I'm trying to construct the identity matrix in Julia 1.1. After looking at the documentation I found that I could compute a 4x4 Identity matrix as follows:
julia> Id4 =1* Matrix(I, 4, 4)
4×4 Array{Int64,2}:
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
Is this the most julianic way of coding it or is there a better/shorter way, as it is an often used matrix?
Given using LinearAlgebra, the most julianic way of expressing the identity matrix is:
I
This answer may seem trite, but it is also kind of profound. The whole point of the operator I is that in the vast majority of cases where users want an identity matrix, it is not necessary to actually instantiate that matrix.
Let's say you want a 1000x1000 identity matrix. Why waste time building the entire matrix, when you could just use I, noting that sizeof(I) evaluates to 1 (ie the size of the object is 1 byte). All functions in base Julia (including LinearAlgebra) understand what I is, and can use it appropriately without having to waste time building the actual matrix it represents first.
Now, it may be the case that for some reason you need to specify the type of the elements of your identity matrix. Note:
julia> I
UniformScaling{Bool}
true*I
so in this case, you are using a notional identity matrix with a diagonal of true and off-diagonal of false. This is sufficient in many cases, even if your other matrices are Int or Float64. Internally, Julia will use methods that specialize on the types. However, if you want to specify your identity matrix to contain integers or floats, use:
julia> 1I
UniformScaling{Int64}
1*I
julia> 1.0I
UniformScaling{Float64}
1.0*I
Note that sizeof(1I) evaluates to 8, indicating the notional Int64 types of the members of that matrix.
Also note that you can use e.g. 5I if you want a notional matrix with 5 on the diagonal and 0 elsewhere.
In some cases (and these cases are much rarer than many might think), you may need to actually build the matrix. In this case, you can use e.g.:
Matrix(1I, 3, 3) # Identity matrix of Int type
Matrix(1.0I, 3, 3) # Identity matrix of Float64 type
Matrix(I, 3, 3) # Identity matrix of Bool type
Bogumił has also pointed out in the comments that if you are uncomfortable with implying the type of the output in the first argument of the constructors above, you can also use the (slightly more verbose):
Matrix{Int}(I, 3, 3) # Identity matrix of Int type
Matrix{Float64}(I, 3, 3) # Identity matrix of Float64 type
Matrix{Bool}(I, 3, 3) # Identity matrix of Bool type
and specify the type explicitly.
But really, the only times you would probably need to do this are as follows:
When you want to input an identity matrix into a function in a package written in such a way that the input must be a concrete matrix type.
When you want to start out with an identity matrix but then mutate it in place into something else via one or several transformations.
Related
Let's say I have a vector V, and I want to either turn this vector into multiple m x n matrices, or get multiple m x n matrices from this Vector V.
For the most basic example: Turn V = collect(1:75) into 3 5x5 matrices.
As far as I am aware this can be done by first using reshape reshape(V, 5, :) and then looping through it. Is there a better way in Julia without using a loop?
If possible, a solution that can easily change between row-major and column-major results is preferrable.
TL:DR
m, n, n_matrices = 4, 2, 5
V = collect(1:m*n*n_matrices)
V = reshape(V, m, n, :)
V = permutedims(V, [2,1,3])
display(V)
From my limited knowledge about Julia:
When doing V = collect(1:m*n), you initialize a contiguous array in memory. From V you wish to create a container of m by n matrices. You can achieve this by doing reshape(V, m, n, :), then you can access the first matrix with V[:,:,1]. The "container" in this case is just another array (thus you have a three dimensional array), which in this case we interpret as "an array of matrices" (but you could also interpret it as a box). You can then transpose every matrix in your array by swapping the first two dimensions like this: permutedims(V, [2,1,3]).
How this works
From what I understand; n-dimensional arrays in Julia are contiguous arrays in memory when you don't do any "skipping" (e.g. V[1:2:end]). For example the 2 x 4 matrix A:
1 3 5 7
2 4 6 8
is in memory just 1 2 3 4 5 6 7 8. You simply interpret the data in a specific way, where the first two numbers makes up the first column, then the second two numbers makes the next column so on so forth. The reshape function simply specifies how you want to interpret the data in memory. So if we did reshape(A, 4, 2) we basically interpret the numbers in memory as "the first four values makes the first column, the second four values makes the second column", and we would get:
1 5
2 6
3 7
4 8
We are basically doing the same thing here, but with an extra dimension.
