So essentially I have two tables. Table 1 has a list of all attendants to a certain event. Table 2 has a list of all members of an organization that attended said event. I'm trying to copy a list of all non-members that attended the event. So the logic in my head is trying to loop through Table 2 and see if the value also exists in Table 1. If it does not, I'm trying to copy it into a list.
var attendants = currentS.getRange("M2:M").getValues(); //this is the list of all members that attended an event
for (var x = 2; x <= checkLast; x++) {
newcheck = currentS.getRange(x,5).getValue(); //this is getting the name of the attendants
if (attendants.indexOf(newcheck) == -1) {
var columnM = currentS.getRange("M1:M").getValues(); //trying to see if name of attendants is in the list of all members that attended the event.
var columnMlast = columnM.filter(String).length;
var final = currentS.getRange(columnMlast+1,13);
final.setValue(currentS.getRange(x,5).getValue()); //if it attendant is not in the list of members that attended, copies the name into a new list.
Whenever I run the code, I end up just getting the whole list of attendants without anything being filtered out. I hope I'm clear, and thanks in advance!!
Explanation:
You can use the filter method to find the items of attendants that are not included in the list of members and that will result in the list of new_members that will be appended at the bottom of members.
Using this solution you don't need for loops and most importantly you don't need to use setValue and getValue in a loop which is computationally expensive.
Solution:
I can't use your code because you have variables that aren't defined in the code snippet you provided.
I will show you an example (sheet and code) that you can use to adjust your current solution.
Example Script:
function myFunction() {
const ss = SpreadsheetApp.getActive();
const currentS = ss.getSheetByName("Sheet1");
const members = currentS.getRange("M2:M").getValues().flat().filter(r=>r!='');
const attendants = currentS.getRange("N2:N").getValues().flat().filter(r=>r!='');
const new_members = attendants.filter(a=>!members.includes(a)).map(nm=>[nm]);
console.log(new_members) // output: [ [ 'guest1' ], [ 'guest2' ], [ 'guest3' ], [ 'guest4' ] ]
currentS.getRange(members.length+2,13,new_members.length,1).setValues(new_members);
}
Example sheet (Input-Output):
For some reason the countries that seem to be returned are all returning in pairs? How can you change the code so it only returns the countries in 'Europe' once?
function newcountry(continent,country)
local object = {}
object.continent = continent
object.country = country
local list = {}
for i in pairs( object ) do
if object.continent == "Europe" then
table.insert(list, object.country)
print(object.country)
end
end
return object
end
a = newcountry("Africa","Algeria")
b = newcountry("Europe","England")
c = newcountry("Europe","France")
d = newcountry("Europe","Spain")
e = newcountry("Asia","China")
I'm not sure what you are trying to accomplish with this code, but to answer your question:
function newcountry(continent,country)
local object = {}
object.continent = continent
object.country = country
local list = {}
if object.continent == "Europe" then
table.insert(list, object.country)
print(object.country)
end
return object
end
This code will print countries in Europe just once. When there was loop in there, it printed name of the country twice, because it did it for each element of object table (continent and country, hence two times).
Generic for loops in Programming in Lua (first edition).
I would also like to point out that list is quite useless at the moment. It is not being returned and stays local. On top of that, every time you call newcountry there is new list created. They are all unique - country objects are not added to single list. But again - I don't know what you are trying to accomplish.
I'm trying to create two spreadsheets: one tracks student attendance at school, the other tracks their attendance at Track practice. The goal is to write a function, that I can set up as a button, that I can click that will automatically send emails to several people if the student is present at school but is absent for sports without getting excused.
Right now, the whole thing is working pretty well, but I have one issue. I have a column that will read "Good" or "Bad" depending on whether the student meets the above condition. The function turns these into an array. I would like to use the index of the "Bad"'s to find the necessary email addresses, which are stored at the same index point in another array that I make from the spreadsheet. I'm not sure how to save this index point and use it to reference the email addresses. Code below.
function sendEmailsMonday() {
var sheet = SpreadsheetApp.getActiveSpreadsheet().getSheetByName("TrackAttendance");
var dataRange = sheet.getRange("D2:D30");
var data = dataRange.getValues();// Gets array of "Good" and "Bad"
for (i in data) {
if(i = "Bad") {
var place = data.indexOf(i);
var dataRange2 = sheet.getRange("M2:M30");// Gets array of email addresses
var data2 = dataRange2.getValues();
var emailAddress = data2[place];
var message = "This is an automated email informing you that your child/advisee ____ was present at school today, but missed Track without being excused. Feel free to email Mr. # with any questions.";
var subject = "___ missed Track Practice";
MailApp.sendEmail(emailAddress, subject, message);
return;
}
}
}
So, the issue comes in with the index lines. If I get rid of
var place = data.indexOf(i);
and replace
var emailAddress = data[place];
with
var emailAddress = data[28];
or any other number, it will grab the email address and send it. But then it has nothing to do with the values in the other column.
Seems like this should be an easy fix but I'm bad at this.
Very late responding now. I think you are almost there.
Your IF statement should read:
if(i == "Bad") {
And then replace 'place' with i:
var emailAddress = data2[i];
It should work as expected now.
players={
{uuid = "abc",name = "TheEisbaer"},
{uuid= "def",name ="Eisbaer68"},
{uuid= "ghj",name = "kevin"}
}
list={name = "TheEisbaer", name = "Eisbaer68"}
With print(list.name) it prints "Eisbaer68".
How do I get it to print "TheEisbaer"?
