I am making a custom widget in Shopify Dashing with the following code:
<div class="main">
<div id="bar">
Info
</div>
<div id="summary" data-bind="summary"></div>
<div id="info" data-bind="info"></div>
<h2 data-bind="more"></h3>
</div>
When the "Info" tag is clicked it is supposed to toggle the views shown in the widget. This works fine the first time it is clicked and the function toggleDetails is called as expected. When it is clicked a second or third time however nothing happens, the function is never called.
Why is this? I really don't understand.
This is my coffescript class:
1 class Dashing.Toggle extends Dashing.Widget
2
3 ready: ->
4 # This is fired when the widget is done being rendered
5 ##showDetails = false
6 #showSummary = true
7 #initiateView()
8
9 onData: (data) ->
10 # Handle incoming data
11 # You can access the html node of this widget with `#node`
12 # Example: $(#node).fadeOut().fadeIn() will make the node flash each time data comes in.
13
14
15 #accessor 'toggleDetails', ->
16 #switchView()
17 console.log "toggling 1"
18
19 initiateView: ->
20 #summaryView().show()
21 #infoView().hide()
22
23 switchView: ->
24 console.log "Switching"
25 if #showSummary
26 #summaryView().fadeOut #animationLength()
27 #infoView().fadeIn #animationLength()
28 #showSummary = false
29 else
30 #infoView().fadeOut #animationLength()
31 #summaryView().fadeIn #animationLength()
32 #showSummary = true
33
34 summaryView: ->
35 console.log "getting summary view"
36 $(#node).find("#summary")
37
38 infoView: ->
39 console.log "getting info view"
40 $(#node).find("#info")
41
42 animationLength: ->
43 1000
Ok I just solved it myself :P I still don't know why but changing form calling the #accessor method toggleDetails to switchView worked wonders. Does anyone know why?
Related
I have a function that takes as input some of the values in a table and returns a tuple if you will - three separate return values, which I want to transpose into the output of a query. Here's a simplified example of what I want to achieve:
multiplier:{(x*2;x*3;x*3)};
select twoX:multiplier[price][0]; threeX:multiplier[price][1]; fourX:multiplier[price][2] from data;
The above basically works (I think I've got the syntax right for the simplified example - if not then hopefully my intention is clear), but is inefficient because I'm calling the function three times and throwing away most of the output each time. I want to rewrite the query to only call the function once, and I'm struggling.
Update
I think I missed a crucial piece of information in my explanation of the problem which affects the outcome - I need to get other data in the query alongside the output of my function. Here's a hopefully more realistic example:
multiplier:{(x*2;x*3;x*4)};
select average:avg price, total:sum price, twoX:multiplier[sum price][0]; threeX:multiplier[sum price][1]; fourX:multiplier[sum price][2] by category from data;
I'll have a go at adapting your answers to fit this requirement anyway, and apologies for missing this bit of information. The real function if a proprietary and fairly complex algorithm and the real query has about 30 output columns, hence the attempt at simplifying the example :)
If you're just looking for the results themselves you can extract (exec) as lists, create dictionary and then flip the dictionary into a table:
q)exec flip`twoX`threeX`fourX!multiplier[price] from ([]price:til 10)
twoX threeX fourX
-----------------
0 0 0
2 3 4
4 6 8
6 9 12
8 12 16
10 15 20
12 18 24
14 21 28
16 24 32
18 27 36
If you need other columns from the original table too then its trickier but you could join the tables sideways using ,'
q)t:([]price:til 10)
q)t,'exec flip`twoX`threeX`fourX!multiplier[price] from t
An apply # can also achieve what you want. Here data is just a table with 10 random prices. # is then used to apply the multiplier function to the price column while also assigning a column name to each of the three resulting lists:
q)data:([] price:10?100)
q)multiplier:{(x*2;x*3;x*3)}
q)#[data;`twoX`threeX`fourX;:;multiplier data`price]
price twoX threeX fourX
-----------------------
80 160 240 240
24 48 72 72
41 82 123 123
0 0 0 0
81 162 243 243
10 20 30 30
36 72 108 108
36 72 108 108
16 32 48 48
17 34 51 51
for a=1:50; %numbers 1 through 50
for b=1:50;
c=sqrt(a^2+b^2);
if c<=50&c(rem(c,1)==0);%if display only if c<=50 and c=c/1 has remainder of 0
pyth=[a,b,c];%pythagorean matrix
disp(pyth)
else c(rem(c,1)~=0);%if remainder doesn't equal to 0, omit output
end
end
end
answer=
3 4 5
4 3 5
5 12 13
6 8 10
7 24 25
8 6 10
8 15 17
9 12 15
9 40 41
10 24 26
12 5 13
12 9 15
12 16 20
12 35 37
14 48 50
15 8 17
15 20 25
15 36 39
16 12 20
16 30 34
18 24 30
20 15 25
20 21 29
21 20 29
21 28 35
24 7 25
24 10 26
24 18 30
24 32 40
27 36 45
28 21 35
30 16 34
30 40 50
32 24 40
35 12 37
36 15 39
36 27 45
40 9 41
40 30 50
48 14 50
This problem involves the Pythagorean theorem but we cannot use the built in function so I had to write one myself. The problem is for example columns 1 & 2 from the first two rows have the same numbers. How do I code it so it only deletes one of the rows if the columns 1 and 2 have the same number combination? I've tried unique function but it doesn't really delete the combinations. I have read about deleting duplicates from previous posts but those have confused me even more. Any help on how to go about this problem will help me immensely!
