I have been developing a graphics editor for SVG's to put in online for my users to access it through their web browsers. It's based on SVG-edit and it written in Javascript.
The application at the moment lacks something called ''boolean operations'', that is the ability for a user to select 2 or more shapes and join them together.
I have found a C++ library that is called LIB2GEOM, and its supposed to handle these operations, i believe this is what Inkscape uses as well.
So is their a possibility to link this library with my application since its not written in Javascript?
<div id="svgcontainer"></div>
<script>
function path2poly()
{
var subj_polygons = [[{X:10,Y:10},{X:110,Y:10},{X:110,Y:110},{X:10,Y:110}],
[{X:20,Y:20},{X:20,Y:100},{X:100,Y:100},{X:100,Y:20}]];
var clip_polygons = [[{X:50,Y:50},{X:150,Y:50},{X:150,Y:150},{X:50,Y:150}],
[{X:60,Y:60},{X:60,Y:140},{X:140,Y:140},{X:140,Y:60}]];
var scale = 100;
subj_polygons = scaleup(subj_polygons, scale);
clip_polygons = scaleup(clip_polygons, scale);
var cpr = new ClipperLib.Clipper();
cpr.AddPolygons(subj_polygons, ClipperLib.PolyType.ptSubject);
cpr.AddPolygons(clip_polygons, ClipperLib.PolyType.ptClip);
var subject_fillType = ClipperLib.PolyFillType.pftNonZero;
var clip_fillType = ClipperLib.PolyFillType.pftNonZero;
var clipTypes = [ClipperLib.ClipType.ctUnion];
var clipTypesTexts = "Union";
var solution_polygons, svg, cont = document.getElementById('svgcontainer');
var i;
for(i = 0; i < clipTypes.length; i++) {
solution_polygons = new ClipperLib.Polygons();
cpr.Execute(clipTypes[i], solution_polygons, subject_fillType, clip_fillType);
//console.log(JSON.stringify(solution_polygons));
alert(polys2path(solution_polygons, scale));
}
}
// helper function to scale up polygon coordinates
function scaleup(poly, scale) {
var i, j;
if (!scale) scale = 1;
for(i = 0; i < poly.length; i++) {
for(j = 0; j < poly[i].length; j++) {
poly[i][j].X *= scale;
poly[i][j].Y *= scale;
}
}
return poly;
}
// converts polygons to SVG path string
function polys2path (poly, scale) {
var path = "", i, j;
if (!scale) scale = 1;
for(i = 0; i < poly.length; i++) {
for(j = 0; j < poly[i].length; j++) {
if (!j) path += "M";
else path += "L";
path += (poly[i][j].X / scale) + ", " + (poly[i][j].Y / scale);
}
path += "Z";
}
return path;
}
</script>
You can try using emscripten to cross compile it to JS
Related
I have a TIFF file. I want to slice it automatically (specifying the number of slices in horizontal and vertical) and save them into TIFF files (I don't want to change the format to png or ...)
I know that in photoshop you can choose the slice tool>>right click>>Divide Slice>>Save for web
However, the "Save for web" doesn't offer saving in TIFF Format and also I don't think it can work for large file (which is the case here).
