4) The query inside the execution tab will be similar to this:
execute count_variable
var tabledata
var fn
set tabledata = count_variable.output
set count_variable = value tabledata at 1 1
execute Global
5) You now can use the information of variables just using the already creatde "count_variable"
It's not easy to explain just by typing, but I hope to have been helpful.
Please mark as +1 in this answer if it was relevant one.
I think there is a better, simpler and more effective (yet risky, due to node's requirements to input data) solution to your problem. It is called Transpose node and does exactly that - pivot your table. But just from version 18.1 on. Here's an example:
https://developer.ibm.com/answers/questions/389161/how-does-new-feature-partial-transpose-work-in-sps/
Although it is easy to get the current step or increment number (variables KSTEP and KINC), I can't find an easy way to know the iteration number when inside the subroutine UMAT.
I know the following "if clause" will tell me if I'm in the first iteration of the first increment of the first step:
IF((KINC.EQ.1).AND.(SUM(STRAN+DSTRAN).EQ.0.0D0)) THEN
And I also know that I can get the iteration number writing to external files. However, is it possible to do it just inside the UMAT subroutine?
There is never really a reason to need to know the iteration number in a UMAT. If you think you need to know it, this is often a sign that you there is a better way to achieve what you want to know.
You can use a common block to track how often you enter a umat, and also which iteration you are on. But I really recommend against this. There is no good reason to know the iteration number. Unless your algorithm is perfect it will cause you more problems than it's worth.
Also in your code to check for the first increment - that will not tell you when you are in a real iteration, it will happen in PRE most likely.
Is there a way to display response options that have 0 responses in SPSS frequency output? The default is for SPSS to omit in the frequency table output any response option that is not selected by at least a single respondent. I looked for a syntax-driven option to no avail. Thank you in advance for any assistance!
It doesn't show because there is no one single case in the data is with that attribute. So, by forcing a row of zero you'll need to realize we're asking SPSS to do something incorrect.
Having said that, you can introduce a fake case with the missing category. E.g. if you have Orange, Apple, and Pear, but no one answered they like Pear, the add one fake case that says Pear.
Now, make a new weight variable that consists of only 1. But for the Pear case, make it very very small like 0.00001. Then, go to Data > Weight Cases > Weight cases by and put that new weight variable over. Click OK to apply. Now what happens is that SPSS will treat the "1" with a weight of 1 and the fake case with a weight that is 1/10000 of a normal case. If you rerun the frequency you should see the one with zero count shows up.
If you have purchased the Custom Table module you can also do that directly as well, as far as I can tell from their technical document. That module costs 637 to 3630 depending on license type, so probably only worth a try if your institute has it.
So, I'm a noob with SPSS, I (shame on me) have a cracked version of SPSS 22 and if I understood your question correctly, this is my solution:
double click the Frequency table in Output
right click table, select Table Properties
go to General and then uncheck the Hide empty rows and columns option
Hope this helps someone!
If your SPSS version has no Custom Tables installed and you haven't collected money for that module yet then use the following (run this syntax):
*Note: please use variable names up to 8 characters long.
set mxloops 1000. /*in case your list of values is longer than 40
matrix.
get vars /vari= V1 V2 /names= names /miss= omit. /*V1 V2 here is your categorical variable(s)
comp vals= {1,2,3,4,5,99}. /*let this be the list of possible values shared by the variables
comp freq= make(ncol(vals),ncol(vars),0).
loop i= 1 to ncol(vals).
comp freq(i,:)= csum(vars=vals(i)).
end loop.
comp names= {'vals',names}.
print {t(vals),freq} /cnames= names /title 'Frequency'. /*here you are - the frequencies
print {t(vals),freq/nrow(vars)*100} /cnames= names /format f8.2 /title 'Percent'. /*and percents
end matrix.
*If variables have missing values, they are deleted listwise. To include missings, use
get vars /vari= V1 V2 /names= names /miss= -999. /*or other value
*To exclude missings individually from each variable, analyze by separate variables.
Since the last post is closed due to unclear expression, here is a edited one.
There are in total 20 items from 5 Likert-type scale questions from a questionnaire. I need to add the 20 items from 5 separate questions to create a total scale. I already got the data.
The question is just like the picture above. How can I run the command to add the 20 items from 5 separate questions? What is the command?
Is it something like Transform > Compute variable. Enter a variable name, specify which items to add up, and hey presto (e.g. "V1+V2+V3" etc)?
You can do exactly as you suggested, using the Transform -> Compute variable... function. Simply type in the name of your new scale in the Target variable box and the addition you want in the Numeric variable box.
You will see that the following SPSS syntax command is run:
COMPUTE total=v1 + v2 + v3 + v4.
EXECUTE.
If any of the variables has a missing value, the simply adding them will result in a missing value as well. If you don't want to impute for missing values, using the MEAN command in syntax works well. Also, if the variables are contiguous in the data file, you can make the syntax much more readable by using the TO modifier.
COMPUTE myscore=MEAN(variable1 TO variable5)*5.
The resulting value provides an efficient expected value.