Thank you
welcome to StackOverflow.
The problem in your code seems to be, that pyth only contains 3 values, [a, b, c]. The unique() funcion used in the next line has no effect in that case, because only one row is contained in pyth. another issue is, that the values idx and out are calculated in each loop cycle. This should be placed after the loops. An example code could look like this:
pyth = zeros(0,3);
for a=1:50
for b=1:50
c = sqrt(a^2 + b^2);
if c<=50 && rem(c,1)==0
abc_sorted = sort([a,b,c]);
pyth = [pyth; abc_sorted];
end
end
end
% do final sorting outside of the loop
[~,idx] = unique(pyth, 'rows', 'stable');
out = pyth(idx,:);
disp(out)
a few other tips for writing MATLAB code:
You do not need to end for or if/else stements with a semicolon
else statements cover any other case not included before, so they do not need a condition.
Some performance reommendations:
Due to the symmetry of a and b (a^2 + b^2 = b^2 + a^2) the b loop could be constrained to for b=1:a, which would roughly save you half of the loop cycles.
if you use && for contencation of scalar values, the second part is not evaluated, if the first part already fails (source).
Regards,
Chris
You can also linearize your algorithm (but we're still using bruteforce):
[X,Y] = meshgrid(1:50,1:50); %generate all the combination
C = (X(:).^2+Y(:).^2).^0.5; %sums of two square for every combination
ind = find(rem(C,1)==0 & C<=50); %get the index
res = unique([sort([X(ind),Y(ind)],2),C(ind)],'rows'); %check for uniqueness
Now you could really optimized your algorithm using math, you should read this question. It will be useful if n>>50.
Hi All Fortran Lovers,
I am trying to write to a file which outputs three variables as
program main
integer N, u
parameter(u=20)
open (u, FILE='points.dat', STATUS='new')
do 10 i= 1, 100
write(u,100) i, i*2, i*5
10 continue
100 format (I5, I10, 9X, I10)
close(u)
print *,'COMPLETE!!'
end
Which Gives output (points.dat stripped file content):
1 2 5
2 4 10
3 6 15
4 8 20
5 10 25
6 12 30
7 14 35
8 16 40
9 18 45
10 20 50
11 22 55
12 24 60
...
...
...
...
...
99 198 495
100 200 500
|(This line added by the write statement)
But I want something like this:
1 2 5
2 4 10
3 6 15
4 8 20
5 10 25
6 12 30
7 14 35
8 16 40
9 18 45
10 20 50
11 22 55
12 24 60
...
...
...
...
...
99 198 495
100 200 500|(The cursor stop here)
i.e. No space at start of each line. The last line stops after printing '500'
I tried using Horizontal spacing using '1X' specifier but no success.
Add advance='no' in write statement. If the line is not the last one, write EOL:
do 10 i= 1, 100
write(u,100,advance='no') i, i*2, i*5
if (i.ne.100) write(u,*)
10 continue
Edit: I see it now, it seems that the fortran program will add EOL to the end of file anyway. Then you have to use external programs to truncate your file, see for example https://www.quora.com/How-do-I-chop-off-just-the-last-byte-of-a-file-in-Bash .
If you have the next values:
30
24
21
26
23
19
17
11
6
10
How is it possible to do a peak2peak starting from 4 (thus value 26) untill the last value? So in this case the answer should be 20.
Thanks!
You can simply use only the values starting from 4th value of your vector:
a = [30,24,21,26,23,19,17,11,6,10];
peak2peak(a(4:end))
I'm fairly sure there should be an elegant solution to this (in MATLAB), but I just can't think of it right now.
I have a list with [classIndex, start, end], and I want to collapse consecutive class indices into one group like so:
This
1 1 40
2 46 53
2 55 55
2 57 64
2 67 67
3 68 91
1 94 107
Should turn into this
1 1 40
2 46 67
3 68 91
1 94 107
How do I do that?
EDIT
Never mind, I think I got it - it's almost like fmarc's solution, but gets the indices right
a=[ 1 1 40
2 46 53
2 55 55
2 57 64
2 67 67
3 68 91
1 94 107];
d = diff(a(:,1));
startIdx = logical([1;d]);
endIdx = logical([d;1]);
b = [a(startIdx,1),a(startIdx,2),a(endIdx,3)];
Here is one solution:
Ad = find([1; diff(A(:,1))]~=0);
output = A(Ad,:);
output(:,3) = A([Ad(2:end)-1; Ad(end)],3);
clear Ad
One way to do it if the column in question is numeric:
Build the differences along the id-column. Consecutive identical items will have zero here:
diffind = diff(a(:,1)');
Use that to index your array, using logical indexing.
b = a([true [diffind~=0]],:);
Since the first item is always included and the difference vector starts with the difference from first to second element, we need to prepend one true value to the list.