I am new to JavaFX and ControlsFX.
I am trying to create a very basic SpreadsheetView using the ControlsFX library. Following is the function to populate and create the SpreadsheetView:
private fun spreadSheetFunc() : SpreadsheetView {
val rowCount = 15
val columnCount = 10
val grid = GridBase(rowCount, columnCount)
val rows = FXCollections.observableArrayList<ObservableList<SpreadsheetCell>>()
var list = FXCollections.observableArrayList<SpreadsheetCell>()
list.add(SpreadsheetCellType.STRING.createCell(0, 0, 1, 1, "row0-col0"))
list.add(SpreadsheetCellType.STRING.createCell(0, 1, 2, 1, "row0-col1"))
list.add(SpreadsheetCellType.STRING.createCell(0, 2, 1, 1, "row0-col2"))
rows.add(list)
list = FXCollections.observableArrayList()
list.add(SpreadsheetCellType.STRING.createCell(1, 0, 1, 1, "row1-col0"))
//commenting row1-col1 as row0-col1 has a rowspan of 2
//list.add(SpreadsheetCellType.STRING.createCell(1, 1, 1, 1, "row1-col1"))
list.add(SpreadsheetCellType.STRING.createCell(1, 2, 1, 1, "row1-col2"))
rows.add(list)
list = FXCollections.observableArrayList()
list.add(SpreadsheetCellType.STRING.createCell(2, 0, 1, 1, "row2-col0"))
list.add(SpreadsheetCellType.STRING.createCell(2, 1, 1, 1, "row2-col1"))
list.add(SpreadsheetCellType.STRING.createCell(2, 2, 1, 1, "row2-col2"))
rows.add(list)
list = FXCollections.observableArrayList()
list.add(SpreadsheetCellType.STRING.createCell(3, 0, 1, 1, "row3-col0"))
list.add(SpreadsheetCellType.STRING.createCell(3, 1, 1, 1, "row3-col1"))
list.add(SpreadsheetCellType.STRING.createCell(3, 2, 1, 1, "row3-col2"))
rows.add(list)
grid.setRows(rows)
return SpreadsheetView(grid)
}
On running it, I get the following error:
java.lang.IndexOutOfBoundsException: Index: 2, Size: 2 at
java.util.ArrayList.rangeCheck(ArrayList.java:653)
I know its happening because I am not adding any value for rowIndex=1 colIndex=1 (see the commented out line) ... but that is what I want.
The row0-col1 has a rowspan of 2 which should mean that even if my row1-col1 is absent, there shouldn't be any problem.
Why doesn't ControlsFX automatically take care of this?
If I uncomment that line, I get the following output:
Edit 1:
Also, I found another issue, when a colspan/rowspan occupies the whole column/row in the SpreadsheetView and then when one presses arrow key to navigate to cells you get an error:
The above situation arises when you press the right arrow key (Even though their isn't a cell on the right)
Let me apologize because it is not well documented how span must be made in the SpreadsheetView. I will update the documentation.
If you want to span, you have to put the same cell in every cell inside the span.
So either you build your own cell, and then in every place. In your case, you would add the same cell in row 0 column 1 and in row 1 column 1.
Or you could keep your code, and simply call the method spanRow on the Grid. This method will automatically take your cell and place it accordingly.
Regarding the second issue, please submit it to our issue tracker so we can fix it : https://bitbucket.org/controlsfx/controlsfx/issues?status=new&status=open
If you have other issue regarding the SpreadsheetView, consider posting in our Google group where we will get notifications : http://groups.controlsfx.org
Related
I trained a BERT based encoder decoder model (EncoderDecoderModel) named ed_model with HuggingFace's transformers module.
I used the BertTokenizer named as input_tokenizer
I tokenized the input with:
txt = "Some wonderful sentence to encode"
inputs = input_tokenizer(txt, return_tensors="pt").to(device)
print(inputs)
The output clearly shows that a input_ids is the return dict
{'input_ids': tensor([[ 101, 5660, 7975, 2127, 2053, 2936, 5061, 102]], device='cuda:0'), 'token_type_ids': tensor([[0, 0, 0, 0, 0, 0, 0, 0]], device='cuda:0'), 'attention_mask': tensor([[1, 1, 1, 1, 1, 1, 1, 1]], device='cuda:0')}
But when I try to predict, I get this error:
ed_model.forward(**inputs)
ValueError: You have to specify either input_ids or inputs_embeds
Any ideas ?
Well, apparently this is a known issue, for example: This issue of T5
The problem is that there's probably a renaming procedure in the code, since we use a encoder-decoder architecture we have 2 types of input ids.
The solution is to explicitly specify the type of input id
ed_model.forward(decoder_input_ids=inputs['input_ids'],**inputs)
I wish it was documented somewhere, but now you know :-)
I am trying to create a custom hover tool using which takes the y-value of the plot and maps the value to different value.
