I would like to re-factor various blocks of code throughout a project using ReSharper 7.1 Search / Replace with pattern.
The code blocks are similar to the following simplified example:
someControl.StatusProgressBar.IsIndeterminate = false;
someControl.StatusProgressBar.Visibility = Visibility.Visible;
someControl.StatusProgressBar.Minimum = 0;
someControl.StatusProgressBar.Maximum = 100;
someControl.StatusProgressBar.Value = percentage;
And I would like to change them to:
someControl.StatusProgressBar.Use(p =>
{
p.IsIndeterminate = false;
p.Visibility = Visibility.Visible;
p.Minimum = 0;
p.Maximum = 100;
p.Value = percentage;
});
'Use' is an extension method
This is easy enough if all the blocks of code are setting the same number of properties. The following search and replace patterns will do the job:
SEARCH
$someControl$.$SomeProperty$.$SubProperty1$ = $val1$;
$someControl$.$SomeProperty$.$SubProperty2$ = $val2$;
$someControl$.$SomeProperty$.$SubProperty3$ = $val3$;
$someControl$.$SomeProperty$.$SubProperty4$ = $val4$;
$someControl$.$SomeProperty$.$SubProperty5$ = $val5$;
REPLACE
$someControl$.$SomeProperty$.Use(p=>
{
p.$SubProperty1$ = $val1$;
p.$SubProperty2$ = $val2$;
p.$SubProperty3$ = $val3$;
p.$SubProperty4$ = $val4$;
p.$SubProperty5$ = $val5$;
});
However, if I also have a code block such as:
someControl.StatusProgressBar.IsIndeterminate = false;
someControl.StatusProgressBar.Visibility = Visibility.Visible;
someControl.StatusProgressBar.Minimum = 0;
someControl.StatusProgressBar.Maximum = 100;
someControl.StatusProgressBar.Value = percentage;
someControl.StatusProgressBar.Orientation = Vertical;
Is it possible, with ReSharper, to capture and replace both code blocks with one pattern? The later having one extra property setting but could easily be more than one extra or less.
I am thinking this is not possible. It would require the ability to create some kind of variable pattern and I can't see a way to do that, be it with regular expressions or otherwise.
Any ideas?
Related
I'm trying to filter the last column on a worksheet but I can't seem to get the Index of the column. To be clear, I need the index relative to the worksheet, no the range. I used VisibleView to find the Column, but there may be hidden rows, so my plan is to then load that column via getRangeByIndexes but I need the relative columnIndex to the worksheet.
I've tried a bunch of variations of the below, but I either get Object doesn't support 'getColumn' or columnIndex is undefined
Note: In the below example I've hardcoded 7 as that will be the last column relative to the VisibleView (Columns and rows are already hidden), but I'd like this to by dynamic for other functions and just returnthe "last visible column index".
var ws = context.workbook.worksheets.getActiveWorksheet()
var visible_rng = ws.getUsedRange(true).getVisibleView()
visible_rng.load(["columnCount", "columnIndex"])
await context.sync();
console.log('visible_rng.columnIndex')
console.log(visible_rng.getCell(0,7).columnIndex)
console.log(visible_rng.getColumn(7).columnIndex)
Well this method seems a bit hacky, please share if you know a better way! But, first thing I found was that getVisibleView only metions rows in the Description.
Represents the visible rows of the current range.
I decided to try getSpecialCells and was able to load the address property. I then had to use split and get the last column LETTER and convert this to the Index.
I also wanted the columnCount but this wasn't working w/ getSpecialCells so I polled that from getVisibleView and return an Object relating to Visible Views that I can build on the function later if I need more details.
