Is it possible to use <,> operators with the if any function? Something like this:
select if (any(>10,Q1) AND any(<2,Q2 to Q10))
You definitely need to create an auxiliary variable to do this.
#Jignesh Sutar's solution is one that works fine. However there are often multiple ways in SPSS to accomplish a certain task.
Here is another solution where the COUNT command comes in handy.
It is important to note that the following solution assumes that the values of the variables are integers. If you have float values (1.5 for instance) you'll get a wrong result.
* count occurrences where Q2 to Q10 is less then 2.
COUNT #QLT2 = Q2 TO Q10 (LOWEST THRU 1).
* select if Q1>10 and
* there is at least one occurrence where Q2 to Q10 is less then 2.
SELECT (Q1>10 AND #QLT2>0).
There is also a variant for this sort of solution that deals with float variables correctly. But I think it is less intuitive though.
* count occurrences where Q2 to Q10 is 2 or higher.
COUNT #QGE2 = Q2 TO Q10 (2 THRU HIGHEST).
* select if Q1>10 and
* not every occurences of (the 9 variables) Q2 to Q10 is two or higher.
SELECT IF (Q1>10 AND #QGE2<9).
Note: Variables beginning with # are temporary variables. They are not stored in the data set.
I don't think you can (would be nice if you could - you can do something similar in Excel with COUNTIF & SUMIF IIRC).
You've have to construct a new variable which tests the multiple ANY less than condition, as per below example:
input program.
loop #j = 1 to 1000.
compute ID=#j.
vector Q(10).
loop #i = 1 to 10.
compute Q(#i) = trunc(rv.uniform(-20,20)).
end loop.
end case.
end loop.
end file.
end input program.
execute.
vector Q=Q2 to Q10.
loop #i=1 to 9 if Q(#i)<2.
compute #QLT2=1.
end loop if Q(#i)<2.
select if (Q1>10 and #QLT2=1).
exe.
Related
I have 5 variables for one questionnaire about social support. I want to define the group with low vs. high support. According to the authors low support is defined as a sum score <= 18 AND two items scoring <= 3.
It would be great to get a dummy variable which shows which people are low vs high in support.
How can I do this in the syntax?
Thanks ;)
Assuming your variables are named Var1, Var2 .... Var5, and that they are consecutive in the dataset, this should work:
recode Var1 to Var5 (1 2 3=1)(4 thr hi=0) into L1 to L5.
compute LowSupport = sum(Var1 to Var5) <= 18 and sum(L1 to L5)>=2.
execute.
New variable LowSupport will have value 1 for rows that have the parameters you defined and 0 for other rows.
Note: If your variables are not consecutive you'll have to list all of them instead of using Var1 to var5.
I have around 300 variables and I am calculating their Skewness and Kurtosis. Now, I want to create a new varaible which will consist of the list of all those variables whose Skewness and Kurtosis are within a certain range. The idea is to select only those variables which are satisfying a condition and perform normalization on all the other variables.
To calcualte Skewness i am using;
Descriptives A TO Z
/Statistics Skewness.
Execute.
I know this is not a valid Syntax but i Need something like this:
Compute x= if(Skewness(A TO Z)>1)
Please help me out with an SPSS Syntax for this.
There are multiple ways to approach this, so there might be an easier way.
you just need to change the 'var1 TO varN' to your list of variables and whatever criteria you want for Skewness & Kurtosis on the two COMPUTE lines that create the flags, and this will do it for you.
If I were doing this I would go a step further and build the normalization into the syntax using WRITE OUT = ".sps" /CMD. INSERT FILE = ".sps", but that isn't what you asked for.
DATASET DECLARE DistributionSyntax.
OMS
/SELECT TABLES
/IF SUBTYPES=["Descriptives"] INSTANCES=[1]
/DESTINATION FORMAT=SAV OUTFILE = 'DistributionSyntax'.
EXAMINE VARIABLES=var1 TO varN
/PLOT NONE
/STATISTICS DESCRIPTIVES
/CINTERVAL 95
/MISSING PAIRWISE
/NOTOTAL.
OMSEND.
DATASET ACTIVATE DistributionSyntax.
USE ALL.
FILTER OFF.
SELECT IF ANY(Var2,'Skewness','Kurtosis').
EXECUTE.
STRING VarName (A64).
COMPUTE SkewnessFlag = (Var2 = 'Skewness' AND ABS(Statistic) > 2).
COMPUTE KurtosisFlag = (Var2 = 'Kurtosis' AND ABS(Statistic) > 2).
COMPUTE VarName = CHAR.SUBSTR(Var1,1,CHAR.INDEX(Var1,' ')-1).
EXECUTE.
USE ALL.
COMPUTE filter_$=(SkewnessFlag = 1).
VALUE LABELS filter_$ 0 'Not Selected' 1 'Selected'.
FORMATS filter_$ (f1.0).
FILTER BY filter_$.
EXECUTE.