The code I could come up with so far to achieve this functionality is
from bokeh.models import HoverTool
import holoviews as hv
df = pd.DataFrame(
{
"zero": [0, 0, 0, 0, 0, 0, 0],
"one": [1, 1, 1, 1, 1, 1, 1],
"two": [2, 2, 2, 2, 2, 2, 2],
}
)
mapping = {i: c for i, c in enumerate(df.columns)}
def col_mapping(num):
return mapping[int(num)]
hover = HoverTool(tooltips=[("x", "$x"), ("y", "$y")])
img = hv.Image((df.index, np.arange(df.shape[1]), df.T)).opts(tools=[hover])
img
x and y will be float values. So the idea is to map the y coordinates to its corresponding value in the mapping dictionary
Let me know how I can get a new value in the hover tool so that when the value is b/w 0 and 1 it will be
Thanks
Here's how I'd do it:
code = f"return ({json.dumps(mapping)})[Math.floor(special_vars.y)];"
hover = HoverTool(tooltips=[("x", "$x"), ("y", "$y"), ('mapped_y', '$y{0}')],
formatters={'$y': CustomJSHover(code=code)})
If you need a some more complicated code than that of col_mapping, then you'd have to use a ColumnDataSource and just add to it the fully transformed column.
While reading shallow copy. It says that copy.copy(x) create shallow copy. But I don't see shallow copy behavior in case of the single dimensional list.
Example:
>> new = [1,2,3,4,5,6]
>> original = copy.copy(new)
>> new
[1, 2, 3, 4, 5, 6]
>> original
[1, 2, 3, 4, 5, 6]
>> id(new)
65022912
>> id(original)
65022512
>> new[2]=13
>> new
[1, 2, 13, 4, 5, 6]
>> original
[1, 2, 3, 4, 5, 6]
So here I assume updating "new" list should update "original" list but it is not happening.
In the case of the multidimensional list, the shallow copy is working properly.
Example:
>> parent_list = [1, 2, [3,4], [5,6]]
>> child_list = copy.copy(parent_list)
>> parent_list[2][1] = "Python"
>> parent_list
[1, 2, [3, 'Python'], [5, 6]]
>> child_list
[1, 2, [3, 'Python'], [5, 6]]
Please guide me, why the shallow copy is not working in case of a one-dimensional list.
Thanks.
There are actually 3 "Levels" to what you want.
1) Create a new reference to the same list. This aliasing is a trivial operation and would look like "original = new" or something like that. original[1] =x will update new[1]; This is the equivalent of copying a pointer in a pointer based language.
original = new
original[1] = x
new[1] will now be updated to x
This should be obvious but for completeness, If you follow the above with:
new = other
original is NOT affected at all.
2) Create a "Copy" of the list. It will allocate a new area and copy elements of the list. This is a "Shallow" copy. The children copied will be references, but the list itself will contain copies. original[1] =x will NOT update new[1], but original[1].childValue =x will update new[1].childValue
original = shallow copy of new
original[1].value = x
new[1].value WILL change to x
original[1] = y
new[1] will NOT be affected
3) Create a deep copy of the list. This will allocate a new area and shallow copy the list, but then will recurse and copy each child referenced in the list. No updates to original will modify new, or vice-a-versa.
original = deep copy of new
original[1].value = x
new[1].value will NOT be affected
original[1] = y
new[1] will NOT be affected
A shallow copy of a list is not usually what you want because your list is left in a hybrid state with some members referencing other lists and some not which will lead to unpredictable behavior, but it may be necessary if your tree is fairly deep and/or you never modify the child nodes.
original[1].value = x is giving error :AttributeError: 'int' object has no attribute 'values'
After writing the below code:
new = [1,2,3,4,5,6]
original = new.copy()
original[1] = 10
print("new:",new , "\n " ,"original:", original)
the output I am getting is :
new: [1, 2, 3, 4, 5, 6]
original: [1, 10, 3, 4, 5, 6]
changes done in the original are not getting reflected for a new list where an index is the same or a shallow copy
When I enter the following commands directly into the R console
library("xts")
mySeries <- xts(c(1.0, 2.0, 3.0, 5.0, 6.0), order.by=c(ISOdatetime(2001, 1, 1, 0, 0, 0), ISOdatetime(2001, 1, 2, 0, 0, 0), ISOdatetime(2001, 1, 3, 0, 0, 0), ISOdatetime(2001, 1, 4, 0, 0, 0), ISOdatetime(2001, 1, 5, 0, 0, 0)))
resultingSeries <- to.monthly(mySeries)
resultingSeries
I will get an output like this
mySeries.Open mySeries.High mySeries.Low mySeries.Close
Jan 2001 1 6 1 6
When I look into the attributes, I see the following output
attributes(resultingSeries)
$dim
[1] 1 4
$dimnames
$dimnames[[1]]
NULL
$dimnames[[2]]
[1] "mySeries.Open" "mySeries.High" "mySeries.Low" "mySeries.Close"
$index
[1] 978307200
attr(,"tclass")
[1] "yearmon"
$tclass
[1] "POSIXct" "POSIXt"
$tzone
[1] ""
$class
[1] "xts" "zoo"
$.indexCLASS
[1] "yearmon"
This is the same I get in Java. I'm wondering where the magic happens so that I see the nice output I get in R. I have no access to the event loop, since I'm using JRI like this (since, it's the recommended way and simplifies error handling):
REngine engine = REngine.engineForClass("org.rosuda.REngine.JRI.JRIEngine");
REXP result = engine.parseAndEval(...)