-
Can you do something like print(players[3].uuid) or how would one do that?
If I understand (and believe me, it's not easy) what you're trying to do:
First of all:
list={name = "TheEisbaer", name = "Eisbaer68"}
is equivalent to
list={name = "Eisbaer68"}
or, more clearly, perhaps:
list={}
list.name = "TheEisbaer"
list.name = "Eisbaer68"
As you use the same key twice in the same table, the most recent key value overwrites the previous one.
It seems you want to achieve the equivalent of a 'join' in databases. If you want to display (or, in general, filter) those 'records' from players that have their 'name' appear also in list, maybe something like this:
players =
{
{uuid= 'abc', name = 'TheEisbaer' },
{uuid= 'def', name = 'Eisbaer68' },
{uuid= 'ghj', name = 'kevin' },
}
list = { TheEisbaer=true, Eisbaer68=true }
for _,t in ipairs(players) do
if list[t.name] then --if key exists in list table ...
print(t.name .. ' => ' .. t.uuid)
end
end
Here the list has keys to make it simpler to lookup. You could also do it differently but it would probably require a nested FOR loop to go through all possible combinations between the two tables.
I have been working on a project and Stack Overflow has helped me with a few problems so far, so I am very thankful!
My question is this:
I have an array like this:
var records:Object = {};
var arr:Array = [
records["nh"] = { medinc:66303, statename:"New Hampshire"},
records["ct"] = { medinc:65958, statename:"Connecticut"},
records["nj"] = { medinc:65173, statename:"New Jersey"},
records["md"] = { medinc:64596, statename:"Maryland"},
etc... for all 50 states. And then I have the array sorted reverse numerically (descending) like this:
arr.sortOn("medinc", Array.NUMERIC);
arr.reverse();
Can I call the name of the record (i.e. "nj" for new jersey) and then get the value from the numeric position above and below the record in the array?
Basically, medinc is medium income of US states, and I am trying to show a ranking system... a user would click Texas for example, and it would show the medinc value for Texas, along with the state the ranks one position below and the state that ranks one position above in the array.
Thanks for your help!
If you know the object, you can use the array.indexOf().
var index:int = records.indexOf(records["nj"]);
var above:Object;
var below:Object;
if(index + 1 < records.length){ //make sure your not already at the top
above = records[index+1];
}
if(index > 0){ //make sure your not already at the bottom
below = records[index-1];
}
I think this is the answer based on my understanding of your data.
var index:int = arr.indexOf(records["nh"]);
That will get you the index of the record that was clicked on and then for find the ones below and above just:
var clickedRecord:Object = arr[index]
var higherRecord:Object = arr[index++]
var lowerRecord:Object = arr[index--]
Hope that answers your question
Do you really need records to be hash?
If no, you can simply move key to record field and change records to simple array:
var records: Array = new Array();
records.push({ short: "nh", medinc:66303, statename:"New Hampshire"}),
records.push({ short: "ct", medinc:65958, statename:"Connecticut"}),
....
This gives you opportunity to create class for State, change Array to Vector and make all of this type-safe, what is always good.
If you really need those keys, you can add objects like above (with "short" field) in the same way you are doing it now (maybe using some helper function which will help to avoid typing shortname twice, like addState(records, data) { records[data.short] = data }).
Finally, you can also keep those records in two objects (or an object and an array or whatever you need). This will not be expensive, if you will create state object once and keep references in array/object/vector. It would be nice idea if you need states sorted on different keys often.
This is not really a good way to have your data set up - too much typing (you are repeating "records", "medinc", "statename" over and over again, while you definitely could've avoided it, for example:
var records:Array = [];
var states:Array = ["nh", "ct", "nj" ... ];
var statenames:Array = ["New Hampshire", "Connecticut", "New Jersey" ... ];
var medincs:Array = [66303, 65958, 65173 ... ];
var hash:Object = { };
function addState(state:String, medinc:int, statename:String, hash:Object):Object
{
return hash[state] = { medinc: medinc, statename: statename };
}
for (var i:int; i < 50; i++)
{
records[i] = addState(states[i], medincs[i], statenames[i], hash);
}
While you have done it already the way you did, that's not essential, but this could've saved you some keystrokes, if you haven't...
Now, onto your search problem - first of all, true, it would be worth to sort the array before you search, but if you need to search an array by the value of the parameter it was sorted on, there is a better algorithm for that. That is, if given the data in your example, your specific task was to find out in what state the income is 65958, then, knowing that array is sorted on income you could employ binary search.
Now, for the example with 50 states the difference will not be noticeable, unless you do it some hundreds of thousands times per second, but in general, the binary search would be the way to go.
If the article in Wiki looks too long to read ;) the idea behind the binary search is that at first you guess that the searched value is exactly in the middle of the array - you try that assumption and if you guessed correct, return the index you just found, else - you select the interval containing the searched value (either one half of the array remaining) and do so until you either find the value, or check the same index - which would mean that the value is not found). This reduces asymptotic complexity of the algorithm from O(n) to O(log n).
Now, if your goal was to find the correspondence between the income and the state, but it wasn't important how that scales with other states (i.e. the index in the array is not important), you could have another hash table, where the income would be the key, and the state information object would be the value, using my example above:
function addState(state:String, medinc:int, statename:String,
hash:Object, incomeHash:Object):Object
{
return incomeHash[medinc] =
hash[state] = { medinc: medinc, statename: statename };
}
Then incomeHash[medinc] would give you the state by income in O(1) time.