I want to create a CoffeeScript range (like [4...496]) but using a length instead of an end range. This can be done with a loop like
myNum = getBigNumber()
newArray = ( n + myNum for n in [0...50] )
but I'm wondering if there is range-related shortcut that I'm missing. Is there something like
[getBigNumber()...].length(50) available in CoffeeScript?
You can just do
range = [myNum...myNum + 50]
Edit: As mu points out in the comments, CoffeeScript will add some complexity whether you use the snippet above or the original code. If performance is an issue, it might be better to drop down to plain JS for the loop (using backticks in the CoffeeScript code).
Assuming you want an ascending (i.e. low to high) range, you can do:
myNum = getBigNumber()
length = 50
range = new Array length
i = 0
`for(; i < length ; i++) { range[i] = i + myNum }` # raw, escaped JS
It's a lot faster than CoffeeScript's way of doing things, but note that CoffeeScript's range syntax also supports creating descending ranges by just flipping the boundary values. So CoffeeScript is (as always) easier on the eyes and simpler to work with, but raw JS is 3.5x faster in my test.
Related
Consider the following JavaScript function, which performs a computation over several lines to clearly indicate the programmer's intent:
function computation(first, second) {
const a = first * first;
const b = second - 4;
const c = a + b;
return c;
}
computation(12, 3)
//143
computation(-3, 2.6)
//7.6
I have tried using do notation to solve this with PureScript but I seem to be just short of understanding some key concept. The do notation examples in the documentation only covers do notation when the value being bound is an array (https://book.purescript.org/chapter4.html#do-notation), but in my example I would like the values to be simple values of the Int or Number type.
While it is possible to perform this computation in one line, it makes the code harder to debug and does not scale to many operations.
How would the computation method be written correctly in PureScript so that...
If computation involved 1000 intermediate steps, instead of 3, the code would not suffer from excessive indenting but would be as readable as possible
Each step of the computation is on its own line, so that, for example, the code could be reviewed line by line by a supervisor, etc., for quality
You don't need the do notation. The do notation is intended for computations happening in a monad, whereas your computation is naked.
To define some intermediate values before returning result, use the let .. in construct:
computation first second =
let a = first * first
b = second - 4
c = a + b
in c
But if you really want to use do, you can do that as well: it also supports naked computations just to give you some choice. The difference is that within a do you can have multiple lets on the same level (and they work the same as one let with multiple definitions) and you don't need an in:
computation first second = do
let a = first * first -- first let
b = second - 4
let c = a + b -- second let
c -- no in
I'm trying to implement the 'Sport Scheduling Problem' (with a Round-Robin approach to break symmetries). The actual problem is of no importance. I simply want to declare the value at x[1,1] to be the set {1,2} and base the sets in the same column upon the first set. This is modelled as in the code below. The output is included in a screenshot below it. The problem is that the first set is not printed as a set but rather some sort of range while the values at x[2,1] and x[3,1] are indeed printed as sets and x[4,1] again as a range. Why is this? I assume that in the declaration of x that set of 1..n is treated as an integer but if it is not, how to declare it as integers?
EDIT: ONLY the first column of the output is of importance.
int: n = 8;
int: nw = n-1;
int: np = n div 2;
array[1..np, 1..nw] of var set of 1..n: x;
% BEGIN FIX FIRST WEEK $
constraint(
x[1,1] = {1, 2}
);
constraint(
forall(t in 2..np) (x[t,1] = {t+1, n+2-t} )
);
solve satisfy;
output[
"\(x[p,w])" ++ if w == nw then "\n" else "\t" endif | p in 1..np, w in 1..nw
]
Backend solver: Gecode
(Here's a summarize of my comments above.)
The range syntax is simply a shorthand for contiguous values in a set: 1..8 is a shorthand of the set {1,2,3,4,5,6,7,8}, and 5..6 is a shorthand for the set {5,6}.
The reason for this shorthand is probably since it's often - and arguably - easier to read the shorthand version than the full list, especially if it's a long list of integers, e.g. 1..1024. It also save space in the output of solutions.
For the two set versions, e.g. {1,2}, this explicit enumeration might be clearer to read than 1..2, though I tend to prefer the shorthand version in all cases.
I have a table in Matlab with some columns representing 128 bit hashes.
I would like to match rows, to one or more rows, based on these hashes.
Currently, the hashes are represented as hexadecimal strings, and compared with strcmp(). Still, it takes many seconds to process the table.
What is the fastest way to compare two hashes in matlab?
I have tried turning them into categorical variables, but that is much slower. Matlab as far as I know does not have a 128 bit numerical type. nominal and ordinal types are deprecated.
Are there any others that could work?
