A loop within a loop in JavaScript - node.js

The solution to this may be simple but I cannot seem to think of an easier solution at present. I am using Excel.js and want to add values from an array to a specific set of cells. In this example E6:E25
The problem I face is that I cannot seem to select specific cells, only all of them in a column.
var array = ["4","3","3","3","3","8","2","4","5","2","2","6","4","3","8","5","4","3","2","2"]
workbook.xlsx.readFile('myFile.xlsx')
.then(function() {
var worksheet = workbook.getWorksheet(1);
var column = worksheet.getColumn(5) // Selecting specific column
column.eachCell(function(cell, rownumber){
// eachCell count is 122, ideally not wanting to iterate 122 times
// when only needing 20
switch (cell['_address']) {
case 'E6':
worksheet.getCell(cell['_address']).value = array[0];
break;
case 'E7':
worksheet.getCell(cell['_address']).value = array[1];
break;
case 'E8':
worksheet.getCell(cell['_address']).value = array[2];
break;
// and so on 20 times
default:
}
});
return workbook.xlsx.writeFile('newFile.xlsx');
})
This is awful I know; I am looking to refactor this and make it a lot cleaner and efficient.

Doesn't a simple for loop like this do the same?
var array = ["4","3","3","3","3","8","2","4","5","2","2","6","4","3","8","5","4","3","2","2"];
for (i=0; i<array.length; i++) {
col = i+6;
worksheet.getCell('E'+col).value = array[i];
}

Related

Get Last Column in Visible Views Index - Excel - Office-JS

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;
}

I cannot find out why this code keeps skipping a loop

Some background on what is going on:
We are processing addresses into standardized forms, this is the code to take addresses scored by how many components found and then rescore them using a levenshtein algorithm across similar post codes
The scores are how many components were found in that address divided by the number missed, to return a ratio
The input data, scoreDict, is a dictionary containing arrays of arrays. The first set of arrays is the scores, so there are 12 arrays because there are 12 scores in this file (it adjusts by file). There are then however many addresses fit that score in their own separate arrays stored in that. Don't ask me why I'm doing it that way, my brain is dead
The code correctly goes through each score array and each one is properly filled with the unique elements that make it up. It is not short by any amount, nothing is duplicated, I have checked
When we hit the score that is -1 (this goes to any address where it doesn't fit in some rule so we can't use its post code to find components so no components are found) the loop specifically ONLY DOES EVERY OTHER ADDRESS IN THIS SCORE ARRAY
It doesn't do this to any other score array, I have checked
I have tried changing the number to something else like 99, same issue except one LESS address got rescored, and the rest stayed at the original failing score of 99
I am going insane, can anyone find where in this loop something may be going wrong to cause it to only do every other line. The index counter of line and sc come through in the correct order and do not skip over. I have checked
I am sorry this is not professional, I have been at this one loop for 5 hours
Rescore: function Rescore(scoreDict) {
let tempInc = 0;
//Loop through all scores stored in scoreDict
for (var line in scoreDict) {
let addUpdate = "";
//Loop through each line stored by score
for (var sc in scoreDict[line.toString()]) {
console.log(scoreDict[line.toString()].length);
let possCodes = new Array();
const curLine = scoreDict[line.toString()][sc];
console.log(sc);
const curScore = curLine[1].split(',')[curLine[1].split(',').length-1];
switch (true) {
case curScore == -1:
let postCode = (new RegExp('([A-PR-UWYZ][A-HK-Y]?[0-9][A-Z0-9]?[ ]?[0-9][ABD-HJLNP-UW-Z]{2})', 'i')).exec(curLine[1].replace(/\\n/g, ','));
let areaCode;
//if (curLine.split(',')[curLine.split(',').length-2].includes("REFERENCE")) {
if ((postCode = (new RegExp('(([A-Z][A-Z]?[0-9][A-Z0-9]?(?=[ ]?[0-9][A-Z]{2}))|[0-9]{5})', 'i').exec(postCode))) !== null) {
for (const code in Object.keys(addProper)) {
leven.LoadWords(postCode[0], Object.keys(addProper)[code]);
if (leven.distance < 2) {
//Weight will have adjustment algorithms based on other factors
let weight = 1;
//Add all codes that are close to the same to a temp array
possCodes.push(postCode.input.replace(postCode[0], Object.keys(addProper)[code]).split(',')[0] + "(|W|)" + (leven.distance/weight));
}
}
let highScore = 0;
let candidates = new Array();
//Use the component script from cityprocess to rescore
for (var i=0;i<possCodes.length;i++) {
postValid.add([curLine[1].split(',').slice(0,curLine[1].split(',').length-2) + '(|S|)' + possCodes[i].split("(|W|)")[0]]);
if (postValid.addChunk[0].split('(|S|)')[postValid.addChunk[0].split('(|S|)').length-1] > highScore) {
candidates = new Array();
highScore = postValid.addChunk[0].split('(|S|)')[postValid.addChunk[0].split('(|S|)').length-1];
candidates.push(postValid.addChunk[0]);
} else if (postValid.addChunk[0].split('(|S|)')[postValid.addChunk[0].split('(|S|)').length-1] == highScore) {
candidates.push(postValid.addChunk[0]);
}
}
score.Rescore(curLine, sc, candidates[0]);
}
//} else if (curLine.split(',')[curLine.split(',').length-2].contains("AREA")) {
// leven.LoadWords();
//}
break;
case curScore > 0:
//console.log("That's a pretty good score mate");
break;
}
//console.log(line + ": " + scoreDict[line].length);
}
}
console.log(tempInc)
score.ScoreWrite(score.scoreDict);
}
The issue was that I was calling the loop on the array I was editing, so as each element got removed from the array (rescored and moved into a separate array) it got shorter by that element, resulting in an issue that when the first element was rescored and removed, and then we moved onto the second index which was now the third element, because everything shifted up by 1 index
I fixed it by having it simply enter an empty array for each removed element, so everything kept its index and the array kept its length, and then clear the empty values at a later time in the code

swift/parse: incrementing strings

The Swift part of the question:
So what I mean by incrementing strings is that say we start off with var string = "title" I want to be able to increment numbers to the end of that like "title1", "title2", "title3...". Should I use a for loop to do this? If so, how? Or another method?
for var i = 1; i < 6; i = i + 1 {
//increment the strings here
}
The Parse part of the question:
I want to have my objectForKey use the many different titles and numbers we will produce above so that the objectForKey will be "title1", "title2", "title3"... I would make multiple columns on Parse with names " title1, title2, title3 and the cells in the tableview would correspond to that data. So cell1 would use title1's data, cell2 will use title2's data and so on. Will it work like this?
var output1 = object.objectForKey(i) as! String
A loop in Swift is like for i in 1...5, and then you can use string interpolation to get the correct string like this:
for i in 1...5 {
let title = "title\(i)"
print(title)
}
Also read Dan's answer.
There are a few ways of looping in Swift, but you should keep in mind that and as of Swift 3,
this will no longer be one of them:
for var i = 0; i <6; i++ {
let string = "title\(i+1)"
}
source : Swift Evolution
Swift's preferred way of general looping is, as GvS stated:
for i in 1...5 {
let title = "title\(i)"
}
However, you are also welcome to use Swifts higher order functions to loop:
(1...5).forEach { i in
let title = "title \(i)"
}
or
(1...5).forEach { let title = "title \($0)" }

Grabbing text from webpage and storing as variable

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;
}

Is it possible to do a Levenshtein distance in Excel without having to resort to Macros?

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)

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