I'm trying to plot the same shape from the 15 min onto the Daily as well. This is the code to plot a shape on the 15 min which works fine;
if crossover(s3K,s3D) and s3K<25 and (s4K-s4D<3 and s4K-s4D>-3) and s4K<35//or s4D-s4K>0 and s4D-s4K<1 and s4K<50 and s1K<40
rwCross:=true
plotshape(rwCross, style = shape.arrowup, location = location.belowbar, color=color.yellow, size=size.small)
But to plot it on the daily i've tried;
rwCrossDaily = security(syminfo.tickerid,'D', rwCross)
plotshape(rwCrossDaily, style = shape.arrowup, location = location.belowbar, color=color.yellow, size=size.small)
Which gives me the mutable variable error. So i tried using a function to get around it;
rwCross_func() =>
if crossover(s3K,s3D) and s3K<25 and (s4K-s4D<3 and s4K-s4D>-3) and s4K<35//or s4D-s4K>0 and s4D-s4K<1 and s4K<50 and s1K<40
rwCross:=true
rwCrossDaily = security(syminfo.tickerid,'D', rwCross_func())
But now it tells me I 'Cannot modify global variable 'rwCross' in function.'
Help please!
Best solution and cleanest solution here is to just make a bool out of your condition in its simplest form:
rwCross = crossover(s3K,s3D) and s3K<25 and (s4K-s4D<3 and s4K-s4D>-3) and s4K<35//or s4D-s4K>0 and s4D-s4K<1 and s4K<50 and s1K<40
rwCross will naturally become true without the if. This way we do not need to have anything mutable, although there are more solutions for situations that we must...
Cheers!
Related
I'm trying to do some re-factoring on my charts to make them re-usable using this as a guide: http://bost.ocks.org/mike/chart/
I'm having problems drawing the lines in my multi-line graph though - specifically passing the data to the x and y values. If I hard code the element names it works, but if I try to use the xValue and yValue objects this does not work. I'm assuming that this is because I'm trying to call a function within the parameter of an other object, but I'm not sure how to get around this. In the exmaple Mike uses d[0] and d[1], but this won't work with JSON data (or I'm not sure how to make it work).
I've posted this JSFiddle so you can see the code. The problem lines are 125 to 131 which in turn is being called from line 165.
var main_line = d3.svg.line()
.interpolate("cardinal")
// Hard coding the elements works
//.x(function(d) { return main_x(d.date); })
//.y(function(d) { return main_y(d.buildFixTime); });
// Passing xValue and yValue does not work
.x(function(d) { return main_x(xValue); })
.y(function(d) { return main_y(yValue); });
http://jsfiddle.net/goodspeedj/fDyLY/
Thank you in advance.
You need to redefine your accessor method within .x() and .y(). The accessor method defines the way that a datum is pulled out of the data that is bound to the selection that you call the line generator on.
Suppose you have a relatively flat data structure such as the following.
data = [{x : 1, y : 2}, {x:1, y:3}, {x:4, y:5}];
You then bind the data to a selection with the following statement
d3.select("body").datum(data).append("path").attr("d",lineGenerator);
Quite a bit is going on underneath this statement. I'll give you a bit more of a walkthrough after showing you a commonly used example.
The important aspect to understand is that similarly to other calls in d3 such as
var exampleRectangles = d3.select("body")
.data(data).enter()
.append("rect")
.attr("width",2)
.attr("height", 3)
.attr("x",function(datum){return datum.x}) // pay attention to this line
.attr("y",0);
d3 is implicitly iterating over each element in your data. For each datum in your data array, in this case there is a total of three datum, you are going to add a rectangle to the dom.
In the line that I tell you to pay attention to you notice that you're defining an anonymous (unnamed) function. What is that datum parameter coming from? It's implicitly being passed to your anonymous function.