From my observations it also seems to be that permutedims in this case reallocates memory. Also, feel free to correct me if I am wrong.
Old answer:
I don't know much about Julia, but in Python using NumPy I would have done something like this:
reshape(V, :, m, n)
EDIT: As #BatWannaBe states, the result is technically one array (but three dimensional). You can always interpret a three dimensional array as a container of 2D arrays, which from my understanding is what you ask for.
The SymTridiagonal data type in Julia is not letting me assign non-diagonal values to anything other than zero. I get this error: ArgumentError: cannot set off-diagonal entry (2, 1).
I need to assign non-diagonal values because I am trying to implement the ImplicitSymmetricQRStep algorithm which needs to do that in the process.
It is indeed not possible to set the off diagonal values of SymTridiagonal matrix - why this decision was taken I cannot say.
I see now two alternatives:
1) In Julia the fields of a structure are not hidden, so it is possible to change the value that way. This is dangerous though, as the internal structure of that matrix might change in future versions without any warnings. Here is an example of how you would do that:
using LinearAlgebra: SymTridiagonal
a = SymTridiagonal([1 2 0; 2 1 2; 0 2 1)] # 1 on diagonal, 2 on off diagonals
a.ev[1] = 4 # a[1, 2] == 4 and a[2, 1] == 4
2) You could also use the Tridiagonal matrix type, that is also in the LinearAlgebra package; this type allows one to set the off diagonal entries. Then you just have to make sure yourself that you don't violate the symmetric properties of that matrix i.e if you set a[i, j] then you also have to set a[j, i] to the same value.
Per DICOM specification, a UID is defined by: 9.1 UID Encoding Rules. In other words the following are valid DICOM UIDs:
"1.2.3.4.5"
"1.3.6.1.4.35045.103501438824148998807202626810206788999"
"1.2.826.0.1.3680043.2.1143.5028470438645158236649541857909059554"
while the following are illegal DICOM UIDs:
".1.2.3.4.5"
"1..2.3.4.5"
"1.2.3.4.5."
"1.2.3.4.05"
"12345"
"1.2.826.0.1.3680043.2.1143.50284704386451582366495418579090595540"
Therefore I know that the string is at most 64 bytes, and should match the following regex [0-9\.]+. However this regex is really a superset, since there are a lot less than (10+1)^64 (=4457915684525902395869512133369841539490161434991526715513934826241L) possibilities.
How would one computes precisely the number of possibilities to respect the DICOM UID rules ?
Reading the org root / suffix rule clearly indicates that I need at least one dot ('.'). In which case the combination is at least 3 bytes (char) in the form: [0-9].[0-9]. In which case there are 10x10=100 possibilities for UID of length 3.
Looking at the first answer, there seems to be something unclear about:
The first digit of each component shall not be zero unless the
component is a single digit.
What this means is that:
"0.0" is valid
"00.0" or "1.01" are not valid
Thus I would say a proper expression would be:
(([1-9][0-9]*)|0)(\.([1-9][0-9]*|0))+
Using a simple C code, I could find:
f(0) = 0
f(1) = 0
f(2) = 0
f(3) = 100
f(4) = 1800
f(5) = 27100
f(6) = 369000
f(7) = 4753000
f(8) = 59049000
The validation of the Root UID part is outside the scope of this question. A second validation step could take care of rejecting some OID that cannot possibly be registered (some people mention restriction on first and second arc for example). For simplicity we'll accept all possible (valid) Root UID.
While my other answer takes good care of this specific application, here is a more generic approach. It takes care of situations where you have a different regular expression describing the language in question. It also allows for considerably longer string lengths, since it only requires O(log n) arithmetic operations to compute the number of combinations for strings of length up to n. In this case the number of strings grows so quickly that the cost of these arithmetic operations will grow dramatically, but that may not be the case for other, otherwise similar situations.
Build a finite state automaton
Start with a regular expression description of your language in question. Translate that regular expression into a finite state automaton. In your case the regular expression can be given as
(([1-9][0-9]*)|0)(\.([1-9][0-9]*|0))+
The automaton could look like this:
Eliminate ε-transitions
This automaton usually contains ε-transitions (i.e. state transitions which do not correspond to any input character). Remove those, so that one transition corresponds to one character of input. Then add an ε-transition to the accepting state(s). If the accepting states have other outgoing transitions, don't add ε-loops to them, but instead add an ε-transition to an accepting state with no outgoing edges and then add the loop to that. This can be seen as padding the input with ε at its end, without allowing ε in the middle. Taken together, this transformation ensures that performing exactly n state transitions corresponds to processing an input of n characters or less. The modified automaton might look like this:
Note that both the construction of the first automaton from the regular expression and the elimination of ε-transitions can be performed automatically (and perhaps even in a single step. The resulting automata might be more complicated than what I constructed here manually, but the principle is the same.
Ensuring unique paths
You don't have to make the automaton deterministic in the sense that for every combination of source state and input character there is only one target state. That's not the case in my manually constructed one either. But you have to make sure that every complete input has only one possible path to the accepting state, since you'll essentially be counting paths. Making the automaton deterministic would ensure this weaker property, too, so go for that unless you can ensure unique paths without this. In my example the length of each component clearly dictates which path to use, so I didn't make it deterministic. But I've included an example with a deterministic approach at the end of this post.
Build transition matrix
Next, write down the transition matrix. Associate the rows and columns with your states (in order a, b, c, d, e, f in my example). For each arrow in your automaton, write the number of characters included in the label of that arrow in the column associated with the source state and the row associated with the target state of that arrow.
⎛ 0 0 0 0 0 0⎞
⎜ 9 10 0 0 0 0⎟
⎜10 10 0 10 10 0⎟
⎜ 0 0 1 0 0 0⎟
⎜ 0 0 0 9 10 0⎟
⎝ 0 0 0 10 10 1⎠
Read result off that matrix
Now applying this matrix with a column vector once has the following meaning: if the number of possible ways to arrive in a given state is encoded in the input vector, the output vector gives you the number of ways one transition later. Take the 64th power of that matrix, concentrate on the first column (since ste start situation is encoded as (1,0,0,0,0,0), meaning only one way to end up in the start state) and sum up all the entries that correspond to accepting states (only the last one in this case). The bottom left element of the 64th power of this matrix is
1474472506836676237371358967075549167865631190000000000000000000000
which confirms my other answer.
Compute matrix powers efficiently
In order to actually compute the 64th power of that matrix, the easiest approach would be repeated squaring: after squaring the matrix 6 times you have an exponent of 26 = 64. If in some other scenario your exponent (i.e. maximal string length) is not a power of two, you can still perform exponentiation by squaring by multiplying the relevant squares according to the bit pattern of the exponent. This is what makes this approach take O(log n) arithmetic operations to compute the result for string length n, assuming a fixed number of states and therefore fixed cost for each matrix squaring.
Example with deterministic automaton
If you were to make my automaton deterministic using the usual powerset construction, you'd end up with
and sorting the states as a, bc, c, d, cf, cef, f one would get the transition matrix
⎛ 0 0 0 0 0 0 0⎞
⎜ 9 10 0 0 0 0 0⎟
⎜ 1 0 0 0 0 0 0⎟
⎜ 0 1 1 0 1 1 0⎟
⎜ 0 0 0 1 0 0 0⎟
⎜ 0 0 0 9 0 10 0⎟
⎝ 0 0 0 0 1 1 1⎠
and could sum the last three elements of the first column of its 64th power to obtain the same result as above.
Single component
Start by looking for ways to form a single component. The corresponding regular expression for a single component is
0|[1-9][0-9]*
so it is either zero or a non-zero digit followed by arbitrary many zero digits. (I had missed the possible sole zero case at first, but the comment by malat made me aware of this.) If the total length of such a component is to be n, and you write h(n) to denote the number of ways to form such a component of length exactly n, then you can compute that h(n) as
h(n) = if n = 1 then 10 else 9 * 10^(n - 1)
where the n = 1 case allows for all possible digits, and the other cases ensure a non-zero first digit.
One or more components
Subsection 9.1 only writes that a UID is a bunch of dot-separated number components, as outlined above. So in regular expressions that would be
(0|[1-9][0-9]*)(\.(0|[1-9][0-9]*))*
Suppose f(n) is the number of ways to write a UID of length n. Then you have
f(n) = h(n) + sum h(i) * f(n-i-1) for i from 1 to n-2
The first term describes the case of a single component, while the sum takes care of the case where it consists of more than one component. In that case you have a first component of length i, then a dot which accounts for the -1 in the formula, and then the remaining digits form one or more components which is expressed via the recursive use of f.
Two or more components
As the comment by cneller indicates, the part of section 9 before subsection 9.1 indicates that there has to be at least two components. So the proper regular expression would be more like
(0|[1-9][0-9]*)(\.(0|[1-9][0-9]*))+
with a + at the end indicating that we want at least one repetition of the parenthesized expression. Deriving an expression for this simply means leaving out the one-component-only case in the definition of f:
g(n) = sum h(i) * f(n-i-1) for i from 1 to n-2
If you sum all the g(n) for n from 3 (the minimal possible UID length) through 64 you get the number of possible UIDs as
1474472506836676237371358967075549167865631190000000000000000000000
or approximately 1.5e66. Which is considerably less than the 4.5e66 you get from your computation, in terms of absolute difference, although it's definitely on the same order of magnitude. By the way, your estimate doesn't explicitely mention UIDs shorter than 64, but you can always consider padding them with dots in your setup. I did the computation using a few lines of Python code:
f = [0]
g = [0]
h = [0, 10] + [9 * (10**(n-1)) for n in range(2, 65)]
s = 0
for n in range(1, 65):
x = 0
if n >= 3:
for i in range(1, n - 1):
x += h[i] * f[n-i-1]
g.append(x)
f.append(x + h[n])
s += x
print(h)
print(f)
print(g)
print(s)
Is there any kind of object class for piecewise / noncontiguous ranges in Julia? For instance, I can create a regular range:
a = UnitRange(1:5)
But, if I wanted to combine this with other ranges:
b = UnitRange([1:5, 8:10, 4:7])
I cannot currently find an object or method. There is a PiecewiseIncreasingRanges module (https://github.com/simonster/PiecewiseIncreasingRanges.jl) that would be just what I want in this situation, except that it, as the name implies, requires the ranges be monotonically increasing.
The context for this is that I am looking for a way to create a compressed, memory efficient version of the SparseMatrixCSC type for sparse matrices with repeating rows. The RLEVectors module will work well to save space on the nonzerovalue vector in the sparse matrix class. Now though I am trying to find something to save space for the rowvalue vector that also defines the sparse matrix, since series of repeating rows will result in ranges of values in that vector (e.g. if the first 10 rows, or even certain columns in the first ten rows, of a sparse matrix are identical, then there will be a lot of 1:10 patterns in the row value vector).
More generally, I'd like a range such as the b object that I try to create above over which I could do an iterated loop, getting:
for (idx, item) in enumerate(hypothetical_object)
println("idx: $idx, item: $item")
end
idx: 1, item: 1
idx: 2, item: 2
...
idx: 5, item: 5
idx: 6, item: 8
idx: 7, item: 9
idx: 8, item: 10
idx: 9, item: 4
idx: 10, item: 5
...
Update: One thing I'm considering, and will probably try implementing if I don't hear other suggestions here, will be to just create an array of PiecewiseIncreasingRange objects, one for each column in my sparse matrix. (I would probably also then break the nonzero value vector into an array of separate pieces, one for each column of my sparse matrix as well). This would at least be relatively simple to implement. I don't have a good sense off the bat how this would compare in terms of computational efficiency to the kind of object I am searching for in this question. I suspect that memory requirements would be about the same.
To loop over a sequence of ranges (or other iterators), you can use the chain function in the Iterators.jl package.
For example:
using Iterators
b = chain(1:5, 8:10, 4:7)
for i in b
println(i)
end
outputs the elements of each range.