The code below is analogous to what I am doing:
nodetype = { 'type1'; 'type2'; 'type1'; 'type2' };
hash = {'d285e87940fb9383ec5e983041f8d7a6'; 'd285e87940fb9383ec5e983041f8d7a6'; 'ec9add3cf0f67f443d5820708adc0485'; '5dbdfa232b5b61c8b1e8c698a64e1cc9' };
entries = table(categorical(nodetype),hash,'VariableNames',{'type','hash'});
%nodes to match. filter by type or some other way so rows don't match to
%themselves.
A = entries(entries.type=='type1',:);
B = entries(entries.type=='type2',:);
%pick a node/row with a hash to find all counterparts of
row_to_match_in_A = A(1,:);
matching_rows_in_B = B(strcmp(B.hash,row_to_match_in_A.hash),:);
% do stuff with matching rows...
disp(matching_rows_in_B);
The hash strings are faithful representations of what I am using, but they are not necessarily read or stored as strings in the original source. They are just converted for this purpose because its the fastest way to do the comparison.
Optimization is nice, if you need it. Try it out yourself and measure the performance gain for relevant test cases.
Some suggestions:
Sorted arrays are easier/faster to search
Matlab's default numbers are double, but you can also construct integers. Why not use 2 uint64's instead of the 128bit column? First search for the upper 64bit, then for the lower; or even better: use ismember with the row option and put your hashes in rows:
A = uint64([0 0;
0 1;
1 0;
1 1;
2 0;
2 1]);
srch = uint64([1 1;
0 1]);
[ismatch, loc] = ismember(srch, A, 'rows')
> loc =
4
2
Look into the compare functions you use (eg edit ismember) and strip out unnecessary operations (eg sort) and safety checks that you know in advance won't pose a problem. Like this solution does. Or if you intend do call a search function multiple times, sort in advance and skip the check/sort in the search function later on.
I would like to have short-hand form about many parameters which I just need to keep fixed in Matlab 2016a because I need them in many places, causing many errors in managing them separately.
Code where the signal is 15x60x3 in dimensions
signal( 1:1 + windowWidth/4, 1:1 + windowWidth,: );
Its pseudocode
videoParams = 1:1 + windowWidth/4, 1:1 + windowWidth,: ;
signal( videoParams );
where you cannot write videoParams as string but should I think write ":" as string and everything else as integers.
There should be some way to do the pseudocode.
Output of 1:size(signal,3) is 3 so it gives 1:3. I do not get it how this would replace : in the pseudocode.
Extension for horcler's code as function
function videoParams = fix(k, windowWidth)
videoParams = {k:k + windowWidth/4, k:k + windowWidth};
end
Test call signal( fix(1,windowWidth){:}, : ) but still unsuccessful giving the error
()-indexing must appear last in an index expression.
so I am not sure if such a function is possible.
How can you make such a int-string-int input for the matrix?
This can be accomplished via comma-separated lists:
signal = rand(15,60,3); % Create random data
windowWidth = 2;
videoParams = {1:1+windowWidth/4, 1:1+windowWidth, 1:size(signal,3)};
Then use the comma-separated list as such:
signal(videoParams{:})
which is equivalent to
signal(1:1+windowWidth/4, 1:1+windowWidth, 1:size(signal,3))
or
signal(1:1+windowWidth/4, 1:1+windowWidth, :)
The colon operator by itself is shorthand for the entirety of a dimension. However, it is only applicable in a direct context. The following is meaningless (and invalid code) as the enclosing cell has no defined size for its third element:
videoParams = {1:1+windowWidth/4, 1:1+windowWidth, :};
To work around this, you could of course use:
videoParams = {1:1+windowWidth/4, 1:1+windowWidth};
signal(videoParams{:},:)
I've been given the task of translating a piece of MATLAB code into IDL and have
hit a roadblock when I came across the MATLAB function accumarry(). The
function, described here
is used to sum elements in one array, based on indices given in another. Example
1 perhaps explains this better than the actual function description at the top
of the page. In trying to reproduce Example 1 in IDL, I haven't been able to avoid a for loop, but I'm confident that it is possible. My best attempt is the following:
vals = [101,102,103,104,105]
subs = [0,1,3,1,3]
n = max(subs)+1
accum = make_array(n)
for i = 0, n-1 do begin
wVals = where(subs eq i,count)
accum[i] = count eq 0 ? 0 : total(vals[wVals])
endfor
print,accum
; 101.000 206.000 0.00000 208.000
Any advice on improving this would be greatly appreciated! I expected IDL to have a similar built-in function, but haven't been able to track one down. Perhaps some magic with histogram binning?
I found a number of possible solutions to this problem on Coyote's IDL site (not surprisingly.)
http://www.idlcoyote.com/code_tips/drizzling.html
I ended up using the following, as a compromise between performance and readability:
function accumarray,data,subs
mx = max(subs)
accum = fltarr(mx+1)
h = histogram(subs,reverse_indices=ri,OMIN=om)
for j=0L,n_elements(h)-1 do if ri[j+1] gt ri[j] then $
accum[j+om] = total(vals[ri[ri[j]:ri[j+1]-1]])
return,accum