FRE VarName.
USE ALL.
COMPUTE filter_$=(KurtosisFlag= 1).
VALUE LABELS filter_$ 0 'Not Selected' 1 'Selected'.
FORMATS filter_$ (f1.0).
FILTER BY filter_$.
EXECUTE.
FRE VarName.
USE ALL.
FILTER OFF.
EXECUTE.
If you omit the select data blocks after you compute the flags and replace it with this, it will calculate normalized versions of the variables that meet your criteria. This calculates new variables, and you will want to add a file location for the syntax file (replace the "~/" in the WRITE and INSERT commands), and change the name of the dataset referenced as 'RAWDATA' to whatever your dataset name is:
USE ALL.
FILTER OFF.
SELECT IF ANY(1,SkewnessFlag,KurtosisFlag).
EXECUTE.
STRING CMD (A250).
COMPUTE CMD = CONCAT("COMPUTE ",RTRIM(VarName),".Norm = ln(",RTRIM(VarName),").").
EXECUTE.
DATA LIST /CMD 1-250 (A).
BEGIN DATA
EXECUTE.
END DATA.
DATASET NAME EXE WINDOW = FRONT.
DATASET ACTIVATE DistributionSyntax.
ADD FILES /FILE = *
/FILE = 'EXE'.
EXECUTE.
DATASET CLOSE EXE.
DATASET ACTIVATE DistributionSyntax.
WRITE OUT="~\Normalize Variables.sps" /CMD.
DATASET CLOSE DistributionSyntax.
DATASET ACTIVATE RAWDATA.
INSERT FILE="~\Normalize Variables.sps".
I am working on programming a Markov chain in Lua, and one element of this requires me to uniformly generate random numbers. Here is a simplified example to illustrate my question:
example = function(x)
local r = math.random(1,10)
print(r)
return x[r]
end
exampleArray = {"a","b","c","d","e","f","g","h","i","j"}
print(example(exampleArray))
My issue is that when I re-run this program multiple times (mash F5) the exact same random number is generated resulting in the example function selecting the exact same array element. However, if I include many calls to the example function within the single program by repeating the print line at the end many times I get suitable random results.
This is not my intention as a proper Markov pseudo-random text generator should be able to run the same program with the same inputs multiple times and output different pseudo-random text every time. I have tried resetting the seed using math.randomseed(os.time()) and this makes it so the random number distribution is no longer uniform. My goal is to be able to re-run the above program and receive a randomly selected number every time.
You need to run math.randomseed() once before using math.random(), like this:
math.randomseed(os.time())
From your comment that you saw the first number is still the same. This is caused by the implementation of random generator in some platforms.
The solution is to pop some random numbers before using them for real:
math.randomseed(os.time())
math.random(); math.random(); math.random()
Note that the standard C library random() is usually not so uniformly random, a better solution is to use a better random generator if your platform provides one.
Reference: Lua Math Library
Standard C random numbers generator used in Lua isn't guananteed to be good for simulation. The words "Markov chain" suggest that you may need a better one. Here's a generator widely used for Monte-Carlo calculations:
local A1, A2 = 727595, 798405 -- 5^17=D20*A1+A2
local D20, D40 = 1048576, 1099511627776 -- 2^20, 2^40
local X1, X2 = 0, 1
function rand()
local U = X2*A2
local V = (X1*A2 + X2*A1) % D20
V = (V*D20 + U) % D40
X1 = math.floor(V/D20)
X2 = V - X1*D20
return V/D40
end
It generates a number between 0 and 1, so r = math.floor(rand()*10) + 1 would go into your example.
(That's multiplicative random number generator with period 2^38, multiplier 5^17 and modulo 2^40, original Pascal code by http://osmf.sscc.ru/~smp/)
math.randomseed(os.clock()*100000000000)
for i=1,3 do
math.random(10000, 65000)
end
Always results in new random numbers. Changing the seed value will ensure randomness. Don't follow os.time() because it is the epoch time and changes after one second but os.clock() won't have the same value at any close instance.
There's the Luaossl library solution: (https://github.com/wahern/luaossl)
local rand = require "openssl.rand"
local randominteger
if rand.ready() then -- rand has been properly seeded
-- Returns a cryptographically strong uniform random integer in the interval [0, n−1].
randominteger = rand.uniform(99) + 1 -- randomizes an integer from range 1 to 100
end
http://25thandclement.com/~william/projects/luaossl.pdf
I work with SPSS and have difficulty finding/generating a syntax for counting cases.
I have about 120 cases and five variables. I need to know the count /proportion of cases where just one, more than one, or all of the cases have a value of 1 (dichotomous variable). Then I need to compute a new variable that shows the number / proportion of cases which include all of the aforementioned cases (also dichotomous).
For example case number one: var1=1, var2=1, var3=1, var4=0, var5=0 --> newvariable=1.
Case number two: var1=0, var2=0, var3=0, var4=0, var5=0 --> newvariable=1.