However, it seems like the problem in this case is that the data entry process has dummy coded all of the items, producing 20 separate variables instead of 5, where each block of 4 variables has a value of 0 or 1 but represents the values 1 to 4. In this case, you can use the following syntax:
COMPUTE mycounter=1.
COMPUTE myscore=0.
EXECUTE.
DO REPEAT a=variable1 TO variable20.
COMPUTE myscore=myscore+mycounter*a.
COMPUTE mycounter=mycounter+1.
IF (mycounter=5) mycounter=1.
END REPEAT.
EXECUTE.
Note that the variables from variable1 to variable20 must have each set of dummy codes from the original items clustered together in ascending order.
I am trying to run this code but it keeps crashing:
log10(x):=log(x)/log(10);
char(x):=floor(log10(x))+1;
mantissa(x):=x/10**char(x);
chop(x,d):=(10**char(x))*(floor(mantissa(x)*(10**d))/(10**d));
rnd(x,d):=chop(x+5*10**(char(x)-d-1),d);
d:5;
a:10;
Ibwd:[[30,rnd(integrate((x**60)/(1+10*x^2),x,0,1),d)]];
for n from 30 thru 1 step -1 do Ibwd:append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd);
Maxima crashes when it evaluates the last line. Any ideas why it may happen?
Thank you so much.
The problem is that the difference becomes negative and your rounding function dies horribly with a negative argument. To find this out, I changed your loop to:
for n from 30 thru 1 step -1 do
block([],
print (1/(2*n-1)-a*last(first(Ibwd))),
print (a*last(first(Ibwd))),
Ibwd: append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd),
print (Ibwd));
The last difference printed before everything fails miserably is -316539/6125000. So now try
rnd(-1,3)
and see the same problem. This all stems from the fact that you're taking the log of a negative number, which Maxima interprets as a complex number by analytic continuation. Maxima doesn't evaluate this until it absolutely has to and, somewhere in the evaluation code, something's dying horribly.
I don't know the "fix" for your specific example, since I'm not exactly sure what you're trying to do, but hopefully this gives you enough info to find it yourself.
If you want to deconstruct a floating point number, let's first make sure that it is a bigfloat.
say z: 34.1
You can access the parts of a bigfloat by using lisp, and you can also access the mantissa length in bits by ?fpprec.
Thus ?second(z)*2^(?third(z)-?fpprec) gives you :
4799148352916685/140737488355328
and bfloat(%) gives you :
3.41b1.
If you want the mantissa of z as an integer, look at ?second(z)
Now I am not sure what it is that you are trying to accomplish in base 10, but Maxima
does not do internal arithmetic in base 10.
If you want more bits or fewer, you can set fpprec,
which is linked to ?fpprec. fpprec is the "approximate base 10" precision.
Thus fpprec is initially 16
?fpprec is correspondingly 56.
You can easily change them both, e.g. fpprec:100
corresponds to ?fpprec of 335.
If you are diddling around with float representations, you might benefit from knowing
that you can look at any of the lisp by typing, for example,
?print(z)
which prints the internal form using the Lisp print function.
You can also trace any function, your own or system function, by trace.
For example you could consider doing this:
trace(append,rnd,integrate);
If you want to use machine floats, I suggest you use, for the last line,
for n from 30 thru 1 step -1 do :
Ibwd:append([[n-1,rnd(1/(2.0*n- 1.0)-a*last(first(Ibwd)),d)]],Ibwd);
Note the decimal points. But even that is not quite enough, because integration
inserts exact structures like atan(10). Trying to round these things, or compute log
of them is probably not what you want to do. I suspect that Maxima is unhappy because log is given some messy expression that turns out to be negative, even though it initially thought otherwise. It hands the number to the lisp log program which is perfectly happy to return an appropriate common-lisp complex number object. Unfortunately, most of Maxima was written BEFORE LISP HAD COMPLEX NUMBERS.
Thus the result (log -0.5)= #C(-0.6931472 3.1415927) is entirely unexpected to the rest of Maxima. Maxima has its own form for complex numbers, e.g. 3+4*%i.
In particular, the Maxima display program predates the common lisp complex number format and does not know what to do with it.
The error (stack overflow !!!) is from the display program trying to display a common lisp complex number.
How to fix all this? Well, you could try changing your program so it computes what you really want, in which case it probably won't trigger this error. Maxima's display program should be fixed, too. Also, I suspect there is something unfortunate in simplification of logs of numbers that are negative but not obviously so.
This is probably waaay too much information for the original poster, but maybe the paragraph above will help out and also possibly improve Maxima in one or more places.
It appears that your program triggers an error in Maxima's simplification (algebraic identities) code. We are investigating and I hope we have a bug fix soon.
In the meantime, here is an idea. Looks like the bug is triggered by rnd(x, d) when x < 0. I guess rnd is supposed to round x to d digits. To handle x < 0, try this:
rnd(x, d) := if x < 0 then -rnd1(-x, d) else rnd1(x, d);
rnd1(x, d) := (... put the present definition of rnd here ...);
When I do that, the loop runs to completion and Ibwd is a list of values, but I don't know what values to expect.