im trying to export series data in excel from teechart editor.
on exporting the excel gets series data as x and y columns in format of two decimal places.
I want is to export data in scientific notation.
I had partial success by providing the format in series tab ->general here i can provide format for the values. But this changes the format for both columns of the series.
what i want is normal default format for x values of the series and scientific format for the Y series in the exported excel.
Please someone guide me how to proceed
From here:
Since the ValueFormat is a Series property, it is applied to both X and Y ValueLists. The only way around I can think on would be to use a temporal chart splitting each series into two series, one for the X values (with the default ValueFormat) and one for the Y values (with the modified ValueFormat). Ie:
Private Sub Form_Load()
TChart1.Header.Text.Text = TChart1.Version
TChart1.Aspect.View3D = False
TChart1.AddSeries scLine
TChart1.AddSeries scLine
TChart1.Series(0).ValueFormat = "0.######0e-0"
TChart1.Series(1).ValueFormat = "0.######0e-0"
Dim i As Integer
For i = 1 To 5
TChart1.Series(0).AddXY i, Rnd * 100, "", clTeeColor
TChart1.Series(1).AddXY i, Rnd * 100, "", clTeeColor
Next i
End Sub
Private Sub Command1_Click()
Dim tmpChart
Set tmpChart = CreateObject("TeeChart.TChart")
Dim i, y As Integer
For i = 0 To TChart1.SeriesCount - 1
tmpChart.AddSeries scPoint
tmpChart.AddSeries scPoint
For j = 0 To TChart1.Series(i).Count - 1
tmpChart.Series(tmpChart.SeriesCount - 2).Add TChart1.Series(i).XValues.Value(j), "", clTeeColor
tmpChart.Series(tmpChart.SeriesCount - 1).Add TChart1.Series(i).YValues.Value(j), "", clTeeColor
Next j
tmpChart.Series(tmpChart.SeriesCount - 2).YValues.Name = "X"
tmpChart.Series(tmpChart.SeriesCount - 1).ValueFormat = TChart1.Series(i).ValueFormat
Next i
TChart1.Export.asXLS.IncludeHeader = True
TChart1.Export.asXLS.UseSeriesFormat = True
TChart1.Export.asXLS.SaveToFile "C:\tmp\axtest.xls"
End Sub
Worth to note here the export to .xlsx files by code, supporting Excel > 2010, hasn't been implemented yet in TeeChart ActiveX: #1670.
Related
I'm using grails with pdfbox plugin. I'd like to print checkboxes in pdf some are checked and some are not.
To print checkbox I did not a direct way(Even by using PDCheckbox class). So I've used the other way to print the checkbox with tick mark using the below code:
public static writeInputFieldToPDFPage( PDPage pdPage, PDDocument document, Float x, Float y, Boolean ticked) {
PDFont font = PDType1Font.HELVETICA
PDResources res = new PDResources()
String fontName = res.addFont(font)
String da = ticked?"/" + fontName + " 10 Tf 0 0.4 0 rg":""
COSDictionary acroFormDict = new COSDictionary()
acroFormDict.setBoolean(COSName.getPDFName("NeedAppearances"), true)
acroFormDict.setItem(COSName.FIELDS, new COSArray())
acroFormDict.setItem(COSName.DA, new COSString(da))
PDAcroForm acroForm = new PDAcroForm(document, acroFormDict)
acroForm.setDefaultResources(res)
document.getDocumentCatalog().setAcroForm(acroForm)
PDGamma colourBlack = new PDGamma()
PDAppearanceCharacteristicsDictionary fieldAppearance =
new PDAppearanceCharacteristicsDictionary(new COSDictionary())
fieldAppearance.setBorderColour(colourBlack)
if(ticked) {
COSArray arr = new COSArray()
arr.add(new COSFloat(0.89f))
arr.add(new COSFloat(0.937f))
arr.add(new COSFloat(1f))
fieldAppearance.setBackground(new PDGamma(arr))
}