So each rectangle has it's own corresponding datum {x : 1, y : 2}, {x:1, y:3}, {x:4, y:5} respectively. Each rectangle's x coordinate is defined by the respective datum.x attribute. Under the sheets, d3 is implicitly looping over the data array that you've defined. A similar approach to the example d3 code could be written as above.
for (var i = 0; i < data.length; i++)
{
d3.select("body").append("rect")
.attr("width",2)
.attr("height", 3)
.attr("x",data[i].x)
.attr("y",0);
}
This follows from the notion of data driven documents (d3). For each item added (a rectangle in the above example a piece of data is tied to it. In the above example you see that there is something kind of similar to your .x() and .y() accessor functions :
.attr("x",function(datum){return datum.x})
This function is telling d3 how to filter over the total datum that's being passed to the .attr() accessor method.
So, you need to determine which data you need to get a hold of to make your .attr("d", lineGenerator)call make sense. The difference between your.datum(data)call and the typical.data(data)call is that instead of parceling the data that's being passed to.data(data)`, the whole array is given as a single piece of data to the line generator function (similar to main_line(data), wherein it will again implicitly loop over the points to construct your path.
So, what you need to do is determine what a single datum will be defined as for your function to operate on.
I'm not going to define that as I don't seem to know quite which information you are operating on, but I would hazard a guess at something like.
.x(xAccessor)
.y(yAccessor)
function xAccessor(datum)
{
return xScale(datum._id.month);
}
function yAccessor(datum)
{
return yScale(datum.buildFixTime);
}
The way you have it set up, xValue and yValue are functions; you have to actually execute them on something to get a value back.
.x(function(d) { return main_x( xValue(d) ); })
.y(function(d) { return main_y( yValue(d) ); });
If you weren't using a scale, you could use
.x(xValue)
.y(yValue);
but only because if you pass in a function d3 executes it for you with the data as a parameter. And that only works for d3 methods that expect functions as possible input -- the scale functions expect data values as input.
I wrote a long piece work for another user last week that you may find useful, explaining methods that accept functions as parameters.
i am just not able to read a csv file .I want to display a graph for it.
I am getting error:
TypeError: n is undefined
please help me out!!
d3.csv("example.csv", function(dataset){
var svg=d3.select("body").append("svg").attr("width",w).attr("height",h);
var xScale=d3.scale.ordinal().domain(d3.range(dataset.length)).rangeRoundBands([0,w],0.05);
var yScale=d3.scale.linear().domain([0,d3.max(dataset.value)]).range([0,h]);
svg.selectAll("rect").data(dataset).enter().append("rect").attr({x:function(d,i) {return xScale(i);}, y:function(d){
return h-yScale(d);}, width:xScale.rangeBand(),height:function(d){return yScale(d);},fill:function(d){return "rgb(0,0,"+(d.value*10)+")";}});
d3.select("svg").selectAll("text").data(dataset).enter().append("text").text(function(d) {return d.value;}).attr("x",function(d,i){
return xScale(i)+xScale.rangeBand()/2;}).attr("y",function(d){return h-yScale(d)+14;}).attr("font-family","sans-serif").attr("font-size","10px").attr
("fill","white").attr("text-anchor","middle");
d3.select("p").on("click",function(){
var numValues=dataset.length;
dataset=[];
for(var i=0;i<numValues;i++)
{var newNumber=Math.floor(Math.random()*25);
dataset.push(newNumber,newNumber);
}
yScale.domain([0,d3.max(dataset)]);
svg.selectAll("rect").data(dataset).transition().delay(function(d,i){return i/dataset.length*1000;})
.duration(500).attr("y",function(d){return h-yScale(d); }).attr("height",function(d) {return yScale(d);}).attr("fill",function(d){
return "rgb(0,0,"+(d.value*10)+")";});
svg.selectAll("text").data(dataset).transition().delay(function(d,i){return i/dataset.length*1000;}).duration(500).text(function(d){return d.value;})
.attr("x",function(d,i){return xScale(i)+xScale.rangeBand()/2;}).attr("y",function(d) {return h-yScale(d)+14;})
.attr("font-family","sans-serif").attr("font- size","10px").attr("fill","white").attr("text-anchor","middle")
;});
});
here is my csv file
names,value
john,78
brad,105
amber,103
james,2
dean,74
pat,45
matt,6
andrew,18
ashley,15
As mentioned by cuckovic there are quite a few errors in your code, not to mention that you seem to be using different styles to achieve similar things which is confusing. Anyway there are 3 fundamental things that are casuing you trouble, the first is your dataset. The value column of the csv is being read as a string. You need to convert it to a number by:
dataset.forEach(function (d,i) {
d.value = +d.value;
});
The next issue you have is the yScale where you have set the range to .range([0,h]);. This is the wrong way round for an svg viewport in which the y direction starts from top and increases towards the bottom. So you need to swap the 0 and h around in the range.