Anything that can help (script, plugin) is welcome
With the help of some online codes, I created the script below which is capable of automatically slicing a big TIFF into smaller Tiff files:
#target photoshop
function main(){
if(!documents.length) return;
var dlg=
"dialog{text:'Script Interface',bounds:[100,100,380,290],"+
"panel0:Panel{bounds:[10,10,270,180] , text:'' ,properties:{borderStyle:'etched',su1PanelCoordinates:true},"+
"title:StaticText{bounds:[60,10,220,40] , text:'File Chop' ,properties:{scrolling:undefined,multiline:undefined}},"+
"panel1:Panel{bounds:[10,40,250,130] , text:'' ,properties:{borderStyle:'etched',su1PanelCoordinates:true},"+
"statictext1:StaticText{bounds:[10,10,111,30] , text:'Accross' ,properties:{scrolling:undefined,multiline:undefined}},"+
"statictext2:StaticText{bounds:[140,10,230,27] , text:'Down' ,properties:{scrolling:undefined,multiline:undefined}},"+
"across:DropDownList{bounds:[10,30,100,50]},"+
"down:DropDownList{bounds:[140,30,230,50]},"+
"saveFiles:Checkbox{bounds:[10,60,230,80] , text:'Save and Close new files?'}},"+
"button0:Button{bounds:[10,140,110,160] , text:'Ok' },"+
"button1:Button{bounds:[150,140,250,160] , text:'Cancel' }}};"
var win = new Window(dlg,'File Chop');
if(version.substr(0,version.indexOf('.'))>9){
win.panel0.title.graphics.font = ScriptUI.newFont("Georgia","BOLD",20);
g = win.graphics;
var myBrush = g.newBrush(g.BrushType.SOLID_COLOR, [1.00, 1.00, 1.00, 1]);
g.backgroundColor = myBrush;
var myPen =g.newPen (g.PenType.SOLID_COLOR, [1.00, 0.00, 0.00, 1],lineWidth=1);
}
win.center();
for(var i=1;i<31;i++){
win.panel0.panel1.across.add ('item', i);
win.panel0.panel1.down.add ('item', i);
}
win.panel0.panel1.across.selection=0;
win.panel0.panel1.down.selection=0;
var done = false;
while (!done) {
var x = win.show();
if (x == 0 || x == 2) {
win.canceled = true;
done = true;
} else if (x == 1) {
done = true;
{
if(!documents.length)return;
var startRulerUnits = preferences.rulerUnits;
preferences.rulerUnits = Units.PIXELS;
doc = app.activeDocument;
app.displayDialogs = DialogModes.NO;
doc.flatten();
var tilesAcross = parseInt(win.panel0.panel1.across.selection.index)+1;
var tilesDown =parseInt(win.panel0.panel1.down.selection.index)+1;
var tileWidth = parseInt(doc.width/tilesAcross);
var tileHeight = parseInt(doc.height/tilesDown);
var SaveFiles = win.panel0.panel1.saveFiles.value;
ProcessFiles(tilesDown,tilesAcross,tileWidth,tileHeight,SaveFiles);
app.preferences.rulerUnits = startRulerUnits;
}
}
}
}
main();
function ProcessFiles(Down,Across,offsetX,offsetY,SaveFiles){
try{
var newName = activeDocument.name.match(/(.*)\.[^\.]+$/)[1];
}catch(e){var newName="UntitledChop"}
var Path='';
try{
Path = activeDocument.path;
}catch(e){Path = "~/Desktop";}
if(SaveFiles){
Path = Folder(decodeURI(Path) +"/FileChop");
if(!Path.exists) Path.create();
}
TLX = 0; TLY = 0; TRX = offsetX; TRY = 0;
BRX = offsetX; BRY = offsetY; BLX = 0; BLY = offsetY;
for(var a = 0; a < Down; a++){
for(var i = 0;i <Across; i++){
var NewFileName = newName +"#"+a+"-"+i;
app.activeDocument.duplicate (NewFileName, true);
activeDocument.selection.select([[TLX,TLY],[TRX,TRY],[BRX,BRY],[BLX,BLY]], SelectionType.REPLACE, 0, false);
executeAction( charIDToTypeID( "Crop" ), undefined, DialogModes.NO );
app.activeDocument.selection.deselect();
if(SaveFiles){
var saveFile = File(decodeURI(Path+"/"+NewFileName+".tiff"));
SaveTIFF(saveFile);
app.activeDocument.close(SaveOptions.DONOTSAVECHANGES);
}
activeDocument = documents[0];
TLX = offsetX * (i+1) ; TRX = TLX + offsetX; BRX = TRX; BLX = TLX;
}
TLX = 0; TLY = offsetY * (a +1); TRX = offsetX; TRY = offsetY * (a +1);
BRX = offsetX; BRY = TRY + offsetY; BLX = 0; BLY = (offsetY * (a +1)+offsetY);
}
if(SaveFiles){
Path.execute()
}
}
function SaveTIFF(saveFile){
tiffSaveOptions = new TiffSaveOptions();
tiffSaveOptions.embedColorProfile = true;
tiffSaveOptions.alphaChannels = true;
tiffSaveOptions.layers = true;
tiffSaveOptions.imageCompression = TIFFEncoding.NONE;
activeDocument.saveAs(saveFile, tiffSaveOptions, true, Extension.LOWERCASE);
}
I have a binary image and a color image of the same size. I need to iterate each blob (white pixel blocks) of the binary image and use it as a mask and find the mean color of this blob region from the color image.
