I have two variables in my pre block, and I need a third (a boolean) that identifies whether certain properties hold of those two:
str = "some string from a datasource";
qty = 15; //Also from a datasource
The third variable, valid, needs to be true when str is not empty and qty is greater than 0. How do I do that?
Oh! I just figured it out:
valid = not (str eq "") && (qty > 0);
I had some syntax errors in the rest of my ruleset; that's where the trouble was coming from.
Related
I'm new to julia, but not so new on Dash; I'm trying to build my first app with Dash for julia, but I can't seem to make a pattern matching callback work properly. Here's the part of the code that's giving me troubles:
callback!(
app,
Output((type= "filter_", index= ALL), "options"),
Input("inputs", "data"),
State((type= "filter_", index= ALL), "value"),
) do inputs, filters
list_outs = []
list_vals = []
for i in 1:length(filters)
push!(list_outs, [(label= input, value= input) for input in inputs])
end
return list_outs
end
What I'm trying to do here is to use the available inputs of the data set, already stored in "inputs", to set the filters' options, creating as many sets of options as there are dropdowns.
The problem here is, I guess, in the format of the output I'm returning: it says "Invalid number of output values for {"index":["ALL"],"type":"filter_"}.options. Expected 3, got 1"
Sadly, I found nothing of use about how to use pattern matching callbacks with julia; I tried passing the output both as an array and as a tuple, but to no avail.
Any help is welcomed, thank you all!
This is the error related to the fact that if the result is a single Output, the callback output is automatically represented as an array of what is returned for uniform further processing. I.e., in your case, as [list_outs]. The fact that the Output with the match pattern is also treated as a single one is my bug, I added the issue and try to fix it in the near future.
Right now you can work around this problem by using Output as an array:
using Dash
using DashHtmlComponents
using DashCoreComponents
app = dash()
app.layout = html_div() do
dcc_input(id = "input", value = "A,B,C"),
dcc_dropdown(id = (type="filter_", index = 1)),
dcc_dropdown(id = (type="filter_", index = 2)),
dcc_dropdown(id = (type="filter_", index = 3)),
dcc_dropdown(id = (type="filter_", index = 4))
end
callback!(
app,
[Output((type= "filter_", index= ALL), "options")], #This is multiple output in explicitly form
Input("input", "value"),
State((type= "filter_", index= ALL), "value"),
) do input, filters
inputs = split(input, ",")
list_outs = []
list_vals = []
for i in 1:length(filters)
push!(list_outs, [(label= input, value= input) for input in inputs])
end
return [list_outs] # Accordingly, we return the result inside an additional array
end
run_server(app, debug = true)
I am a new SystemVerilog user and I have faced a strange (from my point of view) behavior of combination of unique method called for fixed array with with operator.
module test();
int arr[12] = '{1,2,1,2,3,4,5,8,9,10,10,8};
int q[$]
initial begin
q = arr.unique() with (item > 5 ? item : 0);
$display("the result is %p",q);
end
I've expected to get queue {8,9,10} but instead I have got {1,8,9,10}.
Why there is a one at the index 0 ?
You are trying to combine the operation of the find method with unique. Unfortunately, it does not work the way you expect. unique returns the element, not the expression in the with clause, which is 0 for elements 1,2,3,4 and 5. The simulator could have chosen any of those elements to represent the unique value for 0(and different simulators do pick different values)
You need to write them separately:
module test();
int arr[$] = '{1,2,1,2,3,4,5,8,9,10,10,8};
int q[$]
initial begin
arr = arr.find() with (item > 5);
q = arr.unique();
$display("the result is %p",q);
end
Update explaining the original results
The with clause generates a list of values to check for uniqueness
'{0,0,0,0,0,0,0,8,9,10,10,8};
^. ^ ^ ^
Assuming the simulator chooses the first occurrence of a replicated value to remain, then it returns {arr[0], arr[7], arr[8], arr[9]} from the original array, which is {1,8,9,10}
I noticed this interesting thing about the max() and min() functions in SV LRM (1800-2012) 7.12 (Array manipulation methods). I tried out the max() and min() functions in a dummy SV file
int a[3] = {0,5,5};
int q[$];
int b;
q = a.max(); // legal
b = a.max(); // illegal
The illegal statement error was
Incompatible complex type assignment
Type of source expression is incompatible with type of target expression.
Mismatching types cannot be used in assignments, initializations and
instantiations. The type of the target is 'int', while the type of the
source is 'int$[$]'.
So I commented out the illegal statement and tested it. It compiled and ran fine but I was hoping to get some more insight as to why the function returns a queue and not a single element - I printed out the contents of q and the size, but the size is still 1 and 5 is being printed just once. Kind of redundant then to make the max() and min() functions return a queue ?
The "SystemVerilog for Verification" book by Chris Spear and Greg Tumbush has a good explanation on this topic in Chapter 2.6.2, which I am quoting below:
"The array locator methods find data in an unpacked array. At first
you may wonder why these return a queue of values. After all, there
is only one maximum value in an array. However, SystemVerilog needs a
queue for the case when you ask for a value from an empty queue or
dynamic array."
It returns a queue to deal with empty queues and when the with () conditions have no matches. The the empty queue return is a a way to differentiate a true match from no matches.