The next thing to address is the difference between d and d.value. When you bind data through the data() operator you are generally binding an array. In this case it is an array of objects. So after you have bound the data, d refers to each element of that array which in this case is an object containing a name and a value. This can be seen if you console.log your dataset. When d is passed to yScale it doesn't know what to do with it as it is not a number, what you really want to do is to pass d.value to yScale. So replace your d's with d.value.
Finally, the last part of your code starting at d3.select("p") does not seem to add anything.
I'd recommend reading Scott Murray's tutorials, particularly this one if you haven't already.
As a homework assignment, I'm writing a code that uses the bisection method to calculate the root of a function with one variable within a range. I created a user function that does the calculations, but one of the inputs of the function is supposed to be "fun" which is supposed to be set equal to the function.
Here is my code, before I go on:
function [ Ts ] = BisectionRoot( fun,a,b,TolMax )
%This function finds the value of Ts by finding the root of a given function within a given range to a given
%tolerance, using the Bisection Method.
Fa = fun(a);
Fb = fun(b);
if Fa * Fb > 0
disp('Error: The function has no roots in between the given bounds')
else
xNS = (a + b)/2;
toli = abs((b-a)/2);
FxNS = fun(xns);
if FxNS == 0
Ts = xNS;
break
end
if toli , TolMax
Ts = xNS;
break
end
if fun(a) * FxNS < 0
b = xNS;
else
a = xNS;
end
end
Ts
end
The input arguments are defined by our teacher, so I can't mess with them. We're supposed to set those variables in the command window before running the function. That way, we can use the program later on for other things. (Even though I think fzero() can be used to do this)
My problem is that I'm not sure how to set fun to something, and then use that in a way that I can do fun(a) or fun(b). In our book they do something they call defining f(x) as an anonymous function. They do this for an example problem:
F = # (x) 8-4.5*(x-sin(x))
But when I try doing that, I get the error, Error: Unexpected MATLAB operator.
If you guys want to try running the program to test your solutions before posting (hopefully my program works!) you can use these variables from an example in the book:
fun = 8 - 4.5*(x - sin(x))
a = 2
b = 3
TolMax = .001
The answer the get in the book for using those is 2.430664.
I'm sure the answer to this is incredibly easy and straightforward, but for some reason, I can't find a way to do it! Thank you for your help.
To get you going, it looks like your example is missing some syntax. Instead of either of these (from your question):
fun = 8 - 4.5*(x - sin(x)) % Missing function handle declaration symbol "#"
F = # (x) 8-4.5*(x-sin9(x)) %Unless you have defined it, there is no function "sin9"
Use
fun = #(x) 8 - 4.5*(x - sin(x))
Then you would call your function like this:
fun = #(x) 8 - 4.5*(x - sin(x));
a = 2;
b = 3;
TolMax = .001;
root = BisectionRoot( fun,a,b,TolMax );
To debug (which you will need to do), use the debugger.
The command dbstop if error stops execution and opens the file at the point of the problem, letting you examine the variable values and function stack.
