I have a table that is filled with random content that a user enters. I want my users to be able to rapidly search through this table, and one way of facilitating their search is by sorting the table alphabetically. Originally, the table looked something like this:
myTable = {
Zebra = "black and white",
Apple = "I love them!",
Coin = "25cents"
}
I was able to implement a pairsByKeys() function which allowed me to output the tables contents in alphabetical order, but not to store them that way. Because of the way the searching is setup, the table itself needs to be in alphabetical order.
function pairsByKeys (t, f)
local a = {}
for n in pairs(t) do
table.insert(a, n)
end
table.sort(a, f)
local i = 0 -- iterator variable
local iter = function () -- iterator function
i = i + 1
if a[i] == nil then
return nil
else
return a[i], t[a[i]]
end
end
return iter
end
After a time I came to understand (perhaps incorrectly - you tell me) that non-numerically indexed tables cannot be sorted alphabetically. So then I started thinking of ways around that - one way I thought of is sorting the table and then putting each value into a numerically indexed array, something like below:
myTable = {
[1] = { Apple = "I love them!" },
[2] = { Coin = "25cents" },
[3] = { Zebra = "black and white" },
}
In principle, I feel this should work, but for some reason I am having difficulty with it. My table does not appear to be sorting. Here is the function I use, with the above function, to sort the table:
SortFunc = function ()
local newtbl = {}
local t = {}
for title,value in pairsByKeys(myTable) do
newtbl[title] = value
tinsert(t,newtbl[title])
end
myTable = t
end
myTable still does not end up being sorted. Why?
Lua's table can be hybrid. For numerical keys, starting at 1, it uses a vector and for other keys it uses a hash.
For example, {1="foo", 2="bar", 4="hey", my="name"}
1 & 2, will be placed in a vector, 4 & my will be placed in a hashtable. 4 broke the sequence and that's the reason for including it into the hashtable.
For information on how to sort Lua's table take a look here: 19.3 - Sort
Your new table needs consecutive integer keys and needs values themselves to be tables. So you want something on this order:
SortFunc = function (myTable)
local t = {}
for title,value in pairsByKeys(myTable) do
table.insert(t, { title = title, value = value })
end
myTable = t
return myTable
end
This assumes that pairsByKeys does what I think it does...
Related
Disclaimer: This is Glua (Lua used by Garry's Mod)
I just need to compare tables between them and return the difference, like if I was substrating them.
TableOne = {thing = "bob", 89 = 1, 654654 = {"hi"}} --Around 3k items like that
TableTwo = {thing = "bob", 654654 = "hi"} --Same, around 3k
function table.GetDifference(t1, t2)
local diff = {}
for k, dat in pairs(t1) do --Loop through the biggest table
if(!table.HasValue(t2, t1[k])) then --Checking if t2 hasn't the value
table.insert(diff, t1[k]) --Insert the value in the difference table
print(t1[k])
end
end
return diff
end
if table.Count(t1) != table.Count(t2) then --Check if amount is equal, in my use I don't need to check if they are exact.
PrintTable(table.GetDifference(t1, t2)) --Print the difference.
end
My problem being that with only one of difference between the two tables, this returns me more than 200 items. The only item I added was a String. I tried many other functions like this one but they usually cause a stack overflow error because of the table's length.
Your problem is with this line
if(!table.HasValue(t2, t1[k])) then --Checking if t2 hasn't the value
Change it to this:
if(!table.HasValue(t2, k) or t1[k] != t2[k]) then --Checking if t2[k] matches
Right now what is happening is that you're looking at an entry like thing = "bob" and then you're looking to see whether t2 has "bob" as a key. And it doesn't. But neither did t1 so that shouldn't be regarded as a difference.
I have a question about the way I put together this piece of Lua code.
