Hi I am new here and want to solve this problem:
do k=1,31
Data H(1,k)/0/
End do
do l=1,21
Data H(l,1)/0.5*(l-1)/
End do
do m=31,41
Data H(17,m)/0/
End do
do n=17,21
Data H(n,41)/0.5*(n-17)/
End do
I get error for l and n saying that it is a syntax error in DATA statement. Anyone know how to solve this problem?
You have three problems here, and not just with the "l" and "n" loops.
The first problem is that the values in a data statement cannot be arbitrary expressions. In particular, they must be constants; 0.5*(l-1) is not a constant.
The second problem is that the bounds in the object lists must also be constant (expressions); l is not a constant expression.
For the first, it's also worth noting that * in a data value list has a special meaning, and it isn't the multiplication operator. * gives a repeat count, and a repeat count of 0.5 is not valid.
You can fix the second point quite simply, by using such constructions as
data H(1,1:31) /31*0./ ! Note the repeat count specifier
outside a loop, or using an implied loop
data (H(1,k),k=1,31) /31*0./
To do something for the "l" loop is more tedious
data H(1:21,1) /0., 0.5, 1., 1.5, ... /
and we have to be very careful about the number of values specified. This cannot be dynamic.
The third problem is that you cannot specify explicit initialization for an element more than once. Look at your first two loops: if this worked you'd be initializing H(1,1) twice. Even though the same value is given, this is still invalid.
Well, actually you have four problems. The fourth is related to the point about dynamic number of values. You probably don't want to be doing explicit initialization. Whilst it's possible to do what it looks like you want to do, just use assignment where these restrictions don't apply.
do l=1,21
H(l,1) = 0.5*(l-1)
End do
Yes, there are times when complicated explicit initialization is a desirable thing, but in this case, in what I assume is new code, keeping things simple is good. An "initialization" portion of your code which does the assignments is far more "modern".
Related
I need writing a function which takes as input
a = [12,39,48,36]
and produces as output
b=[4,4,4,13,13,13,16,16,16,12,12,12]
where the idea is to repeat one element three times or two times (this should be variable) and divided by 2 or 3.
I tried doing this:
c=[12,39,48,36]
a=size(c)
for i in a
repeat(c[i]/3,3)
end
You need to vectorize the division operator with a dot ..
Additionally I understand that you want results to be Int - you can vectorizing casting to Int too:
repeat(Int.(a./3), inner=3)
Przemyslaw's answer, repeat(Int.(a./3), inner=3), is excellent and is how you should write your code for conciseness and clarity. Let me in this answer analyze your attempted solution and offer a revised solution which preserves your intent. (I find that this is often useful for educational purposes).
Your code is:
c = [12,39,48,36]
a = size(c)
for i in a
repeat(c[i]/3, 3)
end
The immediate fix is:
c = [12,39,48,36]
output = Int[]
for x in c
append!(output, fill(x/3, 3))
end
Here are the changes I made:
You need an array to actually store the output. The repeat function, which you use in your loop, would produce a result, but this result would be thrown away! Instead, we define an initially empty output = Int[] and then append! each repeated block.
Your for loop specification is iterating over a size tuple (4,), which generates just a single number 4. (Probably, you misunderstand the purpose of the size function: it is primarily useful for multidimensional arrays.) To fix it, you could do a = 1:length(c) instead of a = size(c). But you don't actually need the index i, you only require the elements x of c directly, so we can simplify the loop to just for x in c.
Finally, repeat is designed for arrays. It does not work for a single scalar (this is probably the error you are seeing); you can use the more appropriate fill(scalar, n) to get [scalar, ..., scalar].
I am reading data structures and I am following book "Fundamentals of data structures in C++" - E. Horowitz, S. Sahni & D. Mehta.
While reading rules for step counts in time complexity, I am stuck at the following statement.
Can anyone please explain me the following paragraph especially bold literals.
The assignment statement variable = expression has a step count equal to that expression unless the size of variable is a function of instance characteristics.
The non-bold part is straight-forward: the code of an assignment is at a minimum the cost of computing what is to be assigned. The bold part is simply saying that if, once you have finished that calculation, the amount of work to assign that value is not a constant (that is, it is a function of the problem size), then you have to take that into account as well.
For example, how long it takes to assign a value to an integer variable would be a constant, while copying a string won't be if its length is determined by the size of the problem.
I am trying to access a data block, the way it is define is as follows
DATA NAME /'X1','X2','X3','X4','X5','X6','X7','X8','X9','10','11',00028650
1'12','13','14','15','16','17','18','19','20','21','22','23','24'/ 00028660
The code is on paper. Note this is an old code, the only thing i am trying to do is understand how the array is being indexed. I am not trying to compile it.
The way it is accessed is as follows
I = 0
Loop
I = I + 1
write (06,77) (NAME(J,I),J=1,4) //this is inside a write statement.
end loop //77 is a format statement.
Not sure how it is being indexed, if you guys can shed some light that would be great.
The syntax (expr, intvar=int1,int2[,int3]) widely refers to an implied DO loop. There are several places where such a thing may occur, and an input/output statement is one such place.
An implied DO loop evaluates the expression expr with the do control integer variable intvar sequentially taking the values initially int1 in steps of int3 until the value int2 is reached/passed. This loop control is exactly as one would find in a do loop statement.