Here it is:
async function Get_Visible_View_Details_Obj(context, ws) {
var visible_rng = ws.getUsedRange(true).getSpecialCells("Visible");
visible_rng.load("address")
var visible_view_rng = ws.getUsedRange(true).getVisibleView()
visible_view_rng.load("columnCount")
await context.sync();
var Filter_Col_Index = visible_rng.address
var Filter_Col_Index = Filter_Col_Index.split(",")
var Filter_Col_Index = Filter_Col_Index[Filter_Col_Index.length - 1]
var Filter_Col_Index = Filter_Col_Index.split("!")[1]
if (Filter_Col_Index.includes(":") == true) {
var Filter_Col_Index = Filter_Col_Index.split(":")[1]
}
var Filter_Col_Index = Get_Alpha_FromString(Filter_Col_Index)
var Filter_Col_Index = Get_Col_Index_From_Letters(Filter_Col_Index)
var Filter_Col_Index_Obj = {
"last_col_ws_index": Filter_Col_Index,
"columnCount": visible_view_rng.columnCount,
}
return Filter_Col_Index_Obj
}
Helper Funcs:
function Get_Alpha_FromString(str) {
return str.replace(/[^a-z]/gi, '');
}
function Get_Col_Index_From_Letters(str) {
str = str.toUpperCase();
let out = 0, len = str.length;
for (pos = 0; pos < len; pos++) {
out += (str.charCodeAt(pos) - 64) * Math.pow(26, len - pos - 1);
}
return out - 1;
}
Question: Is it possible to adjust the above to produce all possible results instead of an optimized one.
Details: Given a data set of armor, I want to produce a set of combinations where my constraints are fulfilled.
Variable helm1 = model.addVariable("Helm 1").binary();
Variable helm2 = model.addVariable("Helm 2").binary();
Variable helm3 = model.addVariable("Helm 3").binary();
Variable arm1 = model.addVariable("Arm 1").binary();
Variable arm2 = model.addVariable("Arm 2").binary();
Variable arm3 = model.addVariable("Arm 3").binary();
Expression statA = model.addExpression().lower(0).weight(1);
Expression statB = model.addExpression().lower(0).weight(1);
Expression statC = model.addExpression().lower(0).weight(1);
//Lower Limit set for desired stat
Expression statD = model.addExpression().lower(2).weight(1);
// Limit number of helms you can equip
model.addExpression().upper(1).set(helm1,1).set(helm2,1).set(helm3,1);
model.addExpression().upper(1).set(arm1,1).set(arm2,1).set(arm3,1);
statA.set(arm1, 1);
statB.set(helm2, 1);
statB.set(helm3, 1);
statB.set(arm2, 1);
statC.set(helm1, 1);
statC.set(arm2, 1);
statC.set(arm3, 1);
statD.set(helm3, 3);
statD.set(arm1, 1);
Optimisation.Result result = model.maximise();
BasicLogger.debug(result);
Note: Before recommending libraries, please not that the library must be compatible with Android.
Answer: No - ojAlgo will output 1 (the optimal if it can find it) solution.
I'm making an air dictionary and I have a(nother) problem. The main app is ready to go and works perfectly but when I tested it I noticed that it could be better. A bit of context: the language (ancient egyptian) I'm translating from does not use punctuation so a phrase canlooklikethis. Add to that the sheer complexity of the glyph system (6000+ glyphs).
Right know my app works like this :
user choose the glyphs composing his/r word.
app transforms those glyphs to alphanumerical values (A1 - D36 - X1A, etc).
the code compares the code (say : A5AD36) to a list of xml values.
if the word is found (A5AD36 = priestess of Bast), the user gets the translation. if not, s/he gets all the possible words corresponding to the two glyphs (A5A & D36).
If the user knows the string is a word, no problem. But if s/he enters a few words, s/he'll have a few more choices than hoped (exemple : query = A1A5AD36 gets A1 - A5A - D36 - A5AD36).
What I would like to do is this:
query = A1A5AD36 //word/phrase to be translated;
varArray = [A1, A5A, D36] //variables containing the value of the glyphs.
Corresponding possible words from the xml : A1, A5A, D36, A5AD36.
Possible phrases: A1 A5A D36 / A1 A5AD36 / A1A5A D36 / A1A5AD36.
Possible phrases with only legal words: A1 A5A D36 / A1 A5AD36.
I'm not I really clear but to things simple, I'd like to get all the possible phrases containing only legal words and filter out the other ones.
(example with english : TOBREAKFAST. Legal = to break fast / to breakfast. Illegal = tobreak fast.
I've managed to get all the possible words, but not the rest. Right now, when I run my app, I have an array containing A1 - A5A - D36 - A5AD36. But I'm stuck going forward.
