# String of Markdown in DB
beforeMark = #content
# Render string of markdown to html string
afterMark = marked(beforeMark)
# Parse the html to HTML to extract 0, 2, 4th children node (elements)
finalMark = $.parseHTML(afterMark)
# Help needed HERE
# Get 0 2 4th elements if they exist.
# ex) if array has 4 keys, return 0, 2th
# if array has 7 keys, return 0, 2, 4th
# if array has 3 keys, return 0, 2th
# if array has 1 key, return 0th
stringMark = $(finalMark[0]).prop('outerHTML') + $(finalMark[2]).prop('outerHTML') + $(finalMark[2]).prop('outerHTML')
I have the above coffeescript I wrote to truncate a markdown string into html of 3 elements.
I need the last part to be more efficient and proper so that it returns the 0, 2, 4th keys of arrays but only if they exist.
I am new to coffee and I need help!!
If I understand the problem correctly, I believe you want something like this:
stringMark = ''
for data, index in finalMark when index in [0, 2, 4]
stringMark += data.prop 'outerHTML'
Or if you like a little code golf:
stringMark = (v.prop 'outerHTML' for v, i in finalMark when i in [0, 2, 4]).join ''
Related
I'm trying to solve this problem:
There is a grid with with r rows and c columns. A robot sitting in top left cell can only move in 2 directions, right and down. But certain cells have to be avoided and the robot cannot step on them. Find a path for the robot from the top left to the bottom right.
The problem specifically asks for a single path, and that seems straight forward:
Having the grid as boolean[][], the pseudocode I have is
List<String> path = new ArrayList<String>()
boolean found = false
void getPath(r, c){
if (!found) {
if ( (r or c is outofbounds) || (!grid[r][c]) )
return
if (r==0 AND c==0) // we reached
found = true
getPath(r-1, c)
getPath(r, c-1)
String cell = "(" + r + ", " + c + ")"
path.add(cell)
}
}
Though I was wondering how can I get all the possible paths (NOT just the count, but the path values as well). Note that it has r rows and c columns, so its not a nxn grid. I'm trying to think of a DP/recursive solution but unable to come up with any and stuck. It's hard to think when the recursion goes in two ways.
Any pointers? And also any general help on how to "think" about such problems would be appreciated :).
Any pointers? And also any general help on how to "think" about such problems would be appreciated :).
Approach to the problem:
Mentally construct graph G of the problem. In this case the vertices are cells in the grid and directed edges are created where a valid robot move exist.
Search for properties of G. In this case G is a DAG (Directed Acyclic Graph).
Use such properties to come up with a solution. In this case (G is a DAG) its common to use topological sort and dynamic programming to find the amount of valid paths.
Actually you don't need to construct the graph since the set of edges is pretty clear or to do topological sort as usual iteration of the matrix (incremental row index and incremental column index) is a topological sort of this implicit graph.
The dynamic programming part can be solved by storing in each cell [x][y] the amount of valid paths from [0][0] to [x][y] and checking where to move next.
Recurrence:
After computations the answer is stored in dp[n - 1][m - 1] where n is amount of rows and m is amount of columns. Overall runtime is O(nm).
How about find all possible valid paths:
Usual backtracking works and we can speed it up by applying early pruning. In fact, if we calculate dp matrix and then we do backtracking from cell [n - 1][m - 1] we can avoid invalid paths as soon the robot enters at a cell whose dp value is zero.
Python code with dp matrix calculated beforehand:
n, m = 3, 4
bad = [[False, False, False, False],
[ True, True, False, False],
[False, False, False, False]]
dp = [[1, 1, 1, 1],
[0, 0, 1, 2],
[0, 0, 1, 3]]
paths = []
curpath = []
def getPath(r, c):
if dp[r][c] == 0 or r < 0 or c < 0:
return
curpath.append((r, c))
if r == 0 and c == 0:
paths.append(list(reversed(curpath)))
getPath(r - 1, c)
getPath(r, c - 1)
curpath.pop()
getPath(n - 1, m - 1)
print(paths)
# valid paths are [[(0, 0), (0, 1), (0, 2), (0, 3), (1, 3), (2, 3)],
# [(0, 0), (0, 1), (0, 2), (1, 2), (1, 3), (2, 3)],
# [(0, 0), (0, 1), (0, 2), (1, 2), (2, 2), (2, 3)]]
Notice that is very similar to your code, there is a need to store all valid paths together and take care that appended lists are a copy of curpath to avoid ending up with an list of empty lists.