Say, there is a function like the one below, containing 200 myTable tables, where the names are ordered alphabetically:
function loadTable(x)
local myTable
if x == "aaron" then myTable = {1,2,3,4,5,6,7,8,9,0}
elseif x == "bobby" then myTable = {1,3,3,4,5,8,7,8,9,1}
elseif x == "cory" then myTable = {1,2,3,3,3,6,7,8,9,2}
elseif x == "devin" then myTable = {1,2,3,4,5,2,3,4,9,0}
...
else
print("table not available")
end
return myTable
end
and now I want to find the table corresponding to x == "zac" (which happens to be somewhere at the end). I use this line of code:
local foundTable = loadTable("zac")
Isnt this like not efficient at all? If it has to find the table at the very end of the function it has to go through all the previous lines of code. Is there some way to code this more efficiently in lua and find the correct table faster? ?
This can become a lot faster by using... a table!
Simply make a table whose keys are the name of the person and the values are the table you want to load, like this:
local tables = {
john = {1,2,3,4,5,6,7,8,9,0},
peter = {1,3,3,4,5,8,7,8,9,1},
william = {1,2,3,3,3,6,7,8,9,2},
victoria = {1,2,3,4,5,2,3,4,9,0}
--...
}
Then, instead of calling loadTable("richard") simply use tables["richard"] or tables.richard if the key is a valid identifier
Lets say I have a table like so:
{
value = 4
},
{
value = 3
},
{
value = 1
},
{
value = 2
}
and I want to iterate over this and print the value in order so the output is like so:
1
2
3
4
How do I do this, I understand how to use ipairs and pairs, and table.sort, but that only works if using table.insert and the key is valid, I need to be loop over this in order of the value.
I tried a custom function but it simply printed them in the incorrect order.
I have tried:
Creating an index and looping that
Sorting the table (throws error: attempt to perform __lt on table and table)
And a combination of sorts, indexes and other tables that not only didn't work, but also made it very complicated.
I am well and truly stumped.
Sorting the table
This was the right solution.
(throws error: attempt to perform __lt on table and table)
Sounds like you tried to use a < b.
For Lua to be able to sort values, it has to know how to compare them. It knows how to compare numbers and strings, but by default it has idea how to compare two tables. Consider this:
local people = {
{ name = 'fred', age = 43 },
{ name = 'ted', age = 31 },
{ name = 'ned', age = 12 },
}
If I call sort on people, how can Lua know what I intend? I doesn't know what 'age' or 'name' means or which I'd want to use for comparison. I have to tell it.
It's possible to add a metatable to a table which tells Lua what the < operator means for a table, but you can also supply sort with a callback function that tells it how to compare two objects.
You supply sort with a function that receives two values and you return whether the first is "less than" the second, using your knowledge of the tables. In the case of your tables:
table.sort(t, function(a,b) return a.value < b.value end)
for i,entry in ipairs(t) do
print(i,entry.value)
end
If you want to leave the original table unchanged, you could create a custom 'sort by value' iterator like this:
local function valueSort(a,b)
return a.value < b.value;
end
function sortByValue( tbl ) -- use as iterator
-- build new table to sort
local sorted = {};
for i,v in ipairs( tbl ) do sorted[i] = v end;
-- sort new table
table.sort( sorted, valueSort );
-- return iterator
return ipairs( sorted );
end
When sortByValue() is called, it clones tbl to a new sorted table, and then sorts the sorted table. It then hands the sorted table over to ipairs(), and ipairs outputs the iterator to be used by the for loop.
To use:
for i,v in sortByValue( myTable ) do
print(v)
end
While this ensures your original table remains unaltered, it has the downside that each time you do an iteration the iterator has to clone myTable to make a new sorted table, and then table.sort that sorted table.
If performance is vital, you can greatly speed things up by 'caching' the work done by the sortByValue() iterator. Updated code:
local resort, sorted = true;
local function valueSort(a,b)
return a.value < b.value;
end
function sortByValue( tbl ) -- use as iterator
if not sorted then -- rebuild sorted table
sorted = {};
for i,v in ipairs( tbl ) do sorted[i] = v end;
resort = true;
end
if resort then -- sort the 'sorted' table
table.sort( sorted, valueSort );
resort = false;
end
-- return iterator
return ipairs( sorted );
end
Each time you add or remove an element to/from myTable set sorted = nil. This lets the iterator know it needs to rebuild the sorted table (and also re-sort it).
Each time you update a value property within one of the nested tables, set resort = true. This lets the iterator know it has to do a table.sort.
Now, when you use the iterator, it will try and re-use the previous sorted results from the cached sorted table.
If it can't find the sorted table (eg. on first use of the iterator, or because you set sorted = nil to force a rebuild) it will rebuild it. If it sees it needs to resort (eg. on first use, or if the sorted table was rebuilt, or if you set resort = true) then it will resort the sorted table.
Is it possible for a table, when referenced without a key, to return a particular value rather than a reference to itself?
Let's say I have the following table:
local person = {
name = "Kapulani",
level = 100,
age = 30,
}
In Lua, I can quite easily refer to "person.name", "person.level", or "person.age" and get the values as expected. However, I have certain cases where I may want to just reference "person" and, instead of getting "table: " I'd like to return the value of "person.name" instead.
In other words, I'd like person.x (or person[x]) to return the appropriate entry from the table, but person without a key to return the value of person.name (or person["name"]). Is there a mechanism for this that I haven't been able to find?
I have had no success with metatables, since __index will only apply to cases where the key does not exist. If I put "person" into a separate table, I can come up with:
local true_person = {
... -- as above
}
local env_mt = {
__index = function(t, k)
if k == 'person' then
return true_person
end
end
}
local env = setmetatable( {}, env_mt )
This lets me use __index to do some special handling, except there's no discernable way for me to tell, from __index(), whether I'm getting a request for env.person (where I'd want to return true_person.name) or env.person[key] (where I'd want to return true_person as a table, so that 'key' can be accessed appropriately).
Any thoughts? I can approach this differently, but hoping I can approach this along these lines.
You can do it when the table is being used as a string by setting the __tostring metatable entry:
$ cat st.lua
local person = {
name = "Kapulani",
level = 100,
age = 30,
}
print(person)
print(person.name)
print(person.age)
setmetatable(person, {__tostring = function(t) return t.name end})
print(person)
$ lua st.lua
lua st.lua
table: 0x1e8478e0
Kapulani
30
Kapulani
I am not sure that what you are asking for is a good idea because it flies in the face of compositionality. Usually one would expect the following two programs to do the same thing but you want them to behave differently
print(person.name)
local p = person
print( p.name )
Its also not very clear how assignment would work. person.age = 10 should change the age but person = otherPerson should change the reference to the perrson, not the age.
If you don't care about compositionality and are onyl reading data, then a more direct way to solve the problem is to have a query function that receives the fields encoded in a string
query("person.age") -- 17
query("person.name") -- "hugomg"
query("person") -- 17; query gets to default to whatever it wants.
To keep the syntax more lightweight you can omit the optional parenthesis
q"person.age"
q"person"
Or you can extend the __index metamethod on the global table, _G
setmetattable(_G, { __index = function(self, key) return query(key) end })
print ( person_age ) -- You will need to use "_" instead of "." for the
-- query to be a valid identifier.
I have a spreadsheet that I update on a regular basis. I also have to re-sort the spreadsheet when finished because of the changes made. I need to sort with multiple criteria like the below settings. I have searched for examples but my Google search skills have failed me.
