I am just not able to figure out how to proceed:
I am trying to build a model:
It would have 4 Inputs ( Boolean i/p)
It would have 1 output (Signed: 8 bit)
It would perform the following:
Based on which input is 1, it would give a corresponding output reflecting the DataRate.
If I have to write in Matlab, I would write something like this :
if (portA==1)
PSDU_Data_Rate=1;
elseif(portB==1)
PSDU_Data_Rate=2;
elseif (portC==1)
PSDU_Data_Rate=5.5;
elseif(portD==1)
PSDU_Data_Rate=11;
end
I am attaching, the part of the model which I am developing for the same functionality:
Any idea on how to proceed or code correction or suggestion on how it can be improved would be really helpful.
Thanks
Since you have 4 distinct inputs instead of a single input carrying an enumerated value, use If - Else, instead of a Case statement. I'm adding screenshots of how this can be done. Note that the If block also lets you have an Else output if you want to select one of the data rates by default (if none of the inputs are non-zero).
If block settings:
Number of inputs: 4
If expression: u1 ~= 0
Elseif expressions: u2 ~=0, u3 ~= 0, u4 ~= 0
The model consists of an If block connected to a set of If Action Subsystem blocks. The outputs of the latter can be combined into a single signal using a Merge block.
Related
I'm relatively new to Z3 and experimenting with it in python. I've coded a program which returns the order in which different actions is performed, represented with a number. Z3 returns an integer representing the second the action starts.
Now I want to look at the model and see if there is an instance of time where nothing happens. To do this I made a list with only 0's and I want to change the index at the times where each action is being executed, to 1. For instance, if an action start at the 5th second and takes 8 seconds to be executed, the index 5 to 12 would be set to 1. Doing this with all the actions and then look for 0's in the list would hopefully give me the instances where nothing happens.
The problem is: I would like to write something like this for coding the problem
list_for_check = [0]*total_time
m = s.model()
for action in actions:
for index in range(m.evaluate(action.number) , m.evaluate(action.number) + action.time_it_takes):
list_for_check[index] = 1
But I get the error:
'IntNumRef' object cannot be interpreted as an integer
I've understood that Z3 isn't returning normal ints or bools in their models, but writing
if m.evaluate(action.boolean):
works, so I'm assuming the if is overwritten in a way, but this doesn't seem to be the case with range. So my question is: Is there a way to use range with Z3 ints? Or is there another way to do this?
The problem might also be that action.time_it_takes is an integer and adding a Z3int with a "normal" int doesn't work. (Done in the second part of the range).
I've also tried using int(m.evaluate(action.number)), but it doesn't work.
Thanks in advance :)
When you call evaluate it returns an IntNumRef, which is an internal z3 representation of an integer number inside z3. You need to call as_long() method of it to convert it to a Python number. Here's an example:
from z3 import *
s = Solver()
a = Int('a')
s.add(a > 4);
s.add(a < 7);
if s.check() == sat:
m = s.model()
print("a is %s" % m.evaluate(a))
print("Iterating from a to a+5:")
av = m.evaluate(a).as_long()
for index in range(av, av + 5):
print(index)
When I run this, I get:
a is 5
Iterating from a to a+5:
5
6
7
8
9
which is exactly what you're trying to achieve.
The method as_long() is defined here. Note that there are similar conversion functions from bit-vectors and rationals as well. You can search the z3py api using the interface at: https://z3prover.github.io/api/html/namespacez3py.html
I'm still new with SPSS, I Have Data For The Following :
Cereals Vegetables Fruit Meat Dairy Fat Sugar Pulses
I Have Also Computed The Variables With This Formula :
Total FCS = (Cereals*2)+(Vegetables)+(Fruits)+(Meat*4)+(Dairy*4)+(Sugar*0.5)+(Pulses*3)
Now I Want To Rank The Data from the Total FCS In One Column In Order To Make Graph From It As Following:
Rank as :
<28 Poor
>28.5 - <42 Borderline
>42.5 Acceptable
What Should I Do ?
I would use a DO IF statement to assign the ranks. Example below.
DO IF FCS < 28.
COMPUTE RankFCS = 1.
ELSE IF FCS <= 42.5.
COMPUTE RankFCS = 2.
ELSE.
COMPUTE RankFCS = 3.
END IF.
VALUE LABELS RankFCS
1 'Poor'
2 'Borderline'
3 'Acceptable'.
There is a command called Recode in SPSS, you can use that command to create this rank variable. Recode command has two options
1). Recode into same variables
2). Recode into Different variables.
I am using 2nd option as you need to create a new Rank variable.
STRING RankFCS (A8).
RECODE FCS (Lowest thru 28='Poor') (28.5 thru 42='Borderline')
(42.5 thru Highest='Acceptable')
INTO RankFCS.
EXECUTE.