COSDictionary cosDict = new COSDictionary()
COSArray rect = new COSArray()
rect.add(new COSFloat(x))
rect.add(new COSFloat(new Float(y-5)))
rect.add(new COSFloat(new Float(x+10)))
rect.add(new COSFloat(new Float(y+5)))
cosDict.setItem(COSName.RECT, rect)
cosDict.setItem(COSName.FT, COSName.getPDFName("Btn")) // Field Type
cosDict.setItem(COSName.TYPE, COSName.ANNOT)
cosDict.setItem(COSName.SUBTYPE, COSName.getPDFName("Widget"))
if(ticked) {
cosDict.setItem(COSName.TU, new COSString("Checkbox with PDFBox"))
}
cosDict.setItem(COSName.T, new COSString("Chk"))
//Tick mark color and size of the mark
cosDict.setItem(COSName.DA, new COSString(ticked?"/F0 10 Tf 0 0.4 0 rg":"/FF 1 Tf 0 0 g"))
cosDict.setInt(COSName.F, 4)
PDCheckbox checkbox = new PDCheckbox(acroForm, cosDict)
checkbox.setFieldFlags(PDCheckbox.FLAG_READ_ONLY)
checkbox.setValue("Yes")
checkbox.getWidget().setAppearanceCharacteristics(fieldAppearance)
pdPage.getAnnotations().add(checkbox.getWidget())
acroForm.getFields().add(checkbox)
}
This code is working fine in my application, this method is adding checkboxes with tick marks also.
But I can see those rectangle checkboxes or tick marks in only pdf readers, not in all other readers(Like chrome default pdf viewer), and even when I try to print the pdf its not printing the checkboxes, rather its printing some random ASCII numbers.
Please let me know if there is any other way to do this or even if I have to refactor the code.
What is wrong
Your AcroForm checkbox field construction is wrong: You treat it as a text field for which a PDF reader should create an appearance based on the default appearance (DA) value of the field in particular if NeedAppearances is true.
Checkboxes are different, though: you do have to supply an appearance stream at least for the on state, cf. the specification ISO 32000-1:
A check box field represents one or more check boxes that toggle between two states, on and off, when manipulated by the user with the mouse or keyboard. Its field type shall be Btn and its Pushbutton and Radio flags (see Table 226) shall both be clear. Each state can have a separate appearance, which shall be defined by an appearance stream in the appearance dictionary of the field’s widget annotation (see 12.5.5, “Appearance Streams”). The appearance for the off state is optional but, if present, shall be stored in the appearance dictionary under the name Off. Yes should be used as the name for the on state.
(ISO 32000-1 section 12.7.4.2.3 "Check Boxes")
Thus, instead of constructing a DA entry you have to construct an AP ("appearances") entry, itself a dictionary with at least a N ("normal appearances") entry, itself a dictionary with at least an entry for the on state appearance which is recommended to be called Yes.
The specification provides an example which shows a typical check box definition:
1 0 obj
<< /FT /Btn
/T (Urgent)
/V /Yes
/AS /Yes
/AP << /N << /Yes 2 0 R /Off 3 0 R>>
>>
endobj
2 0 obj
<< /Resources 20 0 R
/Length 104
>>
stream
q
0 0 1 rg
BT
/ZaDb 12 Tf
0 0 Td
(4) Tj
ET
Q
endstream
endobj
3 0 obj
<< /Resources 20 0 R
/Length 104
>>
stream
q
0 0 1 rg
BT
/ZaDb 12 Tf
0 0 Td
(8) Tj
ET
Q
endstream
endobj
(The resources in 20 0 obj appear to include a font resource named ZaDb referencing ZapfDingbats.)
By the way, you mention that there is a PDF viewer which actually displays a tick for your document as is. You might want to inform their development that they are doing the wrong thing there.