In the case of the question, the expression is name(j,i), the integer variable j is the loop variable, taking values between the bounds 1 and 4. [The step size int3 is not provided so is treated as 1.] The output statement is therefore exactly like
write(6,77) name(1,i), name(2,i), name(3,i), name(4,i)
as we should note that elements of the implied loop are expanded in order. i itself comes from the loop containing this output statement.
name here may refer to a function, but given the presence of a data statement initializing it, it must somehow be declared as a rank-2 (character) array. The initialization is not otherwise important.
I've got experience in a lot of other programming languages, but I'm having a lot of difficulty with Stata syntax. I've got a statement that evaluates with no problem if I put in values, but I can't figure out why it's not evaluating variables like I expect it to.
gen j=5
forvalues i = 1(1)5 {
replace TrustBusiness_local=`i' if TrustBusiness_local2==`j'
replace j=`j'-1
}
If I replace i and j with 1 and 5 respectively, like I'm expecting to happen from the code above, then it works fine, but I get an if not found error otherwise, which hasn't produced meaningful results when Googled. Does anyone see what I don't see? I hate to brute-force something that could so simply be done with a loop.
Easy to understand once you approach it the right way!
Problem 1. You never defined local macro j. That in itself is not an error, but it often leads to errors. Macros that don't exist are equivalent to empty strings, so Stata sees in this example the code
if TrustBusiness_local2==`j'
as
if TrustBusiness_local2==
which is illegal; hence the error message.
Problem 2. There is no connection of principle between a variable you called j and a local macro called j but referenced using single quotes. A variable in Stata is a variable (namely, column) in your dataset; that doesn't mean a variable otherwise in the sense of any programming language. Variables meaning single values can be held in Stata within scalars or within macros. Putting a constant into a variable, Stata sense, is legal, but usually bad style. If you have millions of observations, for example, you now have a column j with millions of values of 5 within it.
Problem 3. You could, legally, go
local j "j"
so that now the local macro j contains the text "j", which depending on how you use it could be interpreted as a variable name. It's hard to see why you would want to do that here, but it would be legal.
Problem 4. Your whole example doesn't even need a loop as it appears to mean
replace TrustBusiness_local= 6 - TrustBusiness_local2 if inlist(TrustBusiness_local2, 1,2,3,4,5)
and, depending on your data, the if qualifier could be redundant. Flipping 5(1)1 to 1(1)5 is just a matter of subtracting from 6.
Problem 5. Your example written as a loop in Stata style could be
local j = 5
forvalues i = 1/5 {
replace TrustBusiness_local=`i' if TrustBusiness_local2==`j'
local j=`j'-1
}
and it could be made more concise, but given Problem 4 that no loop is needed, I will leave it there.
Problem 6. What you talking about are, incidentally, not if statements so far as Stata is concerned, as the if qualifier used in your examples is not the same as the if command.
The problem of translating one language's jargon into another can be challenging. See my comments at http://www.stata.com/statalist/archive/2008-08/msg01258.html After experience in other languages, the macro manipulations of Stata seemed at first strange to me too; they are perhaps best understood as equivalent to shell programming.
I wouldn't try to learn Stata by Googling. Read [U] from beginning to end. (A similar point was made in the reply to your previous question at use value label in if command in Stata but you don't want to believe it!)
I was reading Hacker News and this article came up. It contains a raytracer that the code is written on the back of a business card. I decided it would be a good academic challenge to translate the c++ to python, but there's a few concepts I'm stuck on.
First, this function comes up: i T(v o,v d,f& t,v& n){...} Which is translated to int Tracer(vector o, vector d, float& t, vector& n){...} What does the float& mean? I know that in other places & is used as a == is that the case here? Can you do that in c++?
Second, I noticed these three lines:
for(i k=19;k--;) //For each columns of objects
for(i j=9;j--;) //For each line on that columns
if(G[j]&1<<k){
I know the << is a the bit shift, and I assume the & is ==. Are the for loops just like one for loop in an other?
Finally, this line: v p(13,13,13); I am not quite sure what it does. Does it create a class labeled by p that extends v (vector) with the defaults of 13,13,13?
These are probably dumb questions, but I want to see if I can understand this and my searching didn't come up with anything. Thank you in advance!
What does the float& mean?
Here, & means "reference", so the argument is passed by reference.
I know that in other places & is used as a == is that the case here?
& means various things in various contexts, but it never means ==. In this case, it's not an operator either; it's part of a type specification, meaning that it's a reference type.
I know the << is a the bit shift, and I assume the & is ==
No, it's a bitwise and operator. The result has its bits set where a bit is set in both operands. Here, with 1<<k as one operand, the result is the kth bit of G[j]; so this tests whether that bit is set.
Are the for loops just like one for loop in an other?
Yes. If you don't use braces around a for-loop's body, then the body is a single statement. So in this case, the body of the first loop is the second loop. To make this clear, I would recommend indenting the body of the loop, and using braces whether or not they are strictly necessary. But of course, I don't write (deliberately) obfuscated code.
Finally, this line: v p(13,13,13);
v is a class with a constructor taking three arguments. This declares an variable called p, of type v, initialised using that constructor; i.e. the three co-ordinates are initialised to 13.
When you seeVector& n it is referencing the vector passed into the function. This means that you can change n inside of this function without having to copy it to another Vector or without returning the Vector. This previous answer should be helpful to you.