In Simulink, I have a signal that is a 1-D array.
I want to get the index of the first value other than 0, but the result received is the index of the last non-zero value
I am new to Simulink, is there any way to break FOR loop?
Or what should I do in this case
Any hint will be great.
Thanks.
Rather than using a for loop iterator subsystem, use the while iterator subsystem.
From the properties you can enable the "Show iteration number port" option to give you the index to increment over the items in the array.
The while iterator also has a "cond" or condition input which you can set to exit the while loop when you get to a non-zero value.
Alternatively, use the for loop but work from the end of the array towards the front and output the last non-zero value which will be the first non-zero value if you worked in a forwards direction.
Use the length minus the index value to access the array to work backwards.
Related
Good day,
I have a set in my program, say nums
{5,6,7}->nums
I have a for loop
For i,1,dim(nums)
EndFor
How would I get the ith element from the set? For example, print 5,6,7 from the for loop?
I've tried calling it like a function, as I've seen in many guides, but that gives an error.
Thanks!
:facepalm:
Somehow I didn't even try brackets. For those with the same issue, it is just
nums[i]
if(arr.indexOf(element))
if(arr.indexOf(element) !== -1)
if(arr.indexOf(element) > 0)
I thought that when I use indexOf() in an if statement, those if statements above would be the same, but I got a different result than I expected.
Is there anyone who knows the difference among these?
Assuming this is a JS question, All three are different.
There is no special logic for indexOf within an if statement. It is the same logic as ever. You just need to understand how indexOf() evaluates. The first checks if the index is truthy or falsey, so ONLY the first index would not qualify and all other indexes would pass. The second checks that it is not -1, anything would pass except values not in the array, and the third checks if the index of the element is greater than 0 so all pass except the first index and any element not in the index.
I am new to Jmeter and trying to do a while loop operation with a condition. So please someone provide solution for the below query.
Query: I am trying to do DELETE request for 50 times using the id as reference. So I kept the condition as "${startId}<=${endId}" in the while loop. But the while loop is executing infinitely. Is there any simple mechanism to iterate the loop for 50 times by increment the startId till it reaches endId.
While Controller accepts function or variable. So you need to either:
provide a variable which has value of "true" and becomes "false" somewhere else
provide a function which returns "false" to exit from While loop.
With your condition it won't evaluate your expression hence it will never become "false". The solution is to wrap your statement into i.e. __javaScript function as:
${__javaScript(${startId}<=${endId},)}
For more information on JMeter functions see How to Use JMeter Functions post series.
You may take help of loop controller and pre-processor.
snaps :
I am using
rand(200)
in my Rails application.
When I run it in console it always returns random number, but if I use it in application line:
index = rand(200)
index is always the same number.
Why is that and how to overcome this?
Simple pseudo-random number generators actually generate a fixed sequence of numbers. The particular sequence you get is determined by the initial "seed" value. My suspicion is that you are always getting the first number in the same sequence. Therefore I suggest we try to change the sequence by calling srand every time before calling rand, thus changing the seed value every time. The docs explain that, when called without a parameter, srand generates a new seed based on current circumstances (e.g. the time on the clock). Thus you should get a difference sequence and hence a different random number:
srand
rand(200)
Now, you may ask - why are you always getting the same sequence? I have no idea! As someone else suggested in one of the comments, the behavior you are seeing is the behavior one would expect if you had other code, anywhere, that calls srand with the same, fixed value every time. So it might be good to look for that.
Try Random.rand(). For example
Random.rand(200)
Or if you're working with an array you could use sample.
[*1..200].sample
rand(200) is run once and that value is assigned to your index variable. So 'index' will always be that number.
If you want index to change, you will need to continually run rand on it.
Here's a simple way to do that:
def randomIndex(num)
index = rand(num)
return index
end
randomIndex(200)
=> // this value will change
Out of curiosity, I'm beginning to learn how to program my TI-83+ calculator. Part of my latest program involves storing numbers in a list. How can I add items to a list on a TI-83+ and how can I loop over them/access them?
Well, if you want to add something to the end, you need the length of the list. Let's say your using L1 as a list and variable A as the value you are trying to add to the list.
Here's what you would do:
:A->L1(1+dim(L1))
Here's how that works. The dim command has 1 parameter. That parameter is a list. When you use the dim command, it returns the length of the list in the parameters. When you want to refer to a specific place in a list, you use the syntax: list_name(location). So This line of code takes the value of variable A and stores it in the location of L1 which is 1 more than the length of L1, therefore appending variable A to the end of L1.
If you want to access a value in list, again use the syntax: list_name(location). On the other hand, if you don't know the location of the value you are looking for, or you are cycling through the list and doing something with each value, you can use a for statement.
Like this:
:FOR(A, 0, dim(L1))
::L1(A)->B
::"do whatever you want with the value of L1(A) here"
:END
Or like this:
:FOR(A, 0, dim(L1))
::if(L1(A) == "insert value being searched for here"):THEN
:::A->B
:::dim(L1)+1->A
::END
:END
The for loop works like this: at the beginning of the loop, 0 is stored to variable A. Then the loop continues until variable A is greater than dim(L1). After each time the loop resets, the value of variable A is increased by 1.
In the first example, the program loops through each value of L1 and does whatever you want to do with each value.
In the second example, the program loops through each value of L1. When the value of L1 matches the value you are looking for, the location of the value is stored in variable B to be used for whatever you want later. Then, the value of variable A is set to 1 more than the length of L1. Since the value of variable A is greater than dim(L1), the for loop is terminated.
An element can be added to the end of a list of unknown length like this:
0→L1(1+dim(L1
Under normal condition, attempting to set the value of an index greater than the length of the list results in ERR: INVALID DIM; however, if the index is only 1 greater than the length of the list, the value is appended to the end of the list.
You could use a list or a matrix, but I would suggest a list. You can find information on lists and their commands from this link.
Lists are also better for saving values in between program executions than just using variables, which may be changes by other programs or math.
You need first to define the size of a list like this :
3->dim(L1
(if you forget, you will have an ERR:Invalid Dim)
Press enter and you get a "10" as answer (don't worry it's normal).
You can find dim( in the [Catalog] and -> is "[STO->].
Then you could fill the list with some data like this :
2->L1(1)
3->L1(3)
Now When you print L1 you get :
{2 0 3 0}
First index is L1(1) not 0 (as usual).
You can delete the list by using DelVar :
DelVar L1
You can fill it with Fill, sort it, convert to matrix ....
Simply go to the List menu (2nd + Stat).
You can iterate on the list using a for loop (no foreach, use dim(L1) for the upper bound).
More informations in the guidebook or you could also ask your questions on this calculator questions stack
Hope this helps =)
You can do what Thibault said, fill it, sort it, convert it (Very well said, by the way). However, you can also do:
3->L1(dim(L1))
This will add 3 to the end of L1.