And so on...
Can anybody help me with a syntax?
Help would much appreciated!
Here we can use the sum of the variables to determine your conditions. So using a scratch variable that is the sum, we can see if it is equal to 1, more than 1 or 5 in your example.
compute #sum = SUM(var1 to var5).
compute just_one = (#sum = 1).
compute more_one = (#sum > 1).
compute all_one = (#sum = 5).
Similarly, all_one could be computed using the ANY command to evaluate if any zeroes exist, i.e. compute all_one = ANY(0,var1 to var5).. These code snippets assume that var1 to var5 are contiguous in the data frame, if not they just need to be replaced with var1,var2,var3,var4,var5 in all given instances.
You could read up on the logical function ANY in the Command Syntax Reference manual, if you negated a test for ANY with "0", then that is effectively a test for all "1"s. Use of the COUNT command would be another approach.
I'm puzzling over how to map a set of sequences to consecutive integers.
All the sequences follow this rule:
A_0 = 1
A_n >= 1
A_n <= max(A_0 .. A_n-1) + 1
I'm looking for a solution that will be able to, given such a sequence, compute a integer for doing a lookup into a table and given an index into the table, generate the sequence.
Example: for length 3, there are 5 the valid sequences. A fast function for doing the following map (preferably in both direction) would be a good solution
1,1,1 0
1,1,2 1
1,2,1 2
1,2,2 3
1,2,3 4
The point of the exercise is to get a packed table with a 1-1 mapping between valid sequences and cells.
The size of the set in bounded only by the number of unique sequences possible.
I don't know now what the length of the sequence will be but it will be a small, <12, constant known in advance.
I'll get to this sooner or later, but though I'd throw it out for the community to have "fun" with in the meantime.
these are different valid sequences
1,1,2,3,2,1,4
1,1,2,3,1,2,4
1,2,3,4,5,6,7
1,1,1,1,2,3,2
these are not
1,2,2,4
2,
1,1,2,3,5
Related to this
There is a natural sequence indexing, but no so easy to calculate.
Let look for A_n for n>0, since A_0 = 1.
Indexing is done in 2 steps.
Part 1:
Group sequences by places where A_n = max(A_0 .. A_n-1) + 1. Call these places steps.
On steps are consecutive numbers (2,3,4,5,...).
On non-step places we can put numbers from 1 to number of steps with index less than k.
Each group can be represent as binary string where 1 is step and 0 non-step. E.g. 001001010 means group with 112aa3b4c, a<=2, b<=3, c<=4. Because, groups are indexed with binary number there is natural indexing of groups. From 0 to 2^length - 1. Lets call value of group binary representation group order.
Part 2:
Index sequences inside a group. Since groups define step positions, only numbers on non-step positions are variable, and they are variable in defined ranges. With that it is easy to index sequence of given group inside that group, with lexicographical order of variable places.
It is easy to calculate number of sequences in one group. It is number of form 1^i_1 * 2^i_2 * 3^i_3 * ....
Combining:
This gives a 2 part key: <Steps, Group> this then needs to be mapped to the integers. To do that we have to find how many sequences are in groups that have order less than some value. For that, lets first find how many sequences are in groups of given length. That can be computed passing through all groups and summing number of sequences or similar with recurrence. Let T(l, n) be number of sequences of length l (A_0 is omitted ) where maximal value of first element can be n+1. Than holds:
T(l,n) = n*T(l-1,n) + T(l-1,n+1)
T(1,n) = n
Because l + n <= sequence length + 1 there are ~sequence_length^2/2 T(l,n) values, which can be easily calculated.
Next is to calculate number of sequences in groups of order less or equal than given value. That can be done with summing of T(l,n) values. E.g. number of sequences in groups with order <= 1001010 binary, is equal to
T(7,1) + # for 1000000
2^2 * T(4,2) + # for 001000
2^2 * 3 * T(2,3) # for 010
Optimizations:
This will give a mapping but the direct implementation for combining the key parts is >O(1) at best. On the other hand, the Steps portion of the key is small and by computing the range of Groups for each Steps value, a lookup table can reduce this to O(1).
I'm not 100% sure about upper formula, but it should be something like it.
With these remarks and recurrence it is possible to make functions sequence -> index and index -> sequence. But not so trivial :-)
I think hash with out sorting should be the thing.
As A0 always start with 0, may be I think we can think of the sequence as an number with base 12 and use its base 10 as the key for look up. ( Still not sure about this).
This is a python function which can do the job for you assuming you got these values stored in a file and you pass the lines to the function
def valid_lines(lines):
for line in lines:
line = line.split(",")
if line[0] == 1 and line[-1] and line[-1] <= max(line)+1:
yield line
lines = (line for line in open('/tmp/numbers.txt'))
for valid_line in valid_lines(lines):
print valid_line
Given the sequence, I would sort it, then use the hash of the sorted sequence as the index of the table.