Does anyone have an idea ? Thank you :)
function fnSearch(e: Event): void {
var val: int = sp.length; //sp is an array filled with variables containing the code for each used glyph.
for (var i: int = 0; i < val; i++) { //repeat for every glyph use.
var X: String = ""; //variable created to compare with xml dictionary
for (var i2: int = 0; i2 < val; i2++) { // if it's the first time, use the first glyph-code, else the one after last used.
if (X == "") {
X = sp[i];
} else {
X = X + sp[i2 + i];
}
xmlresult = myXML.mot.cd; //xmlresult = alphanumerical codes corresponding to words from XMLList already imported
trad = myXML.mot.td; //same with traductions.
for (var i3: int = 0; i3 < xmlresult.length(); i3++) { //check if element X is in dictionary
var codeElement: XML = xmlresult[i3]; //variable to compare with X
var tradElement: XML = trad[i3]; //variable corresponding to codeElement
if (X == codeElement.toString()) { //if codeElement[i3] is legal, add it to array of legal words.
checkArray.push(codeElement); //checkArray is an array filled with legal words.
}
}
}
}
var iT2: int = 500 //iT2 set to unreachable value for next lines.
for (var iT: int = 0; iT < checkArray.length; iT++) { //check if the word searched by user is in the results.
if (checkArray[iT] == query) {
iT2 = iT
}
}
if (iT2 != 500) { //if complete query is found, put it on top of the array so it appears on top of the results.
var oldFirst: String = checkArray[0];
checkArray[0] = checkArray[iT2];
checkArray[iT2] = oldFirst;
}
results.visible = true; //make result list visible
loadingResults.visible = false; //loading screen
fnPossibleResults(null); //update result list.
}
I end up with an array of variables containing the glyph-codes (sp) and another with all the possible legal words (checkArray). What I don't know how to do is mix those two to make legal phrases that way :
If there was only three glyphs, I could probably find a way, but user can enter 60 glyphs max.
On the webpage
http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463
It lists prices for a particular item in a game, I wanted to grab the "Current guide price:" of said item, and store it as a variable so I could output it in a google spreadsheet. I only want the number, currently it is "643.8k", but I am not sure how to grab specific text like that.
Since the number is in "k" form, that means I can't graph it, It would have to be something like 643,800 to make it graphable. I have a formula for it, and my second question would be to know if it's possible to use a formula on the number pulled, then store that as the final output?
-EDIT-
This is what I have so far and it's not working not sure why.
function pullRuneScape() {
var page = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var number = page.match(/Current guide price:<\/th>\n(\d*)/)[1];
SpreadsheetApp.getActive().getSheetByName('RuneScape').appendRow([new Date(), number]);
}
Your regex is wrong. I tested this one successfully:
var number = page.match(/Current guide price:<\/th>\s*<td>([^<]*)<\/td>/m)[1];
What it does:
Current guide price:<\/th> find Current guide price: and closing td tag
\s*<td> allow whitespace between tags, find opening td tag
([^<]*) build a group and match everything except this char <
<\/td> match the closing td tag
/m match multiline
Use UrlFetch to get the page [1]. That'll return an HTTPResponse that you can read with GetBlob [2]. Once you have the text you can use regular expressions. In this case just search for 'Current guide price:' and then read the next row. As to remove the 'k' you can just replace with reg ex like this:
'123k'.replace(/k/g,'')
Will return just '123'.
https://developers.google.com/apps-script/reference/url-fetch/
https://developers.google.com/apps-script/reference/url-fetch/http-response
Obviously, you are not getting anything because the regexp is wrong. I'm no regexp expert but I was able to extract the number using basic string manipulation
var page = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var TD = "<td>";
var start = page.indexOf('Current guide price');
start = page.indexOf(TD, start);
var end = page.indexOf('</td>',start);
var number = page.substring (start + TD.length , end);
Logger.log(number);
Then, I wrote a function to convert k,m etc. to the corresponding multiplying factors.
function getMultiplyingFactor(symbol){
switch(symbol){
case 'k':
case 'K':
return 1000;
case 'm':
case 'M':
return 1000 * 1000;
case 'g':
case 'G':
return 1000 * 1000 * 1000;
default:
return 1;
}
}
Finally, tie the two together
function pullRuneScape() {
var page = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var TD = "<td>";
var start = page.indexOf('Current guide price');
start = page.indexOf(TD, start);
var end = page.indexOf('</td>',start);
var number = page.substring (start + TD.length , end);
Logger.log(number);
var numericPart = number.substring(0, number.length -1);
var multiplierSymbol = number.substring(number.length -1 , number.length);
var multiplier = getMultiplyingFactor(multiplierSymbol);
var fullNumber = multiplier == 1 ? number : numericPart * multiplier;
Logger.log(fullNumber);
}
Certainly, not the optimal way of doing things but it works.
