So I have the following array
[true,true,false,false,false,true,false,false,true,true].
How can I use ramda to make the array look as follows
[true,false,true,false,true]
Also if the data was instead
[{t: '2018-10/09', v:true},{t: '2018-10/08', v:true}, {t: '2018-10/07', v:false},{t: '2018-10/06', v:false},{t: '2018-10/05', v:false},{t: '2018-10/04', v:true},{t: '2018-10/03', v:false},{t: '2018-10/02', v:false},{t: '2018-10/01', v:true},{t: '2018-09/09', v:true}]
and I wanted to dropRepeats using v only, then I would use R.dropRepeats(bool.v) ?
Also as a side note, I am using R.pipe to get the input and do the transformations on the data.
Ramda happens to have a built-in function for that, dropRepeats
const bools = [true,true,false,false,false,true,false,false,true,true]
console.log(R.dropRepeats(bools))
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.js"></script>
Update
A comment asked how I would do this on a slightly more complex structure. It would be much the same, using dropRepeatsWith and eqProps:
const bools = [{t: '2018-10/09', v:true},{t: '2018-10/08', v:true}, {t: '2018-10/07', v:false},{t: '2018-10/06', v:false},{t: '2018-10/05', v:false},{t: '2018-10/04', v:true},{t: '2018-10/03', v:false},{t: '2018-10/02', v:false},{t: '2018-10/01', v:true},{t: '2018-09/09', v:true}]
console.log(R.dropRepeatsWith(R.eqProps('v'), bools))
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.js"></script>
I would suggest a solution using groupWith:
You use groupWith to group consecutive identical elements into sub arrays, then you use head in map to take the first element of each sub array.
const prune = R.pipe(R.groupWith(R.equals), R.map(R.head));
console.log(
prune([true,true,false,false,false,true,false,false,true,true]));
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.js"></script>
Related
My apologies if this is a simple basic info that I should be knowing. This is the first time I am trying to use Java 8 streams and other features.
I have two ArrayLists containing same type of objects. Let's say list1 and list2. Let's say the lists has Person objects with a property "employeeId".
The scenario is that I need to merge these lists. However, list2 may have some objects that are same as in list1. So I am trying to remove the objects from list2 that are same as in list1 and get a result list that then I can merge in list1.
I am trying to do this with Java 8 removeIf() and stream() features. Following is my code:
public List<PersonDto> removeDuplicates(List<PersonDto> list1, List<PersonDto> list2) {
List<PersonDto> filteredList = list2.removeIf(list2Obj -> {
list1.stream()
.anyMatch( list1Obj -> (list1Obj.getEmployeeId() == list2Obj.getEmployeeId()) );
} );
}
The above code is giving compile error as below:
The method removeIf(Predicate) in the type Collection is not applicable for the arguments (( list2Obj) -> {})
So I changed the list2Obj at the start of "removeIf()" to (<PersonDto> list2Obj) as below:
public List<PersonDto> removeDuplicates(List<PersonDto> list1, List<PersonDto> list2) {
List<PersonDto> filteredList = list2.removeIf((<PersonDto> list2Obj) -> {
list1.stream()
.anyMatch( list1Obj -> (list1Obj.getEmployeeId() == list2Obj.getEmployeeId()) );
} );
}
This gives me an error as below:
Syntax error on token "<", delete this token for the '<' in (<PersonDto> list2Obj) and Syntax error on token(s), misplaced construct(s) for the part from '-> {'
I am at loss on what I really need to do to make it work.
Would appreciate if somebody can please help me resolve this issue.
I've simplified your function just a little bit to make it more readable:
public static List<PersonDto> removeDuplicates(List<PersonDto> left, List<PersonDto> right) {
left.removeIf(p -> {
return right.stream().anyMatch(x -> (p.getEmployeeId() == x.getEmployeeId()));
});
return left;
}
Also notice that you are modifying the left parameter, you are not creating a new List.