I have tried:
HierarchyIndex[] hierarchy;
Point[][] contours;
binaryImage.FindContours(out contours, out hierarchy, RetrievalModes.List, ContourApproximationModes.ApproxNone);
using (Mat mask = Mat.Zeros(matColor.Size(), MatType.CV_8UC1))
foreach (var bl in contours)
if (Cv2.ContourArea(bl) > 5)
{
mask.DrawContour(bl, Scalar.White, -1);
Rect rect = Cv2.BoundingRect(bl);
Scalar mean = Cv2.Mean(colorImage[rect], mask[rect]);
mask.DrawContour(bl, Scalar.Black, -1);
}
which works for the blobs not having holes. However in my case I have many blob regions having huge holes that affects the mean calculation.
I couldn't figure it out how to solve it using the hierarchy info; or with another approach.
(My code is for OpenCVSharp but answer in any other wrapper or language is wellcome.)
Edit: I've added an example image. The traffic signs part is the problem.
Actually I think I have solved this problem with this method:
using PLine = List<Point>;
using Shape = List<List<Point>>;
internal static IEnumerable<Tuple<PLine, Shape>> FindContoursWithHoles(this Mat mat)
{
Point[][] contours;
HierarchyIndex[] hierarchy;
mat.FindContours(out contours, out hierarchy, RetrievalModes.Tree, ContourApproximationModes.ApproxNone);
Dictionary<int, bool> dic = new Dictionary<int, bool>();
for (int i = 0; i < contours.Length; i++)
if (hierarchy[i].Parent < 0)
dic[i] = true;
bool ok = false;
while (!ok)
{
ok = true;
for (int i = 0; i < contours.Length; i++)
if (dic.ContainsKey(i))
{
bool isParent = dic[i];
var hi = hierarchy[i];
if (hi.Parent >= 0) dic[hi.Parent] = (!isParent);
if (hi.Child >= 0) dic[hi.Child] = (!isParent);
while (hi.Next >= 0)
{
dic[hi.Next] = isParent;
hi = hierarchy[hi.Next];
if (hi.Parent >= 0) dic[hi.Parent] = (!isParent);
if (hi.Child >= 0) dic[hi.Child] = (!isParent);
}
hi = hierarchy[i];
while (hi.Previous >= 0)
{
dic[hi.Previous] = isParent;
hi = hierarchy[hi.Previous];
if (hi.Parent >= 0) dic[hi.Parent] = (!isParent);
if (hi.Child >= 0) dic[hi.Child] = (!isParent);
}
}
else
ok = false;
}
foreach (int i in dic.Keys.Where(a => dic[a]))
{
PLine pl = contours[i].ToList();
Shape childs = new Shape();
var hiParent = hierarchy[i];
if (hiParent.Child >= 0)
{
childs.Add(contours[hiParent.Child].ToList());
var hi = hierarchy[hiParent.Child];
while (hi.Next >= 0)
{
childs.Add(contours[hi.Next].ToList());
hi = hierarchy[hi.Next];
}
hi = hierarchy[hiParent.Child];
while (hi.Previous >= 0)
{
childs.Add(contours[hi.Previous].ToList());
hi = hierarchy[hi.Previous];
}
}
yield return Tuple.Create(pl, childs);
}
}
By drawing the holes as black, we can use each blob as a single mask:
var blobContours = blobs.FindContoursWithHoles().ToList();
using (Mat mask = Mat.Zeros(mat0.Size(), MatType.CV_8UC1))
for (int i = 0; i < blobContours.Count; i++)
{
var tu = blobContours[i];
var bl = tu.Item1;
if (Cv2.ContourArea(bl) > 100)
{
mask.DrawContour(bl, Scalar.White, -1);
foreach (var child in tu.Item2)
mask.DrawContour(child, Scalar.Black, -1);
Rect rect = Cv2.BoundingRect(bl);
Scalar mean = Cv2.Mean(mat0[rect], mask[rect]);
}
}
I think there should be an easier way.
