Currently I am using NSPredicate to search from a string. But I have around 2000 records. So iterating through these records for search using NSPredicate take some time.
I just want to know whether NSPredicate using any search algorithm internally or should I use any search algorithm for fast searching ?
I want my app should search like contacts app (smoothly). What should I do so that my search will be smooth like contacts app.
Thanks.
NSPredicate doesn't use any search algorithm - all NSPredicate does is represent your query.
It's the job of whatever accepts NSPredicate as a option to choose the search algorithm. In your case, it's probably NSArray's filteredArrayUsingPredicate: method?
If you want to search the data faster, you will have to write a better algorithm yourself, I bet the array just runs through each item and matches it to the predicate. Not efficient but it works for anything :)
Try looking at sqlite (or CoreData), they have better indexing than plain arrays.
Or you could make an index yourself - put each contact into a dictionary with the key being the first letter of the contact - that should reduce the amount of records you have to search!
i.e. instead of an array
[ #"Alex", #"Bob", #"Bill", #"Carl" ... ]
Have a dictionary
{
#"A" : [ #"Alex" ],
#"B" : [ #"Bob", #"Bill" ],
#"C" : [ #"Carl" ]
...
}
For the search string #"Bi" you only have to search through the items in the array for #"B".
PS Don't forget to make your dictionary key case / diacritic independent ;)
Related
How to find partial search?
Now Im trying to find
db.content.find({$text: {$search: "Customer london"}})
It finds all records matching customer, and all records matching london.
If I am searching for a part of a word for example lond or custom
db.content.find({$text: {$search: "lond"}})
It returns an empty result. How can I modify the query to get the same result like when I am searching for london?
You can use regex to get around with it (https://docs.mongodb.com/manual/reference/operator/query/regex/). However, it will work for following :
if you have word Cooking, following queries may give you result
cooking(exact matching)
coo(part of the word)
cooked(The word containing the english root of the document word, where cook is the root word from which cooking or cooked are derived)
If you would like to go one step further and get a result document containing cooking when you type vooking (missplled V instead of C), go for elasticsearch.
Elasticsearch is easy to setup, has extremely powerful edge-ngram analyzer which converts each words into smaller weightage words. Hence when you misspell, you will still get a document based on score elasticsearch gives to that document.
You can read about it here : https://www.elastic.co/guide/en/elasticsearch/reference/current/analysis-edgengram-tokenizer.html
it will always return the empty array for partial words like when you are searching for lond to get this type of text london..
Because it take full words and search same as that they are ..
Not achive same results like :-
LO LON LOND LONDO LONDON
Here you may get help from ELASTIC-SEARCH . It is quite good for full text search when implement with mongoDB.
Refrence : ElasticSearch
Thanks
The find all is to an Array
clientDB.collection('details').find({}).toArray().then((docs) =>
I now used the str.StartWith in a for loop to pick out my record.
if (docs[i].name.startsWith('U', 0)) {
return console.log(docs[i].name);
} else {
console.log('Record not found!!!')
};
This may not be efficient, but it works for now
I have a Mongo search set up that goes through my entries based on numerous criteria.
Currently the easiest way (I know it's not performance-friendly due to using wildcards, but I can't figure out a better way to do this due to case insensitivity and users not putting in whole words) is to use regex wildcards in the search. The search ends up looking like this:
{ gender: /Womens/i, designer: /Voodoo Girl/i } // Should return ~200 results
{ gender: /Mens/i, designer: /Voodoo Girl/i } // Should return 0 results
In the example above, both searches are returning ~200 results ("Voodoo Girl" is a womenswear label and all corresponding entries have a gender: "Womens" field.). Bizarrely, when I do other searches, like:
{ designer: /Voodoo Girl/i, store: /Store XYZ/i } // should return 0 results
I get the correct number of results (0). Is this an order thing? How can I ensure that my search only returns results that match all of my wildcarded queries?
For reference, the queries are being made in nodeJS through a simple db.products.find({criteria}) lookup.