You could also use: left.removeAll(right), but you need equals and hashcode for that and it seems you don't have them; or they are based on something else than employeeId.
Another option would be to collect those lists to a TreeSet and use removeAll:
TreeSet<PersonDto> leftTree = left.stream()
.collect(Collectors.toCollection(() -> new TreeSet<>(Comparator.comparing(PersonDto::getEmployeeId))));
TreeSet<PersonDto> rightTree = right.stream()
.collect(Collectors.toCollection(() -> new TreeSet<>(Comparator.comparing(PersonDto::getEmployeeId))));
leftTree.removeAll(rightTree);
I understand you are trying to merge both lists without duplicating the elements that belong to the intersection. There are many ways to do this. One is the way you've tried, i.e. remove elements from one list that belong to the other, then merge. And this, in turn, can be done in several ways.
One of these ways would be to keep the employee ids of one list in a HashSet and then use removeIf on the other list, with a predicate that checks whether each element has an employee id that is contained in the set. This is better than using anyMatch on the second list for each element of the first list, because HashSet.contains runs in O(1) amortized time. Here's a sketch of the solution:
// Determine larger and smaller lists
boolean list1Smaller = list1.size() < list2.size();
List<PersonDto> smallerList = list1Smaller ? list1 : list2;
List<PersonDto> largerList = list1Smaller ? list2 : list1;
// Create a Set with the employee ids of the larger list
// Assuming employee ids are long
Set<Long> largerSet = largerList.stream()
.map(PersonDto::getEmployeeId)
.collect(Collectors.toSet());
// Now remove elements from the smaller list
smallerList.removeIf(dto -> largerSet.contains(dto.getEmployeeId()));
The logic behind this is that HashSet.contains will take the same time for both a large and a small set, because it runs in O(1) amortized time. However, traversing a list and removing elements from it will be faster on smaller lists.
Then, you are ready to merge both lists:
largerList.addAll(smallerList);
I have a helper that loops through jason data till a given value and sends the data back to the template.I also want to show at what location the data is present (the index), is there any way where in i can return the Index value along with the data?
Handlebars.registerHelper('print_range', function(items,count,options)
{
var out = "";
for(var i=0, l=items.length; i<count; i++)
{
out = out + options.fn(items[i]);
}
return out;
});
<script id="template" type="text/x-handlebars-template">
{{#print_range options "2"}}
<h1>index</h1> // this index should correspond to i in the helper function
<h2>{{optionID}}{{nextID}}</h2>
{{/print_range}}
</script>
Thanks in advance.
Handlebars.registerHelper('print_range', function(items,count,options)
{{#print_range options "2"}}
Not sure 100% what you're asking, but i do see a problem in your code. You specify 3 parameters in the function, but only give it 2 in your statement call to the helper.
{{#print_range options "2"}}
^ fn name ^items ^ count
Where's the value for your options variable? In handlebars, the parameters for the function go in order after the helper name you are calling
I know variants of this question have been asked before (even by me), but I still don't understand a thing or two about this...
It was my understanding that one could retrieve more documents than the 128 default setting by doing this:
session.Advanced.MaxNumberOfRequestsPerSession = int.MaxValue;
And I've learned that a WHERE clause should be an ExpressionTree instead of a Func, so that it's treated as Queryable instead of Enumerable. So I thought this should work:
public static List<T> GetObjectList<T>(Expression<Func<T, bool>> whereClause)
{
using (IDocumentSession session = GetRavenSession())
{
return session.Query<T>().Where(whereClause).ToList();
}
}
However, that only returns 128 documents. Why?
Note, here is the code that calls the above method:
RavenDataAccessComponent.GetObjectList<Ccm>(x => x.TimeStamp > lastReadTime);
If I add Take(n), then I can get as many documents as I like. For example, this returns 200 documents:
return session.Query<T>().Where(whereClause).Take(200).ToList();
Based on all of this, it would seem that the appropriate way to retrieve thousands of documents is to set MaxNumberOfRequestsPerSession and use Take() in the query. Is that right? If not, how should it be done?