Basically I parse the html page as you did (with corrected regex) and split the string into number part and multiplicator (k = 1000). Finally I return the extracted number. This function can be used in Google Docs.
function pullRuneScape() {
var pageContent = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var matched = pageContent.match(/Current guide price:<.th>\n<td>(\d+\.*\d*)([k]{0,1})/);
var numberAsString = matched[1];
var multiplier = "";
if (matched.length == 3) {
multiplier = matched[2];
}
number = convertNumber(numberAsString, multiplier);
return number;
}
function convertNumber(numberAsString, multiplier) {
var number = Number(numberAsString);
if (multiplier == 'k') {
number *= 1000;
}
return number;
}
Let me explain.
I have to do some fuzzy matching for a company, so ATM I use a levenshtein distance calculator, and then calculate the percentage of similarity between the two terms. If the terms are more than 80% similar, Fuzzymatch returns "TRUE".
My problem is that I'm on an internship, and leaving soon. The people who will continue doing this do not know how to use excel with macros, and want me to implement what I did as best I can.
So my question is : however inefficient the function may be, is there ANY way to make a standard function in Excel that will calculate what I did before, without resorting to macros ?
Thanks.
If you came about this googling something like
levenshtein distance google sheets
I threw this together, with the code comment from milot-midia on this gist (https://gist.github.com/andrei-m/982927 - code under MIT license)
From Sheets in the header menu, Tools -> Script Editor
Name the project
The name of the function (not the project) will let you use the func
Paste the following code
function Levenshtein(a, b) {
if(a.length == 0) return b.length;
if(b.length == 0) return a.length;
// swap to save some memory O(min(a,b)) instead of O(a)
if(a.length > b.length) {
var tmp = a;
a = b;
b = tmp;
}
var row = [];
// init the row
for(var i = 0; i <= a.length; i++){
row[i] = i;
}
// fill in the rest
for(var i = 1; i <= b.length; i++){
var prev = i;
for(var j = 1; j <= a.length; j++){
var val;
if(b.charAt(i-1) == a.charAt(j-1)){
val = row[j-1]; // match
} else {
val = Math.min(row[j-1] + 1, // substitution
prev + 1, // insertion
row[j] + 1); // deletion
}
row[j - 1] = prev;
prev = val;
}
row[a.length] = prev;
}
return row[a.length];
}
You should be able to run it from a spreadsheet with
=Levenshtein(cell_1,cell_2)
While it can't be done in a single formula for any reasonably-sized strings, you can use formulas alone to compute the Levenshtein Distance between strings using a worksheet.
Here is an example that can handle strings up to 15 characters, it could be easily expanded for more:
https://docs.google.com/spreadsheet/ccc?key=0AkZy12yffb5YdFNybkNJaE5hTG9VYkNpdW5ZOWowSFE&usp=sharing
This isn't practical for anything other than ad-hoc comparisons, but it does do a decent job of showing how the algorithm works.
looking at the previous answers to calculating Levenshtein distance, I think it would be impossible to create it as a formula.
Take a look at the code here
Actually, I think I just found a workaround. I was adding it in the wrong part of the code...
Adding this line
} else if(b.charAt(i-1)==a.charAt(j) && b.charAt(i)==a.charAt(j-1)){
val = row[j-1]-0.33; //transposition
so it now reads
if(b.charAt(i-1) == a.charAt(j-1)){
val = row[j-1]; // match
} else if(b.charAt(i-1)==a.charAt(j) && b.charAt(i)==a.charAt(j-1)){
val = row[j-1]-0.33; //transposition
} else {
val = Math.min(row[j-1] + 1, // substitution
prev + 1, // insertion
row[j] + 1); // deletion
}
Seems to fix the problem. Now 'biulding' is 92% accurate and 'bilding' is 88%. (whereas with the original formula 'biulding' was only 75%... despite being closer to the correct spelling of building)