So I am getting some json data and putting it inside of a Mutable List. I have a class with id, listId, and name inside of it. Im trying to sort the output of the list by listId which are just integers and then also the name which has a format of "Item 123". Im doing the following
val sortedList = data.sortedWith(compareBy({ it.listId }, { it.name }))
This sorts the listId correctly but the names is sorted alphabetically so the numbers go 1, 13, 2, 3. How am I able to sort both the categories but make the "name" also be sorted numerically?
I think
val sortedList = data.sortedWith(compareBy(
{ it.listId },
{ it.name.substring(0, it.name.indexOf(' ')) },
{ it.name.substring(it.name.indexOf(' ') + 1).toInt() }
))
will work but it is not computationally efficient because it will call String.indexOf() many times.
If you have a very long list, you should consider making another list whose each item has String and Int names.
Related
I am trying to sort that simple list of users by "created". What am I doing wrong?
val user1 = User("2019-01-01 17:42:34")
val user2 = User("2019-01-02 17:42:34")
val user3 = User("2019-01-03 17:42:34")
val list = listOf(user2, user3, user1)
list.sortedWith(compareBy {
LocalDateTime.parse(
it.created,
DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss")
)
})
data class User(val created: String = "")
Your code works. The problem here is that sortedWith returns a new list with the results of the sorting! Check the documentation:
Returns a list of all elements sorted according to the specified [comparator].
The sort is stable. It means that equal elements preserve their order relative to each other after sorting.
So if you want to sort the collection itself you need to use a MutableList and sortWith:
val list = mutableListOf(User("2019-01-01 17:42:34"), User("2019-01-02 17:42:34"), User("2019-01-03 17:42:34"))
list.sortWith(compareBy {
LocalDateTime.parse(
it.created,
DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss"))
})
sortedWith function returns a new list with sorted elements without modifying the original collection. You probably want to use sortWith function of MutableList that sorts the original collection.
I want to sort a list of strings (with possibly duplicate entries) by using as ordering reference the order of the entries in another list. So, the following list is the list I want to sort
List<String> list = ['apple','pear','apple','x','x','orange','x','pear'];
And the list that specifies the order is
List<String> order = ['orange','apple','x','pear'];
And the output should be
List<String> result = ['orange','apple','apple','x','x','x','pear','pear'];
Is there a clean way of doing this?
I don't understand if I can use list's sort and compare with the following problem. I tried using map, iterable, intersection, etc.
There might be a more efficient way but at least you get the desired result:
main() {
List<String> list = ['apple','pear','apple','x','x','orange','x','pear'];
List<String> order = ['orange','apple','x','pear'];
list.sort((a, b) => order.indexOf(a).compareTo(order.indexOf(b)));
print(list);
}
Try it on DartPad
The closure passed to list.sort(...) is a custom comparer which instead of comparing the passed item, compares their position in order and returns the result.
Using a map for better lookup performance:
main() {
List<String> list = ['apple','pear','apple','x','x','orange','x','pear'];
List<String> orderList = ['orange','apple','x','pear'];
Map<String,int> order = new Map.fromIterable(
orderList, key: (key) => key, value: (key) => orderList.indexOf(key));
list.sort((a, b) => order[a].compareTo(order[b]));
print(list);
}
Try it on DartPad
I have a class
#Sortable(includes = ['date'])
class Item {
// other fields not relevant to this question
Date date
}
If I sort a List of these objects it will sort them in ascending order based on the date field. Is there a way to sort them in descending order instead? I know I could just call reverse() on the result of the ascending sort, but this seems a bit inefficient
Here are a couple of ways:
def items = [
new Item(date: new Date(40000)),
new Item(date: new Date(1000)),
new Item(date: new Date(200000)),
new Item(date: new Date(00100)),
]
items.sort { a, b -> b <=> a }
items.sort(true, Collections.reverseOrder())
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).
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...