And yet there is another problem. In some cases, an individual red part of the sign (which is a seperate white blob) does not found as a parent outside circle and a child inside circle, but a large parent contour outside with two circles as children (ie. hole inside another hole, makes a seperate blob which is not found as a parent). Yes it is hierarchically correct but does not help me. I hope I could make my self clear, sorry for my English.
#Miki thank you very much. I was able to achieve what I want using ConnectedComponents. Its simple and fast:
var cc = Cv2.ConnectedComponentsEx(binaryImage, PixelConnectivity.Connectivity8);
foreach (var bl in cc.Blobs)
using (Mat mask = new Mat())
{
cc.FilterByBlob(binaryImage, mask, bl);
Rect rect = bl.Rect;
Scalar mean = Cv2.Mean(colorImage[rect], mask[rect]);
}
I have a text and I want to train by adding feature using the java API. Looking at the examples the main class to build the training set is the svm_problem. It appear like the svm_node represents a feature (the index is the feature and the value is the weight of the feature).
What I have done is to have a map (just to simplify the problem) that keeps an association between the feature and an index. For each of my weight> example I do create a new node :
svm_node currentNode = new svm_node();
int index = feature.getIndexInMap();
double value = feature.getWeight();
currentNode.index = index;
currentNode.value = value;
Is my intuition correct? What does the svm_problem.y refers to? Does it refer to the index of the label? Is the svm_problem.l just the length of the two vectors?
Your intuition is very close, but svm_node is a pattern not a feature. The variable svm_problem.y is an array that contains the labels of each pattern and svm_problem.l is the size of the training set.
Also, beware that svm_parameter.nr_weight is the weight of each label (useful if you have an unbalanced training set) but if you are not going to use it you must set that value to zero.
Let me show you a simple example in C++:
#include "svm.h"
#include <iostream>
using namespace std;
int main()
{
svm_parameter params;
params.svm_type = C_SVC;
params.kernel_type = RBF;
params.C = 1;
params.gamma = 1;
params.nr_weight = 0;
params.p= 0.0001;
svm_problem problem;
problem.l = 4;
problem.y = new double[4]{1,-1,-1,1};
problem.x = new svm_node*[4];
{
problem.x[0] = new svm_node[3];
problem.x[0][0].index = 1;
problem.x[0][0].value = 0;
problem.x[0][1].index = 2;
problem.x[0][1].value = 0;
problem.x[0][2].index = -1;
}
{
problem.x[1] = new svm_node[3];
problem.x[1][0].index = 1;
problem.x[1][0].value = 1;
problem.x[1][1].index = 2;
problem.x[1][1].value = 0;
problem.x[1][2].index = -1;
}
{
problem.x[2] = new svm_node[3];
problem.x[2][0].index = 1;
problem.x[2][0].value = 0;
problem.x[2][1].index = 2;
problem.x[2][1].value = 1;
problem.x[2][2].index = -1;
}
{
problem.x[3] = new svm_node[3];
problem.x[3][0].index = 1;
problem.x[3][0].value = 1;
problem.x[3][1].index = 2;
problem.x[3][1].value = 1;
problem.x[3][2].index = -1;
}
for(int i=0; i<4; i++)
{
cout << problem.y[i] << endl;
}
svm_model * model = svm_train(&problem, ¶ms);
svm_save_model("mymodel.svm", model);
for(int i=0; i<4; i++)
{
double d = svm_predict(model, problem.x[i]);
cout << "Prediction " << d << endl;
}
/* We should free the memory at this point.