An example
In a comment you asked for sample code and indicated that it was ok if it were for a current 2.0.x version of PDFBox. So I tried it and came up with this code:
PDDocument document = new PDDocument();
PDPage page = new PDPage();
document.addPage(page);
PDAcroForm acroForm = new PDAcroForm(document);
document.getDocumentCatalog().setAcroForm(acroForm);
COSDictionary normalAppearances = new COSDictionary();
PDAppearanceDictionary pdAppearanceDictionary = new PDAppearanceDictionary();
pdAppearanceDictionary.setNormalAppearance(new PDAppearanceEntry(normalAppearances));
pdAppearanceDictionary.setDownAppearance(new PDAppearanceEntry(normalAppearances));
PDAppearanceStream pdAppearanceStream = new PDAppearanceStream(document);
pdAppearanceStream.setResources(new PDResources());
try (PDPageContentStream pdPageContentStream = new PDPageContentStream(document, pdAppearanceStream))
{
pdPageContentStream.setFont(PDType1Font.ZAPF_DINGBATS, 14.5f);
pdPageContentStream.beginText();
pdPageContentStream.newLineAtOffset(3, 4);
pdPageContentStream.showText("\u2714");
pdPageContentStream.endText();
}
pdAppearanceStream.setBBox(new PDRectangle(18, 18));
normalAppearances.setItem("Yes", pdAppearanceStream);
pdAppearanceStream = new PDAppearanceStream(document);
pdAppearanceStream.setResources(new PDResources());
try (PDPageContentStream pdPageContentStream = new PDPageContentStream(document, pdAppearanceStream))
{
pdPageContentStream.setFont(PDType1Font.ZAPF_DINGBATS, 14.5f);
pdPageContentStream.beginText();
pdPageContentStream.newLineAtOffset(3, 4);
pdPageContentStream.showText("\u2718");
pdPageContentStream.endText();
}
pdAppearanceStream.setBBox(new PDRectangle(18, 18));
normalAppearances.setItem("Off", pdAppearanceStream);
PDCheckBox checkBox = new PDCheckBox(acroForm);
acroForm.getFields().add(checkBox);
checkBox.setPartialName("CheckBoxField");
checkBox.setFieldFlags(4);
List<PDAnnotationWidget> widgets = checkBox.getWidgets();
for (PDAnnotationWidget pdAnnotationWidget : widgets)
{
pdAnnotationWidget.setRectangle(new PDRectangle(50, 750, 18, 18));
pdAnnotationWidget.setPage(page);
page.getAnnotations().add(pdAnnotationWidget);
pdAnnotationWidget.setAppearance(pdAppearanceDictionary);
}
// checkBox.setReadOnly(true);
checkBox.check();
// checkBox.unCheck();
document.save(new File(RESULT_FOLDER, "CheckBox.pdf"));
document.close();
(CreateCheckBox test testCheckboxForSureshGoud)
Be sure to use either
checkBox.check();
or
checkBox.unCheck();
as otherwise the state of the box is undefined.
#mkl has a good answer, but I think it can be simplified a little bit in case you already have a Document and just want to add a PDCheckbox (scala):
def addCheckboxField(
doc: PDDocument,
form: PDAcroForm,
name: String,
pg: Int, // page number
x: Float,
y: Float,
width: Float,
height: Float
) = {
val normalAppearances = new COSDictionary()
normalAppearances.setItem(
"Yes", {
val appearanceStream = new PDAppearanceStream(doc)
appearanceStream.setResources(new PDResources())
appearanceStream
}
)
val appearanceDictionary = new PDAppearanceDictionary()
appearanceDictionary.setNormalAppearance(new PDAppearanceEntry(normalAppearances))
appearanceDictionary.setDownAppearance(new PDAppearanceEntry(normalAppearances))
val field = new PDCheckBox(form)
field.setPartialName(name)
val widget = field.getWidgets.get(0)
widget.setAppearance(appearanceDictionary)
form.getFields.add(field)
val page = doc.getPage(pg)
widget.setRectangle(new PDRectangle(x, y, width, height))
widget.setPage(page)
widget.setPrinted(true)
page.getAnnotations().add(widget)
// do what you want with it
field.unCheck()
}
It's likely there are other simplifications that can be made, but this is what worked for me.
PdfBox version: 2.0.21
I'm trying to make a kind of "cloth simulation" by using ROBLOX's new rope constraints and a grid of parts.
Currently, I've made a 10x10 grid of .4x.4x.4 blocks and now I want to connect each one up with rope constraints.
I've named each part in the grid after their row and column (eg: first part in the grid being 1 1, last one being 10 10)
and then I get the parts around each individual grid part using their name and string manipulation.
I then insert 4 attachments into each part and 4 rope constraints.