For my app, I need to retrieve thousands of documents (that have very little data in them). We keep these documents in memory and used as the data source for charts.
** EDIT **
I tried using int.MaxValue in my Take():
return session.Query<T>().Where(whereClause).Take(int.MaxValue).ToList();
And that returns 1024. Argh. How do I get more than 1024?
** EDIT 2 - Sample document showing data **
{
"Header_ID": 3525880,
"Sub_ID": "120403261139",
"TimeStamp": "2012-04-05T15:14:13.9870000",
"Equipment_ID": "PBG11A-CCM",
"AverageAbsorber1": "284.451",
"AverageAbsorber2": "108.442",
"AverageAbsorber3": "886.523",
"AverageAbsorber4": "176.773"
}
It is worth noting that since version 2.5, RavenDB has an "unbounded results API" to allow streaming. The example from the docs shows how to use this:
var query = session.Query<User>("Users/ByActive").Where(x => x.Active);
using (var enumerator = session.Advanced.Stream(query))
{
while (enumerator.MoveNext())
{
User activeUser = enumerator.Current.Document;
}
}
There is support for standard RavenDB queries, Lucence queries and there is also async support.
The documentation can be found here. Ayende's introductory blog article can be found here.
The Take(n) function will only give you up to 1024 by default. However, you can change this default in Raven.Server.exe.config:
<add key="Raven/MaxPageSize" value="5000"/>
For more info, see: http://ravendb.net/docs/intro/safe-by-default
The Take(n) function will only give you up to 1024 by default. However, you can use it in pair with Skip(n) to get all
var points = new List<T>();
var nextGroupOfPoints = new List<T>();
const int ElementTakeCount = 1024;
int i = 0;
int skipResults = 0;
do
{
nextGroupOfPoints = session.Query<T>().Statistics(out stats).Where(whereClause).Skip(i * ElementTakeCount + skipResults).Take(ElementTakeCount).ToList();
i++;
skipResults += stats.SkippedResults;
points = points.Concat(nextGroupOfPoints).ToList();
}
while (nextGroupOfPoints.Count == ElementTakeCount);
return points;
RavenDB Paging
Number of request per session is a separate concept then number of documents retrieved per call. Sessions are short lived and are expected to have few calls issued over them.
If you are getting more then 10 of anything from the store (even less then default 128) for human consumption then something is wrong or your problem is requiring different thinking then truck load of documents coming from the data store.
RavenDB indexing is quite sophisticated. Good article about indexing here and facets here.
If you have need to perform data aggregation, create map/reduce index which results in aggregated data e.g.:
Index:
from post in docs.Posts
select new { post.Author, Count = 1 }
from result in results
group result by result.Author into g
select new
{
Author = g.Key,
Count = g.Sum(x=>x.Count)
}
Query:
session.Query<AuthorPostStats>("Posts/ByUser/Count")(x=>x.Author)();
You can also use a predefined index with the Stream method. You may use a Where clause on indexed fields.
var query = session.Query<User, MyUserIndex>();
var query = session.Query<User, MyUserIndex>().Where(x => !x.IsDeleted);
using (var enumerator = session.Advanced.Stream<User>(query))
{
while (enumerator.MoveNext())
{
var user = enumerator.Current.Document;
// do something
}
}
Example index:
public class MyUserIndex: AbstractIndexCreationTask<User>
{
public MyUserIndex()
{
this.Map = users =>
from u in users
select new
{
u.IsDeleted,
u.Username,
};
}
}
Documentation: What are indexes?
Session : Querying : How to stream query results?
Important note: the Stream method will NOT track objects. If you change objects obtained from this method, SaveChanges() will not be aware of any change.
Other note: you may get the following exception if you do not specify the index to use.
InvalidOperationException: StreamQuery does not support querying dynamic indexes. It is designed to be used with large data-sets and is unlikely to return all data-set after 15 sec of indexing, like Query() does.