But this example is large enough already */
}
How can I compute a MD5 or SHA1 hash of text in a specific cell and set it to another cell in Google Spreadsheet?
Is there a formula like =ComputeMD5(A1) or =ComputeSHA1(A1)?
Or is it possible to write custom formula for this? How?
Open Tools > Script Editor then paste the following code:
function MD5 (input) {
var rawHash = Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, input);
var txtHash = '';
for (i = 0; i < rawHash.length; i++) {
var hashVal = rawHash[i];
if (hashVal < 0) {
hashVal += 256;
}
if (hashVal.toString(16).length == 1) {
txtHash += '0';
}
txtHash += hashVal.toString(16);
}
return txtHash;
}
Save the script after that and then use the MD5() function in your spreadsheet while referencing a cell.
This script is based on Utilities.computeDigest() function.
Thanks to gabhubert for the code.
This is the SHA1 version of that code (very simple change)
function GetSHA1(input) {
var rawHash = Utilities.computeDigest(Utilities.DigestAlgorithm.SHA_1, input);
var txtHash = '';
for (j = 0; j <rawHash.length; j++) {
var hashVal = rawHash[j];
if (hashVal < 0)
hashVal += 256;
if (hashVal.toString(16).length == 1)
txtHash += "0";
txtHash += hashVal.toString(16);
}
return txtHash;
}
Ok, got it,
Need to create custom function as explained in
http://code.google.com/googleapps/appsscript/articles/custom_function.html
And then use the apis as explained in
http://code.google.com/googleapps/appsscript/service_utilities.html
I need to handtype the complete function name so that I can see the result in the cell.
Following is the sample of the code that gave base 64 encoded hash of the text
function getBase64EncodedMD5(text)
{
return Utilities.base64Encode( Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, text));
}
The difference between this solution and the others is:
It fixes an issue some of the above solution have with offsetting the output of Utilities.computeDigest (it offsets by 128 instead of 256)
It fixes an issue that causes some other solutions to produce the same hash for different inputs by calling JSON.stringify() on input before passing it to Utilities.computeDigest()
function MD5(input) {
var result = "";
var byteArray = Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, JSON.stringify(input));
for (i=0; i < byteArray.length; i++) {
result += (byteArray[i] + 128).toString(16) + "-";
}
result = result.substring(result, result.length - 1); // remove trailing dash
return result;
}
to get hashes for a range of cells, add this next to gabhubert's function:
function RangeGetMD5Hash(input) {
if (input.map) { // Test whether input is an array.
return input.map(GetMD5Hash); // Recurse over array if so.
} else {
return GetMD5Hash(input)
}
}
and use it in cell this way:
=RangeGetMD5Hash(A5:X25)
It returns range of same dimensions as source one, values will spread down and right from cell with formulae.
It's universal single-value-function to range-func conversion method (ref), and it's way faster than separate formuleas for each cell; in this form, it also works for single cell, so maybe it's worth to rewrite source function this way.
Based on #gabhubert but using array operations to get the hexadecimal representation
function sha(str){
return Utilities
.computeDigest(Utilities.DigestAlgorithm.SHA_1, str) // string to digested array of integers
.map(function(val) {return val<0? val+256 : val}) // correct the offset
.map(function(val) {return ("00" + val.toString(16)).slice(-2)}) // add padding and enconde
.join(''); // join in a single string
}
Using #gabhubert answer, you could do this, if you want to get the results from a whole row. From the script editor.
function GetMD5Hash(value) {
var rawHash = Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, value);
var txtHash = '';
for (j = 0; j <rawHash.length; j++) {
var hashVal = rawHash[j];
if (hashVal < 0)
hashVal += 256;
if (hashVal.toString(16).length == 1)
txtHash += "0";
txtHash += hashVal.toString(16);
}
return txtHash;
}
function straightToText() {
var ss = SpreadsheetApp.getActiveSpreadsheet().getSheets();
var r = 1;
var n_rows = 9999;
var n_cols = 1;
var column = 1;
var sheet = ss[0].getRange(r, column, n_rows, ncols).getValues(); // get first sheet, a1:a9999
var results = [];
for (var i = 0; i < sheet.length; i++) {
var hashmd5= GetMD5Hash(sheet[i][0]);
results.push(hashmd5);
}
var dest_col = 3;
for (var j = 0; j < results.length; j++) {
var row = j+1;
ss[0].getRange(row, dest_col).setValue(results[j]); // write output to c1:c9999 as text
}
}
And then, from the Run menu, just run the function straightToText() so you can get your result, and elude the too many calls to a function error.