I am using numerical recipes scheme to generate random numbers (ran3, page 7 in this PDF file). I didn't notice anything strange but this time, I got a negative numbers at the "warm up" stage which are larger than MBIG. The code look as if this shouldn't happen. I can easily fix this with changing the if statement to be a while statement at the line that says if(mk.lt.MZ)mk=mk+MBIG but I want to know what are the consequences.
Edit:here is the function
FUNCTION ran3a(idum)
INTEGER idum
INTEGER MBIG,MSEED,MZ
C REAL MBIG,MSEED,MZ
REAL ran3a,FAC
PARAMETER (MBIG=1000000000,MSEED=161803398,MZ=0,FAC=1./MBIG)
C PARAMETER (MBIG=4000000.,MSEED=1618033.,MZ=0.,FAC=1./MBIG)
INTEGER i,iff,ii,inext,inextp,k
INTEGER mj,mk,ma(55)
C REAL mj,mk,ma(55)
SAVE iff,inext,inextp,ma
DATA iff /0/
if(idum.lt.0.or.iff.eq.0)then
iff=1
mj=MSEED-iabs(idum)
mj=mod(mj,MBIG)
ma(55)=mj
mk=1
do 11 i=1,54
ii=mod(21*i,55)
ma(ii)=mk
mk=mj-mk
if(mk.lt.MZ)mk=mk+MBIG
mj=ma(ii)
11 continue
do 13 k=1,4
do 12 i=1,55
ma(i)=ma(i)-ma(1+mod(i+30,55))
if(ma(i).lt.MZ)ma(i)=ma(i)+MBIG
12 continue
13 continue
inext=0
inextp=31
idum=1
endif
inext=inext+1
if(inext.eq.56)inext=1
inextp=inextp+1
if(inextp.eq.56)inextp=1
mj=ma(inext)-ma(inextp)
if(mj.lt.MZ)mj=mj+MBIG
ma(inext)=mj
ran3a=mj*FAC
return
END
I was getting Seg Faults (using gfortran 4.8) because the function was trying to change the input value idum from the negative number to 1. There is no reason for that line (nor anything with iff), so I deleted it and printed out the array ma at several different places and found no negative numbers in the array.
One possibility, though, is if iabs(idum) is larger than MSEED, you might have a problem with the line mj=MSEED - iabs(idum). You should protect from this by using mj=abs(MSEED-abs(idum)) like the book has written.
Had a look at the pdf. What you need to do is
1) Seed it: value = ran3(-1)
2) Use it: value = ran3(0)
Is it possible to use DECODE on a SUM field ?
For example, something like this :
DECODE( SUM("QTA' CONV SUM 1 2 2"),'>0',1,'=0',0)
In your specific case, where you want the result to be 1 if the SUM is greater than zero or 0 if the SUM is zero, you can use the Oracle SIGN function. I'm not sure if there's any special Discoverer syntax, but outside of Discoverer I'd do it something like this:
SIGN(SUM(whatever you're summing))
The "techonthenet" site documents the function here.
An alternative to #EdGibbs, if the values that you're summing are integers:
Greatest(Sum(...),1)
If you had negatives to contend with then maybe:
Least(Greatest(Sum(...),1),0)
The syntax gets a bit gnarly so if you did want to stick with Decode or Case then:
Case Sum(...)
When 0 Then 0
Else 1
End
... or ...
Decode(Sum(...),0,0,Sum(...))
I wouldn't go for a solution that repeats the Sum() though. Keep it DRY
I am working on a very basic WEKA assignment, and I'm trying to use WEKA to preprocess data from the GUI (most current version). I am trying to do very basic if statements and mathematical statements in the expression box when double clicking on MathExpression and I haven't had any success. For example I want to do
if (a5 == 2 || a5 == 0) then y = 1; else y = 0
Many different variations of this haven't worked for me and I'm also unclear on how to refer to "y" or if it needs a reference within the line.
Another example is -abs(log(a7)–3) which I wasn't able to work out either. Any ideas about how to make these statements work?
From javadoc of MathExpression
The 'A'
letter refers to the value of the attribute being processed.
Other attribute values (numeric only) can be accessed through
the variables A1, A2, A3, ...
Your filter applies to all attributes of your dataset. If I load iris dataset and apply following filter.
weka.filters.unsupervised.attribute.MathExpression -E log(A).
your attribute ,sepallength values change as following.
Before Filter After Filter
Minimum 4.3 Minimum 1.459
Maximum 7.9 Maximum 2.067
Mean 5.843 Mean 1.755
StdDev 0.828 StdDev 0.141
Also if you look to javadoc, there is no if else function but ifelse function. Therefore you should write something like
ifelse ( (A == 2 || A == 0), 1,0 )
Also this filter applies to all attributes. If you want to change only one attribute and according to other attribute values ; then you need to use "Ignore range option" and use A1,A2 to refer to other attribute values.
if you need to add new attribute use AddExpression.
An instance filter that creates a new attribute by applying a mathematical expression to existing attributes.