I would like to update specific parts of a Grid dynamically in different ways. Consider the following toy example: I have two rows: one must be updated one-by-one (a, b, c), as these symbols depend on different triggers; the second row depends on one single trigger (show) that allows displaying/hiding some data.
Now I know that I can wrap the whole Grid structure into Dynamic, and even specify which symbols to track, thus this example does what I want:
Checkbox[Dynamic[show]]
test = {0, 0};
Dynamic[Grid[{{Dynamic#a, Dynamic#b, Dynamic#c},
If[show, Prepend[test, "test:"], {}]}, Frame -> All],
TrackedSymbols :> {show}]
Though for certain reasons I would like to have a locally specified Dynamic, that is only applied to the second row of the Grid.
For those who are wondering what ungodly situation would it be, just imagine the followings: show is used in any of a, b or c, and these I do NOT want to update when show is changing, their changes depend on other triggers. Why not remove then show from the symbols of the first row? Imagine, I can't, as show is present in a function that is used in a, b or c, and this function I cannot access easily.
Of course wrapping the first argument of If into Dynamic won't help here, as the Grid itself or any of its cells won't become dynamic:
Grid[{
{Dynamic#a, Dynamic#b, Dynamic#c},
If[Dynamic#show, Prepend[test, "test:"], {}]
}, Frame -> All]
Furthermore, wrapping a row into Dynamic makes the given row invalid, as it does not have head List anymore:
Grid[{
{Dynamic#a, Dynamic#b, Dynamic#c},
Dynamic#If[show, Prepend[test, "test:"], {}]
}, Frame -> All]
Mapping Dynamic over the row does not work either because show is not updated dynamically:
Grid[{
{Dynamic#a, Dynamic#b, Dynamic#c},
Dynamic /# If[show, Prepend[test, "test:"], {}]
}, Frame -> All]
Also, wrapping Dynamic[If[...]] around list members work, but now I have to evaluate If 3 times instead of just 1.
Grid[{
{Dynamic#a, Dynamic#b, Dynamic#c},
Dynamic[If[show, #, ""]] & /# Prepend[test, "test:"]
}, Frame -> All]
Would like to know if there is any solution to overcome this particular problem by locally applying a Dynamic wrapper on a row.
Here is a solution using the Experimental ValueFunction
show = True;
test = {0, 0};
Checkbox[Dynamic[show]]
Now write your own little Dynamic update function on the side
Needs["Experimental`"];
row = {};
updateRow[x_, v_] := row = If[v, Prepend[test, "test:"], {}];
ValueFunction[show] = updateRow;
Now make the Grid, and now can use Dynamic on EACH row, not around the whole Grid, which is what you wanted:
Grid[{
{Dynamic#a, Dynamic#b, Dynamic#c},
{Dynamic#row}
},
Frame -> All
]
ps. I just read a post here by telefunkenvf14 that mentions this package and this function, which I did not know about, and when I saw this function, I remembered this question, and I thought it should be possible to use that function to solve this problem.
ps. I need to work more on placing the grid row correctly....
update(1)
I can't figure how to splice the final row over the columns in the grid. Which is strange, as it has List head, yet it won't go across all the columns. It will only go in the first cell. Tried Sequence, SpanFromLeft, and such, but no luck. May be someone can figure this part out.
Here is my current trial:
Needs["Experimental`"];
row = {};
updateRow[x_, v_] := row = If[v, {"test:", 0, 0}, {}];
ValueFunction[show] = updateRow;
show = False;
Checkbox[Dynamic[show]]
f = Grid[{
{Dynamic#a, Dynamic#b, Dynamic#c},
List#Dynamic[row]
},
Frame -> All
]
It seems it should be doable. I do not see what is the problem now...