Runtime: O((n + m) * (amount of valid paths)) since simulated robot moves belong to valid paths or first step into an invalid path detected using foresight (dp). Warning: This method is exponential as amount of valid paths can be .
I need to make a nested loop with an arbitrary depth. Recursive loops seem the right way, but I don't know how to use the loop variables in side the loop. For example, once I specify the depth to 3, it should work like
count = 1
for i=1, Nmax-2
for j=i+1, Nmax-1
for k=j+1,Nmax
function(i,j,k,0,0,0,0....) // a function having Nmax arguments
count += 1
end
end
end
I want to make a subroutine which takes the depth of the loops as an argument.
UPDATE:
I implemented the scheme proposed by Zoltan. I wrote it in python for simplicity.
count = 0;
def f(CurrentDepth, ArgSoFar, MaxDepth, Nmax):
global count;
if CurrentDepth > MaxDepth:
count += 1;
print count, ArgSoFar;
else:
if CurrentDepth == 1:
for i in range(1, Nmax + 2 - MaxDepth):
NewArgs = ArgSoFar;
NewArgs[1-1] = i;
f(2, NewArgs, MaxDepth, Nmax);
else:
for i in range(ArgSoFar[CurrentDepth-1-1] + 1, Nmax + CurrentDepth - MaxDepth +1):
NewArgs = ArgSoFar;
NewArgs[CurrentDepth-1] = i;
f(CurrentDepth + 1, NewArgs, MaxDepth, Nmax);
f(1,[0,0,0,0,0],3,5)
and the results are
1 [1, 2, 3, 0, 0]
2 [1, 2, 4, 0, 0]
3 [1, 2, 5, 0, 0]
4 [1, 3, 4, 0, 0]
5 [1, 3, 5, 0, 0]
6 [1, 4, 5, 0, 0]
7 [2, 3, 4, 0, 0]
8 [2, 3, 5, 0, 0]
9 [2, 4, 5, 0, 0]
10 [3, 4, 5, 0, 0]
There may be a better way to do this, but so far this one works fine. It seems easy to do this in fortran. Thank you so much for your help!!!
Here's one way you could do what you want. This is pseudo-code, I haven't written enough to compile and test it but you should get the picture.
Define a function, let's call it fun1 which takes inter alia an integer array argument, perhaps like this
<type> function fun1(indices, other_arguments)
integer, dimension(:), intent(in) :: indices
...
which you might call like this
fun1([4,5,6],...)
and the interpretation of this is that the function is to use a loop-nest 3 levels deep like this:
do ix = 1,4
do jx = 1,5
do kx = 1,6
...
Of course, you can't write a loop nest whose depth is determined at run-time (not in Fortran anyway) so you would flatten this into a single loop along the lines of
do ix = 1, product(indices)
If you need the values of the individual indices inside the loop you'll need to unflatten the linearised index. Note that all you are doing is writing the code to transform array indices from N-D into 1-D and vice versa; this is what the compiler does for you when you can specify the rank of an array at compile time. If the inner loops aren't to run over the whole range of the indices you'll have to do something more complicated, careful coding required but not difficult.
Depending on what you are actually trying to do this may or may not be either a good or even satisfactory approach. If you are trying to write a function to compute a value at each element in an array whose rank is not known when you write the function then the preceding suggestion is dead flat wrong, in this case you would want to write an elemental function. Update your question if you want further information.
you can define your function to have a List argument, which is initially empty
void f(int num,List argumentsSoFar){
// call f() for num+1..Nmax
for(i = num+1 ; i < Nmax ; i++){
List newArgs=argumentsSoFar.clone();
newArgs.add(i);
f(i,newArgs);
}
if (num+1==Nmax){
// do the work with your argument list...i think you wanted to arrive here ;)
}
}
caveat: the stack should be able to handle Nmax depth function calls
Yet another way to achieve what you desire is based on the answer by High Performance Mark, but can be made more general:
subroutine nestedLoop(indicesIn)
! Input indices, of arbitrary rank
integer,dimension(:),intent(in) :: indicesIn
! Internal indices, here set to length 5 for brevity, but set as many as you'd like
integer,dimension(5) :: indices = 0
integer :: i1,i2,i3,i4,i5
indices(1:size(indicesIn)) = indicesIn
do i1 = 0,indices(1)
do i2 = 0,indices(2)
do i3 = 0,indices(3)
do i4 = 0,indices(4)
do i5 = 0,indices(5)
! Do calculations here:
! myFunc(i1,i2,i3,i4,i5)
enddo
enddo
enddo
enddo
enddo
endsubroutine nestedLoop
You now have nested loops explicitly coded, but these are 1-trip loops unless otherwise desired. Note that if you intend to construct arrays of rank that depends on the nested loop depth, you can go up to rank of 7, or 15 if you have a compiler that supports it (Fortran 2008). You can now try:
call nestedLoop([1])
call nestedLoop([2,3])
call nestedLoop([1,2,3,2,1])
You can modify this routine to your liking and desired applicability, add exception handling etc.