Here's the code (ab stands for above, be stands for below, etc) :
for i2 = 1, #gParts do
local ab = tostring(tonumber(gParts[i2].Name:match("^(%S+)"))-5).." "..tostring(tonumber(string.sub(gParts[i2].Name,-1))-1)
local be = tostring(tonumber(gParts[i2].Name:match("^(%S+)"))+5).." "..tostring(tonumber(string.sub(gParts[i2].Name,-1))+1)
local le = tostring(tonumber(gParts[i2].Name:match("^(%S+)"))-1).." "..tostring(tonumber(string.sub(gParts[i2].Name,-1)))
local ri = tostring(tonumber(gParts[i2].Name:match("^(%S+)"))+1).." "..tostring(tonumber(string.sub(gParts[i2].Name,-1)))
for i3 = 1, 4 do
local atchm = Instance.new("Attachment",gParts[i2])
local ropeconst = Instance.new("RopeConstraint",gParts[i2])
end
end
Rope constraint has 2 main properties I need to use; attachment 1 and attachment 2.
I've never really messed with the new constraints, but I believe this should work.
Do keep in mind that the constraints are a new Instance in Roblox, and that they are likely still experimental.
X = 10;
Y = 10;
spread = 4;
--Spread is the Length of the Constraint. You may have to increase this, especially if it's stiff.
function createAttachments()
--This is under the assumption that gParts is a table filled with the Part Instances
for i,v in pairs(gParts) do
local atch = Instance.new("Attachment",v);
end;
end;
function connectConstraints(part,x,y)
if x ~= X then
connectRight = x+1.." "..y;
end;
if y ~= Y then
connectDown = x.." "..y+1;
end;
if connectRight ~= nil then
local ropeconst = Instance.new("RopeConstraint",part);
ropeconst.Length = spread;
ropeconst.Attachment0 = part.Attachment;
ropeconst.Attachment1 = connectRight.Attachment;
end;
if connectLeft ~= nil then
local ropeconst = Instance.new("RopeConstraint",part);
ropeconst.Length = spread;
ropeconst.Attachment0 = part.Attachment;
ropeconst.Attachment1 = connectLeft.Attachment;
end
end
createAttachments();
connectConstraints();
If this does not work for you, please let me know. I can contact you from the site itself if needed.
I have a logic problem for an iOS app but I don't want to solve it using brute-force.
I have a set of integers, the values are not unique:
[3,4,1,7,1,2,5,6,3,4........]
How can I get a subset from it with these 3 conditions:
I can only pick a defined amount of values.
The sum of the picked elements are equal to a value.
The selection must be random, so if there's more than one solution to the value, it will not always return the same.
Thanks in advance!
This is the subset sum problem, it is a known NP-Complete problem, and thus there is no known efficient (polynomial) solution to it.
However, if you are dealing with only relatively low integers - there is a pseudo polynomial time solution using Dynamic Programming.
The idea is to build a matrix bottom-up that follows the next recursive formulas:
D(x,i) = false x<0
D(0,i) = true
D(x,0) = false x != 0
D(x,i) = D(x,i-1) OR D(x-arr[i],i-1)
The idea is to mimic an exhaustive search - at each point you "guess" if the element is chosen or not.
To get the actual subset, you need to trace back your matrix. You iterate from D(SUM,n), (assuming the value is true) - you do the following (after the matrix is already filled up):
if D(x-arr[i-1],i-1) == true:
add arr[i] to the set
modify x <- x - arr[i-1]
modify i <- i-1
else // that means D(x,i-1) must be true
just modify i <- i-1
To get a random subset at each time, if both D(x-arr[i-1],i-1) == true AND D(x,i-1) == true choose randomly which course of action to take.
Python Code (If you don't know python read it as pseudo-code, it is very easy to follow).
arr = [1,2,4,5]
n = len(arr)
SUM = 6
#pre processing:
D = [[True] * (n+1)]
for x in range(1,SUM+1):
D.append([False]*(n+1))
#DP solution to populate D:
for x in range(1,SUM+1):
for i in range(1,n+1):
D[x][i] = D[x][i-1]
if x >= arr[i-1]:
D[x][i] = D[x][i] or D[x-arr[i-1]][i-1]
print D
#get a random solution:
if D[SUM][n] == False:
print 'no solution'
else:
sol = []
x = SUM
i = n
while x != 0:
possibleVals = []
if D[x][i-1] == True:
possibleVals.append(x)
if x >= arr[i-1] and D[x-arr[i-1]][i-1] == True:
possibleVals.append(x-arr[i-1])
#by here possibleVals contains 1/2 solutions, depending on how many choices we have.
