I know this is probably a quite specific question but I would like to use the rrcurveto Operator in a PostScript File to draw a Font Glyph. I am using ps2pdf as Interpreter (Because I don't get GhostView to run on my M1 Mac) which might make trouble in this case.
Furthermore I can't use for example the hsbw Operator. Do I have to use a certain Dictionary for these Operators?
I would appreciate help a lot!
I tried replacing the Operators, which isn't easy when it comes to rrcurve. I also tried to use the CharStrings Dictionary but I'm not sure wether I implemented it correctly.
This operator doesn't exist in PostScript, but it is very similar to the rcurveto operator. You can write it like this, adding the various vectors together:
/rrcurveto { % dx1 dy1 dx2 dy2 dx3 dy3 . -
1 dict begin
{dy3 dx3 dy2 dx2 dy1 dx1}{exch def}forall
dx1 dy1
dx1 dx2 add dy1 dy2 add
dx1 dx2 dx3 add add dy1 dy2 dy3 add add
rcurveto
end
} def
Related
I understand that "cm" concatenates two CTMs, however, it's not obvious to me what the specific definition of concatenation is. Reading through to "graphics state operators", in the specification, has not helped me.
Thus far I've looked at a whole bunch of different resources about matrix concatenation. There seems to be a number of different ways concatenation is defined for matrices: some examples seem to show it as:
[1,2; concat [5,6; = [1,2,5,6;
3,4] 7,8]. 3,4,7,8]
... however that would seem to break the transform matrices, so I assume that's not it.
Another option is that they just mean matrix addition:
[1,2; + [5,6; = [6 ,8 ;
3,4] 7,8]. 10,12]
but I feel, if it were just a matrix addition, they would just call it addition/matrix addition.
my last idea is:
[1,2; + [5,6; = [15,26 ;
3,4] 7,8]. 37,48]
but that seems like a bizarre approach, not least because it would have numbers behaving like text.
Thanks in advance
I need to select values from a single column in a Julia dataframe based on multiple criteria sourced from an array. Context: I'm attempting to format the data from a large Julia DataFrame to support a PCA (primary component analysis), so I first split the original data into an anlytical matrix and a label array. This is my code, so far (doesn't work):
### Initialize source dataframe for PCA
dfSource=DataFrame(
colDataX=[0,5,10,15,5,20,0,5,10,30],
colDataY=[1,2,3,4,5,6,7,8,9,0],
colRowLabels=[0.2,0.3,0.5,0.6,0.0,0.1,0.2,0.1,0.8,0.0])
### Extract 1/2 of rows into analytical matrix
matSource=convert(Matrix,DataFrame(dfSource[1:2:end,1:2]))'
### Extract last column as labels
arLabels=dfSource[1:2:end,3]
### Select filtered rows
datGet=matSource[:,arLabels>=0.2 & arLabels<0.7][1,:]
print(datGet)
output> MethodError: no method matching...
At the last line before the print(datGet) statement, I get a MethodError indicating a method mismatch related to use of the & logic. What have I done wrong?
A small example of alternative implementation (maybe you will find it useful to see what DataFrames.jl has in-built):
# avoid materialization if dfSource is large
dfSourceHalf = #view dfSource[1:2:end, :]
lazyFilter = Iterators.filter(row -> 0.2 <= row[3] < 0.7, eachrow(dfSourceHalf))
matFiltered = mapreduce(row -> collect(row[1:2]), hcat, lazyFilter)
matFiltered[1, :]
(this is not optimized for speed, but rather as a showcase what is possible, but still it is already several times faster than your code)
This code works:
dfSource=DataFrame(
colDataX=[0,5,10,15,5,20,0,5,10,30],
colDataY=[1,2,3,4,5,6,7,8,9,0],
colRowLabels=[0.2,0.3,0.5,0.6,0.0,0.1,0.2,0.1,0.8,0.0])
matSource=convert(Matrix,DataFrame(dfSource[1:2:end,1:2]))'
arLabels=dfSource[1:2:end,3]
datGet=matSource[:,(arLabels.>=0.2) .& (arLabels.<0.7)][1,:]
print(datGet)
output> [0,10,0]
Note the use of parenthetical enclosures (arLabels.>=0.2) and (arLabels<0.7), as well as the use of the .>= and .< syntax (which forces Julia to iterate through a container/collection). Finally, and most crucially (since it's the part most people miss), note the use of .& in place of just &. The dot operator makes all the difference!
Anytime you combine two flags or more, you use |. Except Swift, which changed it to use array syntax. But now I am staring at some documentation on local notifications that does this:
requestAuthorizationWithOptions: (UNAuthorizationOptionAlert + UNAuthorizationOptionSound)
Where those are enums. I can't find where the + operator suddenly started working in this manner. I don't see any indicators that | couldn't just be used instead. I realize that maybe the enums themselves could just happen to work by the nature of their bits and how they add, but I doubt my last statement.
So, to codify the discussion in comments above: no, + is not a replacement for |. a + b == a | b if and only if a and b have bitpatterns which share no overlapping bits; in the cases of option enums, this is often true (because it is useful to represent options in this way), but not guaranteed. This is the case that you see above, though good code will be a bit more explicit (and readable) by using bitwise or.
I'm trying to figure out if there's a way in mathematica where I can solve for particular variables when given other variables and a set of equations. Essentially there are 6 variables, and I'm given 3 of them and have to calculate the others using these equations-
Variables-
B,Qs,f0,R,c,L
Equations-
f0=1/(2*Pi*Sqrt[L*c])
Qs=(w*L)/R
w=2*Pi*f0
B=f0/Qs
We are given the values of any 3 of those variables and have to figure out the rest using those values.
I was thinking perhaps using Eliminate but I'm not sure exactly how that would be structured as I've only used it previously with set variables that don't change and a single output.
When using the Solve function with Mathematica, you can specify for what variables you want Solve to specify the solutions. Note that Solve may not be able to find expressions in terms of these variables (if the equations you give it are contradictory or insufficient) or for all values as some functions have no inverse or only partial inverses.
Your question looks a lot like homework in Electromagnetics, but here is an example with your original problem. You will have to adapt these ideas to give to Solve the set of variables you are looking for. Also remember to use == to specify equality testing. A simple = is for immediate assignment to a variable.
Solve[{f0 == 1/(2*Pi*Sqrt[L*c]), Qs == (w*L)/R, w == 2*Pi*f0, B == f0/Qs}, {f0, B, c}]
{{f0->w/(2 [Pi]), B->w/(2 [Pi] Qs), c->L/(Qs^2 R^2)}}
I am working with Wolfram Mathematica 8 and have the following problem. I have an optimization problem under certain constraints and want to have an analytical (symbolical solution). I am maximizing function piA. My input is:
piA[a_, WA1_, WA0_] =
a/(1 + a)*(X - (y*WA1 + 1)^(1/y)) - 1/(1 + a) ((y*WA0 + 1)^(1/y));
Maximize[{piA[a, WA1, WA0], WA0 >= -1/y, WA1 >= -1/y}, WA0]
What I get most of the times is:
Maximize[{-((1 + WA0 y)^((1/y))/(1 + a)) + (
a (X - (1 + WA1 y)^(1/y)))/(1 + a), WA0 >= -(1/y), WA1 >= -(1/y)},a]
Basically, the command does nothing, but outputs itself. Only once I have managed to get the proper output (too long to paste here). I have tested it with simpler functions and it works. Unfortunately, I cannot understand what causes the problem. It is not a syntax problem, since it has worked like that several times. Any help would be very much appreciated.
P.S. Just checked again and my input ALWAYS generates the wrong output. The time it generated the solution was when I accidentally set parameters X and y to certain numbers.
The most likely reason is that given the function and constraints, Mathematica doesn't know how to maximize your function with respect to WA0. Note you also have a free variables X and a in there, and it might not have enough information about the domain of X and a to be able to properly form a solution to your equation.
I've had instances where I tried feeding in some equations and constraints and Mathematica simply couldn't do anything with them because they were too general. This may be the case here as well. Is there a specific problem you're trying to solve, and is there any way you could give Mathematica more context?
I don't think this is a bug at all, but it's unfortunate that sometimes Mathematica will just spit back your input when it doesn't have any rules for solving what you gave it.
The usual reason these things happens seems to be when the expressions given are too general for Mathematica to handle, or when it it's faced with a set of expressions that are ill formed.
Just as an example, I tried passing in fractions into a function I wrote that specifically looked for rational expressions, thinking it would work. It turned out that it needed to handle both Rational[a, b] and Times[a, Power[b, -1]]. It could be the case that Mathematica is not expecting a constraint to be of the form GreaterEqual[a, b].
Mathematica returns an answer if you assign the variable a some value. Maybe you could build your strategy on that? In fact it does provide an answer if you assign a value to any of the variables.
( I would need more background of the problem to go from there... )