update(2)
As a temporary solution, I split the second row by force before hand. This made it possible to do what I want. Not sure if this meets the OP specifications or not (my guess is that it does not), but here it is:
Needs["Experimental`"];
ra = 0;
rb = 0;
rc = 0;
updateRow[x_, v_] :=
row = If[v, ra = "test:"; rb = 0; rc = 0, ra = ""; rb = ""; rc = ""]
ValueFunction[show] = updateRow;
show = False;
Checkbox[Dynamic[show]]
f = Grid[{
{Dynamic#a, Dynamic#b, Dynamic#c},
{Dynamic#ra, Dynamic#rb, Dynamic#rc}
},
Frame -> All]
This is actually a comment on #Nasser's solution and suggested fix to avoid manual splitting of the second row, but because of space limitations in the comment area, I post it as answer. Will be happy to delete it as soon as Nasser confirms that it works and incorporates it into his answer.
The clue to a solution is found in the Possible Issues section of Item in the documentation:
If Item is not the top-most item in the child of a function that supports Item, it will not work.
I use this to modify #Nasser's solution in the following way. First, I need to change the definition of row so that for both values of show the length of row is the same.
Needs["Experimental`"];
row = {"", "", ""};
updateRow[x_, v_] := row = If[v, Prepend[test, "test:"], {"", "", ""}];
Experimental`ValueFunction[show] = updateRow;
The second change needed is to wrap each element of Dynamic#row with Item:
Grid[{{Dynamic#a, Dynamic#b, Dynamic#c},
{Item[Dynamic#row[[1]]], Item[Dynamic#row[[2]]],
Item[Dynamic#row[[3]]]}}, Frame -> All]
Edit: Item wrapper is not really needed; it works just as well without it:
Grid[{{Dynamic#a, Dynamic#b, Dynamic#c},
{Dynamic#row[[1]], Dynamic#row[[2]],
Dynamic#row[[3]]}}, Frame -> All]
I have a Java array such as:
String[] arr = new String[] {"123","doc","projectReport.doc"};
In my opinion the natural way to access would be:
#set($att_id = $arr[0])
#set($att_type = $arr[1])
#set($att_name = $arr[2])
But that it is not working. I have come with this workaround. But it a bit too much code for such an easy task.
#set($counter = 0)
#foreach($el in $arr)
#if($counter==0)
#set($att_id = $el)
#elseif($counter==1)
#set($att_type = $el)
#elseif($counter==2)
#set($att_name = $el)
#end
#set($counter = $counter + 1)
#end
Is there any other way?
You can use use Velocity 1.6: for an array named $array one can simply do $array.get($index).
In the upcoming Velocity 1.7, one will be able to do $array[$index] (as well as $list[$index] and $map[$key]).
You could wrap the array in a List using Arrays.asList(T... a). The new List object is backed by the original array so it doesn't wastefully allocate a copy. Even changes made to the new List will propagate back to the array.
Then you can use $list.get(int index) to get your objects out in Velocity.
If you need to get just one or two objects from an array, you can also use Array.get(Object array, int index)
to get an item from an array.
String[] arr = new String[] {"123", "doc", "projectReport.doc"};
In my opinion the natural way to access would be:
#set($att_id = $arr[0])
#set($att_type = $arr[1])
#set($att_name = $arr[2])
The value for this can be get by using $array.get("arr", 1) because there is no direct way to get the value from array like $att_id = $arr[0] in velocity.
Hope it works :)
Velocity 1.6
$myarray.isEmpty()
$myarray.size()
$myarray.get(2)
$myarray.set(1, 'test')
http://velocity.apache.org/engine/1.7/user-guide.html
there is an implicit counter $velocityCount which starts with value 1 so you do not have to create your own counter.
Brian's answer is indeed correct, although you might like to know that upcoming Velocity 1.6 has direct support for arrays; see the Velocity documentation for more information.
I ended up using the ListTool from the velocity-tools.jar. It has methods to access an array's elements and also get its size.
I has the same question and it got answered on another thread
#set ( $Page = $additionalParams.get('Page') )
#set ( $Pages = [] )
#if ( $Page != $null && $Page != "" )
#foreach($i in $Page.split(";"))
$Pages.add($i)
#end
#end
Array indexing in Confluence / Velocity templates