Clicking on the "-" marks in the editor creates a break point, forcing the function to pause execution at that point, again so that you can examine the contents. Note that you can step through the code line by line using the debug buttons at the top of the editor.
dbquit quits debug mode
dbclear all clears all break points
I have a function in matlab with something like this:
function [ out ] = myFunc(arg1, arg2)
times = [];
for i = 1:arg1
tic
% do some long calculations
times = [times; toc];
end
% Return
out = times;
end
I want to abort the running function now but keep the values of times which are currently already taken. How to do it? When I press strg+c, I simply loose it because it's only a local function variable which is deleted when the function leaves the scope...
Thanks!
Simplest solution would be to turn it from a function to a script, where times would no longer be a local variable.
The more elegant solution would be to save the times variable to a .mat file within the loop. Depending on the time per iteration, you could do this on every loop, or once every ten loops, etc.
Couldn't you use persistent variables to solve your problem, e.g.
function [ out ] = myFunc(arg1, arg2)
persistent times
if nargin == 0
out = times;
return;
end;
times = [];
for i = 1:arg1
tic
% do some long calculations
times = [times; toc];
end
% Return
out = times;
end
I'm not sure whether persistent variables are cleared upon Ctrl-C, but I don't think it should be the case. What this should do: if you supply arguments, it will run as before. When you omit all arguments however, the last value of times should be returned.
onCleanup functions still fire in the presence of CTRL-C, however I don't think that's really going to help because it will be hard for you to connect the value you want to the onCleanup function handle (there are some tricky variable lifetime issues here). You may have more luck using a MATLAB handle object to track your value. For example
x = containers.Map(); x('Value') = [];
myFcn(x); % updates x('Value')
% CTRL-C
x('Value') % contains latest value
Another possible solution is to use the assignin function to send the data to your workspace on each iteration. e.g.
function [ out ] = myFunc(arg1, arg2)
times = [];
for i = 1:arg1
tic
% do some long calculations
times = [times; toc];
% copy the variable to the base workspace
assignin('base', 'thelasttimes', times)
end
% Return
out = times;
end
I'm teaching myself Octave and as a motivational exercise am attempting to create some Bode plots. I'd like to create a plot that has multiple curves for different values of a parameter in a transfer function, for example the time constant of a simple RC filter. I'm trying to do it as follows:
tau = [1,2,3]
for i = tau
g(i) = tf(1,[tau(i),1])
endfor
bode(g(1),g(2),g(3))
But it doesn't work, I get the error
error: octave_base_value::imag (): wrong type argument `struct'
However, it works fine if there are not multiple arguments to the bode command and the last line is simply:
bode(g(1))
Any advice as to where I've gone wrong would be appreciated - is there a better way to do what I want to do?
I was able to do it with the following sequence (with octave 3.2.4 on debian):
bode(g(1))
set (findobj (gcf, "type", "axes"), "nextplot", "add")
bode(g(2))
bode(g(3))
The second command is similar to hold on but it works when there are subplots; I found it here.
Using your own code:
subplot(211), hold on
subplot(212), hold on
tau = [1,2,3]
for i = 1:length(tau),
g(i) = tf(1,[tau(i),1]);
bode(g(i))
endfor
The problem with this solution is that you cannot identify a specific plot. You cannot access figure properties through bode() function directly.
Here then a plausible solution to bring you colorful plots:
colorsplot = ["b","m","g"]
tau = [1,2,3]
g = tf(1,[tau(1),1]);
[mag, ph, w] = bode(g);
subplot(211), semilogx(w,20*log(mag)), hold on
subplot(212), semilogx(w,ph), hold on
for i = 2:length(tau),
g = tf(1,[tau(i),1]);
[mag, ph, waux] = bode(g,w);
subplot(211), semilogx(w,20*log(mag),colorsplot(i))
subplot(212), semilogx(w,ph,colorsplot(i))
endfor