To answer the aside real fast, something like ElasticSearch is a wonderful way to get more powerful, performant searching capabilities in your app.
Now, the reason that your searches are returning results is that "mens" is a substring of "womens"! You probably want either /^Mens/i and /^Womens/i (if Mens starts the gender field), or /\bMens\b/ if it can appear in the middle of the field. The first form will only match the given field from the beginning of the string, while the second form looks for the given word surrounded by word boundaries (that is, not as a substring of another word).
If you can use the /^Mens/ form (note the lack of the /i), it's advisable, as anchored case-sensitive regex queries can use indexes, while other regex forms cannot.
$regex can only use an index efficiently when the regular expression has an anchor for the beginning (i.e. ^) of a string and is a case-sensitive match.
So i need some advice as to what i'm doing incorrectly.
My database is setup up exactly like a file system consisting of folders and files.
It begins with a folder, but can have a relatively infinite number of subfolders and or files.
{
"name":"folder1",
"uniqueID":"zzz0",
"subcontents": [ {"name":"subfolder1", "uniqueID":"zzz1"},
{"name":"subfile1", "uniqueID":"zzz2"},
{"name":"subfile2", "uniqueID":"zzz3"},
{"name":"subfolder2", "subcontents": [...etc...], "uniqueID":"zzz4"},
]
}
Each folder/file document have a uniqueID so that I can reference to it (seen above zzz#). My question is, can I make a mongoDB query to pull out only a single document?
Like say for example db.fileSystemCollection.find({"uniqueID":"zzz4"}) and it would give me the following result? Do i have to use indexes to do this? I've been trying but the query returns empty every time.
intended result ---> {"name":"subfolder2", "subcontents": [...etc...], "uniqueID":"zzz4"}
[EDIT]
Based on the responses below, I will consider an XML database instead on mongoDB. The json structure cant be rearranged to work with MongoDB (too much data).
Short answer is no, as it's stated by Chris.
Your embedded representation of a tree is really good for intuitive understanding (and implementation as well). But if you want to allow effective searches on your tree using indices in MongoDB, you might consider another ways for tree storage. A bunch of ways is listed at http://docs.mongodb.org/manual/tutorial/model-tree-structures/
Please keep in mind that every representation has its own pros and cons depending on your access patterns.
Since for filesystem-like structure it's likely to have the ability to find all the sub contents of a given folder, you may use child references pattern for this:
{
"name":"folder1",
"uniqueID":"zzz0",
"subcontents": [ "zzz1",
"zzz2",
"zzz3",
"zzz4"
]
}
{
"name":"subfolder1",
"uniqueID":"zzz1"
}
...
No; searching for {uniqueID: "zzz4"} will only get you documents whose top-level uniqueID matches.
What you probably want is to maintain an array on the document which lists all the unique IDs in that tree. So your document would be:
{
"name":"folder1",
"uniqueID":"zzz0",
"idList": ["zzz0", "zzz1", "zzz2", "zzz3", "zzz4"],
"subcontents": [ {"name":"subfolder1", "uniqueID":"zzz1"},
{"name":"subfile1", "uniqueID":"zzz2"},
{"name":"subfile2", "uniqueID":"zzz3"},
{"name":"subfolder2", "subcontents": [...etc...], "uniqueID":"zzz4"},
]
}
Then you can index that:
db.fileSystemCollection.ensureIndex({"idList": 1})
Then you can find on it:
db.fileSystemCollection.find({"idList": "zzz4})
That'll return you those documents.
As an aside, if you're trying to store files in Mongo, have you looked at GridFS?
I want to be able to search for my objects by searching for the last 4 characters of the id. How can I do that?
Book.where(_id: params[:q])
Where the param would be something like a3f4, and in this case the actual id for the object that I want to be found would be:
bc313c1f5053b66121a8a3f4
Notice the last for characters are what we searched for. How can I search for just "part" of my objects id? instead of having my user search manually by typing in the entire id?