I was looking for an option that would provide a shorter result. What do you think about this? It only returns 4 characters. The unfortunate part is that it uses i's and o's which can be confused for L's and 0's respectively; with the right font and in caps it wouldn't matter much.
function getShortMD5Hash(input) {
var rawHash = Utilities.computeDigest(Utilities.DigestAlgorithm.MD5, input);
var txtHash = '';
for (j = 0; j < 16; j += 8) {
hashVal = (rawHash[j] + rawHash[j+1] + rawHash[j+2] + rawHash[j+3]) ^ (rawHash[j+4] + rawHash[j+5] + rawHash[j+6] + rawHash[j+7])
if (hashVal < 0)
hashVal += 1024;
if (hashVal.toString(36).length == 1)
txtHash += "0";
txtHash += hashVal.toString(36);
}
return txtHash.toUpperCase();
}
I needed to get a hash across a range of cells, so I run it like this:
function RangeSHA256(input)
{
return Array.isArray(input) ?
input.map(row => row.map(cell => SHA256(cell))) :
SHA256(input);
}
I have a probably pretty simple question but I am still not sure!
Actually I only want to smooth a histogram, and I am not sure which of the following to methods is correct. Would I do it like this:
vector<double> mask(3);
mask[0] = 0.25; mask[1] = 0.5; mask[2] = 0.25;
vector<double> tmpVect(histogram->size());
for (unsigned int i = 0; i < histogram->size(); i++)
tmpVect[i] = (*histogram)[i];
for (int bin = 1; bin < histogram->size()-1; bin++) {
double smoothedValue = 0;
for (int i = 0; i < mask.size(); i++) {
smoothedValue += tmpVect[bin-1+i]*mask[i];
}
(*histogram)[bin] = smoothedValue;
}
Or would you usually do it like this?:
vector<double> mask(3);
mask[0] = 0.25; mask[1] = 0.5; mask[2] = 0.25;
for (int bin = 1; bin < histogram->size()-1; bin++) {
double smoothedValue = 0;
for (int i = 0; i < mask.size(); i++) {
smoothedValue += (*histogram)[bin-1+i]*mask[i];
}
(*histogram)[bin] = smoothedValue;
}
My Questin is: Is it resonable to copy the histogram in a extra vector first so that when I smooth at bin i I can use the original i-1 value or would I simply do smoothedValue += (*histogram)[bin-1+i]*mask[i];, so that I use the already smoothed i-1 value instead the original one.
Regards & Thanks for a reply.
Your intuition is right: you need a temporary vector. Otherwise, you will end up using partly old values, and partly new values, and the result will not be correct. Try it yourself on paper with a simple example.
There are two ways you can write this algorithm:
Copy the data to a temporary vector first; then read from that one, and write to histogram. This is what you did in your first code fragment.
Read from histogram and write to a temporary vector; then copy from the temporary vector back to histogram.
To prevent needless copying of data, you can use vector::swap. This is an extremely fast operation that swaps the contents of two vectors. Using strategy 2 above, this would result in:
vector<double> mask(3);
mask[0] = 0.25; mask[1] = 0.5; mask[2] = 0.25;
vector<double> newHistogram(histogram->size());
for (int bin = 1; bin < histogram->size()-1; bin++) {
double smoothedValue = 0;
for (int i = 0; i < mask.size(); i++) {
smoothedValue += (*histogram)[bin-1+i]*mask[i];
}
newHistogram[bin] = smoothedValue;
}
histogram->swap(newHistogram);