Sort range from A1:E59
[x] Data has header rows
sort by "Priority" A > Z
then by "Open" Z > A
then by "Project" A > Z
Mogsdad's answer works fine if none of your cells have values automatically calculated via a formula. If you do use formulas, though, then that solution will erase all of them and replace them with static values. And even so, it is more complicated than it needs to be, as there's now a built-in method for sorting based on multiple columns. Try this instead:
function onEdit(e) {
var priorityCol = 1;
var openCol = 2;
var projectCol = 3;
var sheet = SpreadsheetApp.getActiveSheet();
var dataRange = sheet.getDataRange();
dataRange.sort([
{column: priorityCol, ascending: true},
{column: openCol, ascending: false},
{column: projectCol, ascending: true}
]);
}
Instead of making a separate function, you can use the built-in onEdit() function, and your data will automatically sort itself when you change any of the values. The sort() function accepts an array of criteria, which it applies one after the other, in order.
Note that with this solution, the first column in your spreadsheet is column 1, whereas if you're doing direct array accesses like in Mogsdad's answer, the first column is column 0. So your numbers will be different.
That is a nice specification, a great place to start!
Remember that Google Apps Script is, to a large extent, JavaScript. If you extend your searching into JavaScript solutions, you'll find plenty of examples of array sorts here on SO.
As it happens, much of what you need is in Script to copy and sort form submission data. You don't need the trigger part, but the approach to sorting can be easily adapted to handle multiple columns.
The workhorse here is the comparison function-parameter, which is used by the JavaScript Array.sort() method. It works through the three columns you've indicated, with ascending or descending comparisons. The comparisons used here are OK for Strings, Numbers and Dates. It could be improved with some cleaning up, or even generalized, but it should be pretty fast as-is.
function sortMySheet() {
var sheet = SpreadsheetApp.getActiveSheet();
var dataRange = sourceSheet.getDataRange();
var data = dataRange.getValues();
var headers = data.splice(0,1)[0]; // remove headers from data
data.sort(compare); // Sort 2d array
data.splice(0,0,headers); // replace headers
// Replace with sorted values
dataRange.setValues(data);
};
// Comparison function for sorting two rows
// Returns -1 if 'a' comes before 'b',
// +1 if 'b' before 'a',
// 0 if they match.
function compare(a,b) {
var priorityCol = 0; // Column containing "Priority", 0 is A
var openCol = 1;
var projectCol = 2;
// First, compare "Priority" A > Z
var result = (a[priorityCol] > b[priorityCol] ) ?
(a[priorityCol] < b[priorityCol] ? -1 : 0) : 1;
if (result == 0) {
// "Priority" matched. Then compare "Open" Z > A
result = (b[openCol] > a[openCol] ) ?
(b[openCol] < a[openCol] ? -1 : 0) : 1;
}
if (result == 0) {
// "Open" matched. Finally, compare "Project" A > Z
result = (a[projectCol] > b[projectCol] ) ?
(a[projectCol] < b[projectCol] ? -1 : 0) : 1;
}
return result;
}
Try this using the Apps Script sort instead of the native JavaScript. I had the same issue with sorting the header row(s) and this solved the issue.
So I think something like this should work:
function onOpen() {
SpreadsheetApp.getActiveSpreadsheet()
.getSheetByName("Form Responses 1").sort(2);
}
Regarding sorting by multiple columns, you can chain that sort() method, with the final sort() having the highest priority, and the first sort() the lowest. So something like this should sort by Start date, then by End date:
function onOpen() {
SpreadsheetApp.getActiveSpreadsheet()
.getSheetByName("Form Responses 1").sort(3).sort(2);
}
Reference link:-
https://support.google.com/docs/thread/16556745/google-spreadsheet-script-how-to-sort-a-range-of-data?hl=en
Not sure if this is still relevant, but you can use the sort() function to define another tab as a sorted version of the original data.
Say your original data is in a tab named Sheet1; I'm also going to act as though your Priority, Open, and Project columns are A, B, and C, respectively.
Create a new tab, and in cell A1 type:
=sort(Sheet1!A1:E59, 1, TRUE, 2, FALSE, 3, TRUE)
The first argument specifies the sheet and range to be sorted, followed by three pairs: the first of each pair specifies the column (A=1, B=2, etc.), and the second specifies ascending (TRUE) or descending (FALSE).