#chose randomly one of them
from random import randint
r = possibleVals[randint(0,len(possibleVals)-1)]
#if decided to add element:
if r != x:
sol.append(x-r)
#modify i and x accordingly
x = r
i = i-1
print sol
P.S.
The above give you random choice, but NOT with uniform distribution of the permutations.
To achieve uniform distribution, you need to count the number of possible choices to build each number.
The formulas will be:
D(x,i) = 0 x<0
D(0,i) = 1
D(x,0) = 0 x != 0
D(x,i) = D(x,i-1) + D(x-arr[i],i-1)
And when generating the permutation, you do the same logic, but you decide to add the element i in probability D(x-arr[i],i-1) / D(x,i)
I'm writing a game using Corona SDK in lua language. I'm having a hard time coming up with a logic for a system like this;
I have different items. I want some items to have 1/1000 chance of being chosen (a unique item), I want some to have 1/10, some 2/10 etc.
I was thinking of populating a table and picking a random item. For example I'd add 100 of "X" item to the table and than 1 "Y" item. So by choosing randomly from [0,101] I kind of achieve what I want but I was wondering if there were any other ways of doing it.
items = {
Cat = { probability = 100/1000 }, -- i.e. 1/10
Dog = { probability = 200/1000 }, -- i.e. 2/10
Ant = { probability = 699/1000 },
Unicorn = { probability = 1/1000 },
}
function getRandomItem()
local p = math.random()
local cumulativeProbability = 0
for name, item in pairs(items) do
cumulativeProbability = cumulativeProbability + item.probability
if p <= cumulativeProbability then
return name, item
end
end
end
You want the probabilities to add up to 1. So if you increase the probability of an item (or add an item), you'll want to subtract from other items. That's why I wrote 1/10 as 100/1000: it's easier to see how things are distributed and to update them when you have a common denominator.
You can confirm you're getting the distribution you expect like this:
local count = { }
local iterations = 1000000
for i=1,iterations do
local name = getRandomItem()
count[name] = (count[name] or 0) + 1
end
for name, count in pairs(count) do
print(name, count/iterations)
end
I believe this answer is a lot easier to work with - albeit slightly slower in execution.
local chancesTbl = {
-- You can fill these with any non-negative integer you want
-- No need to make sure they sum up to anything specific
["a"] = 2,
["b"] = 1,
["c"] = 3
}
local function GetWeightedRandomKey()
local sum = 0
for _, chance in pairs(chancesTbl) do
sum = sum + chance
end
local rand = math.random(sum)
local winningKey
for key, chance in pairs(chancesTbl) do
winningKey = key
rand = rand - chance
if rand <= 0 then break end
end
return winningKey
end
Being fairly new to Stata, I'm having a difficulty figuring out how to do the following:
I have time-series data on selling price (p) and quantity sold (q) for 10 products in a single datafile (i,e., 20 variables, p01-p10 and q01-q10). I am strugling with appropriate stata command that computes sales revenue (pq) time-series for each of these 10 products (i.e., pq01-pq10).
Many thanks for your help.
forval i = 1/10 {
local j : display %02.0f `i'
gen pq`j' = p`j' * q`j'
}
A standard loop over 1/10 won't get you the leading zero in 01/09. For that we need to use an appropriate format. See also
#article {pr0051,
author = "Cox, N. J.",
title = "Stata tip 85: Looping over nonintegers",
journal = "Stata Journal",
publisher = "Stata Press",
address = "College Station, TX",
volume = "10",
number = "1",
year = "2010",
pages = "160-163(4)",
url = "http://www.stata-journal.com/article.html?article=pr0051"
}
(added later) Another way to do it is
local j = string(`i', "%02.0f")
That makes it a bit more explicit that you are mapping from numbers 1,...,10 to strings "01",...,"10".