I found in MongoDB's help docs, that I can provide a regex:
db.x.find({someId : {$regex : "123\\[456\\]"}}) // use "\\" to escape
Is there a way for me to search using the regular mongo ruby driver and not using Mongoid?
Usually, in Mongoid you can search with a regexp like you normally would with a string in your call to where() ie:
Book.where(:title => /^Alice/) # returns all books with titles starting with 'Alice'
However this doesn't work in your case, because the _id field is not stored as a string, but as an ObjectID. However, you could add (and index) a field on your models which could provide this functionality for you, which you can populate in an after_create callback.
<shameless_plug>
Alternatively, if you're just looking for a shorter solution to the default Mongoid IDs, I could suggest something like mongoid_token which makes it pretty easy to add shorter tokens/ids to your Mongoid documents.
</shameless_plug>
I've been trying to see if there is any way I can improve on the performance of the following Predicate:
[NSPredicate predicateWithFormat:#"A=%# and SUBQUERY($self.words,$k,$k.keyword IN %#).#count==%d",
<value for A>,
keywordList,
[keywordList count]];
What I'm trying to do is return all the records of A that have keywords that are ALL contained in the supplied array (keywordList). I have a small database of about 2000 records. However, the keywords for each Entity ranges from 40-300 keywords. The keywords are represented as a to-Many relationship from A to an entity called Keywords. The above query works but takes about 7 seconds to run on my iPhone4. I want to see what I can do to shrink that down to a sub-second response.
Thanks,
Michael
This doesn't quite solve the problem as the results that come back from intersection of the fetchedObjs is not correct. It doesn't guarantee that all the 'A' objects contain all the supplied keywords. In fact, what comes back is an empty list and the entire process actually takes longer.
Something must be wrong with my model or the previous answer. I'm not getting the results I'm expecting. One alternative to returning a list of 'A' would be to return a list of ManagedObjectIDs of 'A'. [mo valueForKey: (NSString *)] returns the object of the relationship.
Note: I originally posted this question anonymously thinking I was logged in, so i can't seem to comment on anybody's answer.
I think you are approaching the problem backwards. If you have the keywords, you should search for the Keyword objects and then walk their A relationships to find the related A objects. Then check for overlap in the relationship sets.
So something like:
NSFetchRequest *fetch=//...set up fetch
[fetch setEntity:Keyword_entity];
NSPredicate *p=[NSPredicate predicateWithFormat:#"keyword IN %#",keywords];
[fetch setPredicate:p];
NSArray *fetchedObj=//... execute fetch
NSMutableSet *inCommon=[NSMutableSet setWithCapacity:[fetchedObjs count]];
for (NSManagedObject *mo in fetchedObjs){
if ([inCommon count]==0){
[inCommon addObjects:[mo valueForKey:#"aRelationshipName"]];
}else{
[inCommon intersectSet:[mo valueForKey:#"aRelationshipName"]];
}
}
... inCommon would then contain a set of all unique A objects that shared all the keywords.
Update:
From OP author:
This doesn't quite solve the problem
as the results that come back from
intersection of the fetchedObjs is not
correct. It doesn't guarantee that all
the 'A' objects contain all the
supplied keywords.
Okay, lets take another tack. Assuming you have a data model that looks something like this:
A {
keywords<<-->>Keyword.as
}
Keyword{
keyword:string
as<<-->>A.keywords
}
Then something like this should work:
NSFetchRequest *keyWordFetch=//...set up fetch
[keyWordFetch setEntity:Keyword_entity];
NSPredicate *p=[NSPredicate predicateWithFormat:#"keyword IN %#",keywords];
[keyWordFetch setPredicate:p];
NSArray *fetchedKeywords=//... execute keyWordFetch
NSFetchRequest *entityAFetch=//...set up fetch
[entityAFetch setEntity:A_entity];
NSPredicate *pred=[NSPredicate predicateWithFormat:#"ALL keywords IN %#", fetchedKeywords);
[entityAFetch setPredicate:pred];
NSArray *fetchedAs=//... execute fetch