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Generalise heterogeneous tuple of variable-length MVector's to a variable number of MVector's

By : Murali Narayanan
Date : September 02 2020, 01:00 PM
fixed the issue. Will look into that further If you don't need the "variadic type arguments" N1...Nk explicitely, you can use a tuple of Vararg of some UnionAll types:
code :
julia> MyType3{T} = Tuple{Vararg{MVector{M, <:T} where M}}
Tuple{Vararg{MArray{Tuple{M},#s14,1,M} where #s14<:T where M,N} where N} where T

julia> (MVector{2}([1,2]), MVector{3}([1,2,3])) isa MyType3{Int}

julia> (MVector{2}([1,2]), MVector{3}([1,2,3])) isa MyType3{Number}

julia> (MVector{2}([1,2]), MVector{3}([1,2,3])) isa MyType3{String}

julia> (MVector{2}([1,2]), MVector{3}([1,2,3])) isa MyType3

julia> (MVector{2}(["a", "b"]), MVector{3}([1,2,3])) isa MyType3
julia> struct MyType7{T, Ns<:Tuple, D}
           function MyType7(vecs::Vararg{MVector{M, T} where M}) where {T}
               Ns = [typeof(v).parameters[4] for v in vecs]
               new{T, Tuple{Ns...}, typeof(vecs)}(vecs)

julia> MyType7(MVector{2}([1,2]), MVector{3}([1,2,3]))
MyType7{Int64,Tuple{2,3},Tuple{MArray{Tuple{2},Int64,1,2},MArray{Tuple{3},Int64,1,3}}}(([1, 2], [1, 2, 3]))

julia> typeof(ans)

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How can I find all of the permutations consisting of 1 element from a variable number of arrays of variable length?

By : Siddharth Sharma
Date : March 29 2020, 07:55 AM
To fix this issue If you're using python, this is part of the standard library: itertools.product. But assuming you're not, here's a pseudocode version.
code :
// Create an initialised array of indexes.
int[] index0(arrays) {
    // We require all arrays to be non-empty.
    for a in arrays {
        assert len(a) != 0;
    return new int[len(arrays)];

// Increment the indices. Returns false when the indices wrap round to the start.
bool next_index(indices, arrays) {
    for (i = len(indices) - 1; i >= 0; --i) {
        indices[i] += 1
        if indices[i] < len(arrays[i]) {
            return true;
        indices[i] = 0;
    return false;
indices = index0(arrays); 
    for (i = 0; i < len(arrays); ++i) {
        print arrays[i][indices[i]];
} while next_index(indices);

Excel - Variable number of leading zeros in variable length numbers?

By : Ze Yad
Date : March 29 2020, 07:55 AM
hope this fix your issue Assuming the data is not corrupt in the database, then try and export from the database to a csv or text file.
The following can then be done to ensure the import is formatted correctly

Ambiguous type variable in length of heterogeneous list

By : Simonliii
Date : March 29 2020, 07:55 AM
With these it helps The problem isn't with HLength, and it's already as polymorphic as it can be. The issue is with HIndex, which is unnecessarily specific in the xs parameter.
From HIndex xs 'Zero Bool we should be able to infer that xs has Bool ': xs' shape for some xs'. Since the HIndex implementation is driven by the Nat class parameter, we can leave the other parameters unspecified in the instance heads, and instead refine them in the instance constraints, enabling GHC to do the mentioned inference:
code :
class HIndex xs i y | xs i -> y where
    hIndex :: HList xs -> HNat i -> y

instance (xs ~ (x ': xs')) => HIndex xs Zero x where
    hIndex (x ::: xs) HZero = x

instance (xs ~ (y ': xs'), HIndex xs' i x) => HIndex xs (Succ i) x where
    hIndex (x ::: xs) (HSucc i) = hIndex xs i
test1 :: (Eq a, Num a) => HList '[HList (Bool : xs') -> a -> Bool] 
> :t hLength test1
hLength test1 :: HNat ('Succ 'Zero)

How to implement a variable-length heterogeneous sequence of structs

By : Revaz Liluashvili
Date : March 29 2020, 07:55 AM
I wish this help you There's 2 structs of different sizes: , I would suggest using union here, to wrap both structs in one:
code :
 union Data {
      struct A A;
      struct B B;
 enum DataType {
      TYPE_A = 0,
 struct DataHolder {
      union Data Data;
      enum DataType DataType;

mypy: How to "convert" variable length tuple back to fixed length tuple?

By : Monty McMont
Date : March 29 2020, 07:55 AM
Does that help You can use cast to get around this.
code :
from typing import cast, Tuple

t = (1,2,3)

t = cast(Tuple[int, int, int], tuple(x+1 for x in t))
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