Integers can be used to store individual numbers, but not mathematical expressions. For example, lets say I have the expression:
6x^2 + 5x + 3
How would I store the polynomial? I could create my own object, but I don't see how I could represent the polynomial through member data. I do not want to create a function to evaluate a passed in argument because I do not only need to evaluate it, but also need to manipulate the expression.
Is a vector my only option or is there a more apt solution?
A simple yet inefficient way would be to store it as a list of coefficients. For example, the polynomial in the question would look like this:
[6, 5, 3]
If a term is missing, place a zero in its place. For instance, the polynomial 2x^3 - 4x + 7 would be represented like this:
[2, 0, -4, 7]
The degree of the polynomial is given by the length of the list minus one. This representation has one serious disadvantage: for sparse polynomials, the list will contain a lot of zeros.
A more reasonable representation of the term list of a sparse polynomial is as a list of the nonzero terms, where each term is a list containing the order of the term and the coefficient for that order; the degree of the polynomial is given by the order of the first term. For example, the polynomial x^100+2x^2+1 would be represented by this list:
[[100, 1], [2, 2], [0, 1]]
As an example of how useful this representation is, the book SICP builds a simple but very effective symbolic algebra system using the second representation for polynomials described above.
A list is not the only option.
You can use a map (dictionary) mapping the exponent to the corresponding coefficient.
Using a map, your example would be
{2: 6, 1: 5, 0: 3}
A list of (coefficient, exponent) pairs is quite standard. If you know your polynomial is dense, that is, all the exponent positions are small integers in the range 0 to some small maximum exponent, you can use the array, as I see Óscar Lopez just posted. :)
You can represent expressions as Expression Trees. See for example .NET Expression Trees.
This allows for much more complex expressions than simple polynomials and those expressions can also use multiple variables.
In .NET you can manipulate the expression tree as a tree AND you can evaluate it as a function.
Expression<Func<double,double>> polynomial = x => (x * x + 2 * x - 1);
double result = polynomial.Compile()(23.0);
An object-oriented approach would say that a Polynomial is a collection of Monomials, and a Monomial encapsulates a coefficient and exponent together.
This approach works when when you have a polynomial like this:
y(x) = x^1000 + 1
An approach that tied a data structure to a polynomial order would be terribly wasteful for this pathological case.
You need to store two things:
The degree of your polynomial (e.g. "3")
A list containing each coefficient (e.g. "{3, 0, 2}")
In standard C++, "std::vector<>" and "std::list<>" can do both.
Vector/array is obvious choice. Depending on type of expressions you may consider some sort of sparse vector type (custom made, i.e. based on dictionary or even linked list if you expressions have 2-3 non-zero coefficients 5x^100+x ).
In either case exposing through custom class/interface would be beneficial as you can replace implementation later. You would likely want to provide standard operations (+, -, *, equals) if you plan to write a lot of expression manipulation code.
Just store the coefficients in an array or vector. For example, in C++ if you are only using integer coefficients, you could use std::vector<int>, or for real numbers, std::vector<double>. Then you just push the coefficients in order and access them by variable exponent number.
For example (again in C++), to store 5*x^3 + 9*x - 2 you might do:
std::vector<int> poly;
poly.push_back(-2); // x^0, acceesed with poly[0]
poly.push_back(9); // x^1, accessed with poly[1]
poly.push_back(0); // x^2, etc
poly.push_back(5); // x^3, etc
If you have large, sparse, polynomials, then maybe you'd want to use a map instead of a vector. If you have fixed sized lengths, then you'd perhaps use an fixed length array instead of a vector.
I've used C++ for examples, but this same scheme can be used in any language.
You can also transform it into reverse Polish notation:
6x^2 + 5x + 3 -> x 2 ^ 6 * x 5 * + 3 +
Where x and numbers are "pushed" onto a stack and operations (^,*,+) take the two top-most values from the stack and replace them with the result of the operation. In the end you get the resultant value on the stack.
In this form it's easy to calculate arbitrarily complex expressions.
This representation is also close to tree representation of expressions where non-leaf tree nodes represent operations and functions and leaf nodes are for constants and variables.
What's good about trees is that you can also easily evaluate expressions and you can also do things like symbolic differentiation on them. Both have recursive nature.