From an OOP approach, each loop could be represented by a "Loop" object - this object would have the ability to be constructed while containing another instance of itself. You could then theoretically nest these as deep as you need to.
Loop1 would execute Loop2 would execute Loop3.. and onwards.
What I am trying to accomplish, is to be able to put some values inside an array, then based on a t (0-1), get a value out of the array based on its stored values.
To make this more clear, here's an example:
Array values = [0, 10]
Now this array would return value 0 for t=1 and value 10 for t=1. So t=.3 will give a value of 3.
Another example:
Array values = [10, 5, 5, 35]
t=.25 will give a value of 5
t=.125 will give a value of 7.5
Im looking for the most efficient formula to get the value at any given t using a given array.
Currently I'm using this (pseudo code)
var t:Number = .25;
var values:Array = [10, 5, 5, 35];
if(t == 1) value = [values.length-1];
else
var offset:Number = 1/values.length;
var startIndex:int = int(t/offset);
var fraction:Number = t % offset;
var roundPart:Number = (values[startIndex+1] - values[startIndex]) * fraction;
var value:Number = values[startIndex] + roundPart;
But i'm sure there's a far more better way of doing this. So i'm calling for the mathematicians on here!
Here is a One Liner in Mathematica. It's doing the same thing you are, only slightly more compact.
Arrays indexes start at 1.
values = {10, 5, 5, 35, 0}
f[a_, x_] := a[[k = IntegerPart[(k1 = (Dimensions[a][[1]] - 2) x)] + 1]] +
FractionalPart[k1] (a[[k + 1]] - a[[k]])
So your interpolation result on:
In[198]:= f[values,1]
Out[198]= 35
Etc.
If you plot changing the x scale:
I am working in classic ASP; using getRows to get multidimension array of rows and column.
while iterating a row; I want to pass that single row into another function to build the column layout.
with C# I can do this:
obj[][] multiDimArray = FunctionCall_To_InitializeArray_4X16();
for (int rowId = 0 ; rowId < 4 ; rowId++)
{
FunctionCall_to_ProcessSingleRow(multiDimArray[rowId][]);
//this function only accept single dimension array
}
How can I do this is asp classic/vbscript:
1. I have a function that accept single dimension array as parameter.
2. Call that function and pass 1 part of 2 dimension array.
Thank you
I think you will need to populate a new array or dictionary object with the single dimension you want to process.
here a piece from working code, should get you going..
aResults = oRst.Getrows
oRst.Close
Set oRst = Nothing
Call SubCloseDatabaseOracle
iRows = UBound(aResults, 2)
iCols = UBound(aResults, 1)
row = 1 'first row
line = ""
separator = ""
FOR col = 0 TO iCols
line = line & separator & cStr(aResults(col,row))
separator = ";"
NEXT
aSingleDimensionArray = split(line,";")
Is there a standard library function or built-in construct to concatenate two sequences in JavaFX?
Here a Sequences.concatenate() function is mentioned, but it is nowhere to be seen in the official API.
Of course one could iterate over each sequence, inserting the values into a new sequence e.g:
function concatenate(seqA: Object[], seqB: Object[]) : Object[] {
for(b in seqB) insert b into seqA;
seqA;
}
..but surely something as basic as concatenation is already defined for us somewhere..
It is very simple, since there cannot be sequence in sequence (it all gets flattened), you can do it like this:
var a = [1, 2];
var b = [3, 4];
// just insert one into another
insert b into a;
// a == [1, 2, 3, 4];
// or create a new seq
a = [b, a];
// a == [3, 4, 1, 2];
Hope that helps.