Consider the below code. to find the minimum value of a that is greater than 5. a has data but none of its entries have above 5. b is empty, so it will return an empty. c will return 7.
int a[3] = '{0,5,5};
int b[$] = '{};
int c[4] = '{0,15,5,7};
int q[$];
q = a.min() with (item > 5); // no items >5, will return an empty queue
q = b.min(); // queue is empty, will return an empty queue
q = c.min() with (item > 5); // will return a queue of size 1 with value 7
I believe the example results as per Greg's answer is not correct.
As per System Verilog Language:
min() returns the element with the minimum value or whose expression evaluates to a minimum.
max() returns the element with the maximum value or whose expression evaluates to a maximum.
So, when with expression is evaluated, the resultant value will be:
a.min() with (item > 5); {0,0,0} -> Minimum is 0 and corresponding item is 5.
c.min() with (item > 5); {0,1,0,1}-> Minimum is 0 and corresponding item is 5.
Since, example demonstrates the usage of min, the result will be:
q = a.min() with (item > 5); // A queue of size 1 with value 5.
q = c.min() with (item > 5); //A queue of size 1 with value 5.
I have a string containing several elements, some identical and some unique. I want my code to check every 2 following elements in my string and if they're equal, it should call a function ShuffleString, where the input variable (randomize) is the string itself, that will re-shuffle the string in a new position. Then, the script should re-check every 2 following elements in the string again until no two identical elements appear next to each other.
I have done the following:
My function file ShuffleString works fine. The input variable randomize, as stated earlier, contains the same elements as MyString but in a different order, as this was needed on an unrelated matter earlier in the script.
function [MyString] = ShuffleString(randomize)
MyString = [];
while length(randomize) > 0
S = randi(length(randomize), 1);
MyString = [MyString, randomize(S)];
randomize(S) = [];
end
The script doesn't work as intended. Right now it looks like this:
MyString = ["Cat" "Dog" "Mouse" "Mouse" "Dog" "Hamster" "Zebra" "Obama"...
"Dog" "Fish" "Salmon" "Turkey"];
randomize = MyString;
while(1)
for Z = 1:length(MyString)
if Z < length(MyString)
Q = Z+1;
end
if isequal(MyString{Z},MyString{Q})
[MyString]=ShuffleString(randomize)
continue;
end
end
end
It just seems to reshuffle the string an infinite amount of times. What's wrong with this and how can I make it work?
You are using an infinite while loop that has no way to break and hence it keeps iterating.
Here is a simpler way:
Use the third output argument of the unique function to get the elements in numeric form for easier processing. Apply diff on it to check if consecutive elements are same. If there is any occurrence of same consecutive elements, the output of diff will give at least one zero which when applied with negated all will return true to continue the loop and vice versa. At the end, use the shuffled indices/numeric representation of the strings obtained after the loop to index the first output argument of unique (which was calculated earlier). So the script will be:
MyString = ["Cat" "Dog" "Mouse" "Mouse" "Dog" "Hamster" "Zebra" "Obama"...
"Dog" "Fish" "Salmon" "Turkey"]; %Given string array
[a,~,c] = unique(MyString);%finding unique elements and their indices
while ~all(diff(c)) %looping until there are no same strings together
c = ShuffleString(c); %shuffling the unique indices
end
MyString = a(c); %using the shuffled indices to get the required string array
For the function ShuffleString, a better way would be to use randperm. Your version of function works but it keeps changing the size of the arrays MyString and randomize and hence adversely affects the performance and memory usage. Here is a simpler way:
function MyString = ShuffleString(MyString)
MyString = MyString(randperm(numel(MyString)));
end
I have created a matrix with row 1 full of strings and 4 other rows with numbers. They are created in a handle class with the object "Projekter".
So in the object "Projekter" row 1, the first value is blank, but the second value is 'Ole'. So I know that 'Ole' is in (1,2). x is the name/string I want to search for, which in this case is 'Ole'.
As you see below it should search row 1 from column 2 untill the last name/string and if i = 'Ole', it should bring me the value 2 because " i " should be equal 2.
A is just a controller if the function works, but at this point it doesn't.
The error it gives is "Undefined function 'eq' for input arguments of type 'cell'."
How do I fix this so it return the " i " value when the statement is correct?
Thank you in advance!
function number(obj,x)
A = [];
for i = 2:size(obj.Projekter,2)
if obj.Projekter(1,i)==x
A = A + 1;
end
end
disp(A)
end
Maybe you have to index the cell content:
your_cell = {'a_string'};
your_string = your_cell{1};
function [returnValue] = number(obj,x)
for i = 2:size(obj.Projekter,2)
if obj.Projekter{1,i}==x
returnValue = i;
return;
end
end
end
Note the change from obj.Projekter(1,i)==x to obj.Projekter{1,i}==x (use curly braces instead of parens). I have then specified that returnValue will hold the value that should be returned by doing function [returnValue] = number(obj,x). We then set returnValue equal to i and return from the function when the condition of the if statement is true.
As suggested in the comments, it is probably better to do:
function [returnValue] = number(obj, x)
returnValue = find(strcmp(x, obj.Projekter) == 1);
strcmp(x, obj.Projektor) will give you an array the length of obj.Projekter with 1's wherever the strings match, and 0's where they don't, you can then find the indices that are set to 1. This has the added benefit of
not using a loop so it's faster
Giving you every occurrence of a match, not just the first one.