Example, http://gdata.youtube.com/feeds/api/users/google/uploads?v=2&start-index=1&max-results=50&alt=json,
"author":[{"name":{"$t":"Google"},"uri":{"$t":"http://gdata.youtube.com/feeds/api/users/Google"},"yt$userId":{"$t":"K8sQmJBp8GCxrOtXWBpyEA"}}]
"media$credit":[{"$t":"google","role":"uploader","scheme":"urn:youtube","yt$display":"Google","yt$type":"partner"}]
Both author and media$credit tag is a list, so is it possible index 1 on the list? If it's impossible, why list?
Thanks
I'm not aware of any scenario in which there would be multiple values returned for those fields. I can't say for certain why it's a list and not a single element.
Related
I realize that one can use ReactiveBase to setup a connection to an index and then, use the components to query and return information from that index.
Is there a way to bind to multiple indices and search on a field across them or return results that are a 'merge' of fields from these multiple indices?
In order to query multiple indices ReactiveBase can accept comma separated values in the app prop. Example:
<ReactiveBase
app="index1,index2"
...
>
...
</ReactiveBase>
thanks to siddharthlatest, it is possible by providing multiple comma-separated indices - reference.
It has been reflected in the ReactiveBase documentation as well.
Suppose you have a list of items (instructions in a function, posts on a blog, episodes in a TV series etc) that need to be kept in order, what is the recommended way to store them in Neo4j? Two possibilities that come to mind:
Assuming the items don't already have a suitable property for sorting by, assign them incrementing sequence numbers.
Use a linked list of nodes.
Which of these is typically recommended? Or is there a third option I'm missing?
Use a linked list.
Sequence numbers still have to be sorted, which is unnecessary overhead. And to do the sort, neo4j has to iterate through every node in the sequence, even if you are only interested in a small part of the sequence.
When are working with various Work Item types (user stories,Tasks, etc) we assign tags to reference the area of work. This makes it easy to filter the Backlogs items view to find any related stories.
I would like to build a query to identify Work Items where the tags have not yet been assigned.
I know this can be achieved using excel and filtering, however I specifically would like to do this using the queries. Is this possible??
Thanks in advance for any assistance.
Firstly, I have to say that it is not possible to create work item query to show work items which don't contain tags. As you see that the Operator for Tags is Contains or Does Not Contain, it is not possible to use these two operators to filter out these non-tagged work items.
Secondly, as you have more than 100 tags, it is not an effective way to use "Does Not Contain" operator to exclude all tagged work items.
So,
How about you adding a 'Null' tag to all non-tagged work items to specify that these work items don't have any tags? With this approach, you can create a work item query with Tags Contains Null to list these non-tagged work items.
If you don't want to take this approach, you need to work with excel just as you mentioned above, or take Dave's advice to work with API.
I have a field "taxonomy" and I would like to pass it to Omniture. This Taxonomy field can have multiple values and I would like to calculate the count for each of the taxonomy separately. Let me explain my question with an example:
Item can be associated with multiple taxonomy. Letz say we are talking about an item called "Item1". This item can be associated with taxonomy "Tx1","Tx2","Tx3".
So the js code will be like,
s.prop1 = "Item1"
s.prop2 = "Tx1, Tx2, Tx3"
I was expecting omniture to store individual buckets/counters for "Tx1" "Tx2" and "Tx3". But Omniture created one bucket for "Tx1, Tx2, Tx3". I hope there is a way to treat them separately. Ideally I would not like to pass the taxonomies separately in separate prop variables.
I would appreciate if someone can point me in the right direction.
Thanks,
Rag
Actually Omniture does support list-style props, aptly called a list prop. You have to talk to Client Care to enable it for your prop, and you will need to specify a delimiter to use (like the comma). Then you would populate it exactly as you are doing, except without the space between the comma and value. ("a,b,c" not "a, b, c").
An alternative is the list eVar as Racheet mentioned. The difference is that a list prop sends in values and each value is within a prop variable scope, vs. the list eVar values are recorded with an eVar scope.
If you decide to do it w/ a list eVar instead of list prop, read up on how Omniture handles the allocation for them..there are some implications to consider with them only being able to be set to first or last allocation...
Omniture does have the functionality you're looking for from version 15 onwards. Unfortunately it won't do it in a prop. You need to use a slightly different variable type. The variable you're looking for is a list eVar.
Every implementation has three of these available:
s.list1
s.list2
s.list3
Here's the blog post where Omniture introduced them.
You can also find information on them in Knowledge Base article 10549 from inside the Adobe Marketing Cloud interface.
Given two linked lists of integers. I was asked to return a linked list which contains the non-common elements. I know how to do it in O(n^2), any way to do it in O(n)?
Use a hash table.
Iterate through the first linked list, entering the values you come across into a hash table.
Iterate through the second linked list, adding any element not found into the hash table into your list of non-common elements.
This solution should be O(n), assuming no collisions in the hash table.
create a new empty list. have a hash table and populate it with elements of both lists. complexity n. then iterate over each list sequentially and while iterating, put those elements in the new list which are not present in the hash table. complexity n. overall complexity=n
If they're unsorted, then I don't believe it is possible to get better than O(n^2). However, you can do better by sorting them... you can sort in reasonably fast time, and then get something like O(nlogn) (I'm not certain that's what it would be, but I think it can be that fast if you use the right algorithm).