/edit
In Java I execute each command from above as follows:
REXP result = engine.parseAndEval("resultingSeries") // or any other command
What I get is
org.rosuda.REngine.REXPDouble#4ac66122+[12]
The payload being doubles: 1, 6, 1, 6
The attributes are the same as specified above.
Now R does some magic to display the output above. Is there a way I can get the same output without having to create it manually by myself? Where's the implementation stored, that R gets the above mentioned output?
Here is a piece of code that will work, here i extracted the first element of the field mySeries.Open from the object resultingSeries (which i converted to a data frame) which is equal to 1, notice that you can't pass all of the resultingSeries object strait into Java, you will need to break it down.
package stackoverflow;
import org.rosuda.JRI.REXP;
import org.rosuda.JRI.Rengine;
/**
*
* #author yschellekens
*/
public class StackOverflow {
public static void main(String[] args) throws Exception {
String[] Rargs = {"--vanilla"};
Rengine rengine = new Rengine( Rargs, false, null);
rengine.eval("library('xts')");
rengine.eval("mySeries <- xts(c(1.0, 2.0, 3.0, 5.0, 6.0), order.by=c(ISOdatetime(2001, 1, 1, 0, 0, 0), ISOdatetime(2001, 1, 2, 0, 0, 0), ISOdatetime(2001, 1, 3, 0, 0, 0), ISOdatetime(2001, 1, 4, 0, 0, 0), ISOdatetime(2001, 1, 5, 0, 0, 0)))");
rengine.eval("resultingSeries <- to.monthly(mySeries)");
rengine.eval("resultingSeries<-as.data.frame(resultingSeries)");
REXP result= rengine.eval("resultingSeries$mySeries.Open");
System.out.println("Greeting from R: "+result.asDouble());
}
}
And the Java output:
run:
Greeting from R: 1.0
I figured out the following workaround. The solution is far from perfect.
R offers a command to save its console output as characters vector.
capture.output( {command} )
We can access the output using
REXPString s = rengine.parseAndEval("capture.output( to.monthly(mySeries))")
String[] output = result.asStrings()
The variable output will contain all output lines
[0] mySeries.Open mySeries.High mySeries.Low mySeries.Close
[1]Jan 2001 1 6 1 6
Alternatively you coud use JRIEngine and attack yourself to the event loop, which it did not want in my case (due to the more complicated error handling).
I have a set of datapoints such as (THIS IS AN EXAMPLE)
val=4; (*this value is calculated before in the program, so it not known a priori*)
x={0,1,2,3};
data=Table[0, {val}];
data[[1]] = {1,5,6,8};
data[[2]] = {9,7,1,3};
data[[3]] = {3,4,5,6};
data[[4]] = {2,2,4,6};
Now I can plot each of these data with ListPlot as
ListPlot[Transpose[{x,data[[1]]}]]
and if I want to plot more than one I can do
ListPlot[{Transpose[{x, data[[1]]}], Transpose[{x, data[[2]]}]}]
but how can I plot all of them in one code single line, by considering that val is calculated before in the program?
Is there a way to do something like
For[i = 1, i < val + 1, i++, ListPlot[Transpose[{x,data[i]}]]......]
having a single graph with all x-y curves?
Indeed I would like a static picture of
Manipulate[ListPlot[Transpose[{x, data[[i]]}]], {i, 1, val,1}]
Thanks
Virgilio
You want to "do the same thing" to every element of a list. That should tell you to think of using Map. Your list is named data and each element is your four element sublist. If you look at the help page for Map it shows you need to think up a function that does what you need to do to each individual sublist. You have already understood that you need to use Transpose with x and your sublist so that tells you your function and you are almost there. The result of Map will be a list of all those results. So
In[1]:= x = {0, 1, 2, 3};
data = {{1, 5, 6, 8}, {9, 7, 1, 3}, {3, 4, 5, 6}, {2, 2, 4, 6}};
ListPlot[Map[Transpose[{x, #}] &, data], Joined -> True]
Out[3]= ...FourOverlaidPlotsSnipped...
Go through that a word at a time until you can really understand the thinking that was done to be able to write that. You will use this idea again and again if you keep using Mathematica.
For the example you give the cleanest method is to use DataRange:
data = {{1, 5, 6, 8}, {9, 7, 1, 3}, {3, 4, 5, 6}, {2, 2, 4, 6}};
ListLinePlot[data, DataRange -> {0, 3}]
Please ask your future questions on the dedicated Mathematica StackExchange site: