How to avoid duplicate entries of column in REALM database? - swift

Can we avoid duplicate entries of the column in REALM database?
Eg. I have entries like 4 4 4 4 4 4 3 3 3 3 4 5 5 5 2 2 2 2 4 3 3 3 3 2 2 6 6 6
I want to show 2 3 4 5 6 only.

As #Ahmad F said, specify if you want to do it after or before storing them.
In case you want to filter them AFTER storing you can go with something like this:
return realm.objects(YourObject.self).value(forKeyPath: "#distinctUnionOfObjects.entryKey")

if using realm version 3.10 you can using
Results..distinct(by: )
if using Old version - before Realm 3.10 try this
func uniq<S : Sequence, T : Hashable>(source: S) -> [T] where S.Iterator.Element == T {
var buffer = [T]()
var added = Set<T>()
for elem in source {
if !added.contains(elem) {
buffer.append(elem)
added.insert(elem)
}
}
return buffer
}
using func
uniq(source: realm.objects(Object.self))

Hmm, I think you should add the primary key in your class definition.
And it will help you to avoid duplicate items.

Related

What sort descriptor will keep the order of NSOrderedSet when using FetchRequest in SwiftUI

I have 2 entities in CoreData. Deck and Flashcard. Deck has an ordered to-many relationship with Flashcard. This means that Deck contains property flashcards: NSOrderedSet containing the Flashcards.
In a DeckView I try to show all Flashcards from a certain Deck in a user-specified order. For that I use FetchRequest.
static func fetchInDeck(deck: Deck) -> NSFetchRequest<Flashcard> {
let request = Flashcard.fetchRequest()
request.predicate = NSPredicate(format: "deck == %#", deck)
request.sortDescriptors = []
return request
}
The flashcards maintain order until I start adding items, new items get added in seemingly random places, but after restarting the app they go where they should. (e.g. after adding 10 items the order was [3 5 7 9 10 8 6 4 2 1], and after restart [10 9 8 7 6 5 4 3 2 1]) For adding I used a method generated automatically by Xcode.
let newFlashcard = Flashcard(context: context)
insertIntoFlashcards(newFlashcard, at: 0)
Is there a sort descriptor that will cause those items to stay in order? If not is there another way to maintain the order?

How do I convert a dictionary of dictionaries into a table?

I've got a dictionary of dictionaries:
`1`2!((`a`b`c!(1 2 3));(`a`b`c!(4 5 6)))
| a b c
-| -----
1| 1 2 3
2| 4 5 6
I'm trying to work out how to turn it into a table that looks like:
1 a 1
1 b 2
1 c 3
2 a 4
2 b 5
2 c 6
What's the easiest/'right' way to achieve this in KDB?
Not sure if this is the shortest or best way, but my solution is:
ungroup flip`c1`c2`c3!
{(key x;value key each x;value value each x)}
`1`2!((`a`b`c!(1 2 3));(`a`b`c!(4 5 6)))
Which gives expected table with column names c1, c2, c3
What you're essentially trying to do is to "unpivot" - see the official pivot page here: https://code.kx.com/q/kb/pivoting-tables/
Unfortunately that page doesn't give a function for unpivoting as it isn't trivial and it's hard to have a general solution for it, but if you search the Kx/K4/community archives for "unpivot" you'll find some examples of unpivot functions, for example this one from Aaron Davies:
unpiv:{[t;k;p;v;f] ?[raze?[t;();0b;{x!x}k],'/:(f C){![z;();0b;x!enlist each (),y]}[p]'v xcol't{?[x;();0b;y!y,:()]}/:C:(cols t)except k;enlist(not;(.q.each;.q.all;(null;v)));0b;()]};
Using this, your problem (after a little tweak to the input) becomes:
q)t:([]k:`1`2)!((`a`b`c!(1 2 3));(`a`b`c!(4 5 6)));
q)`k xasc unpiv[t;1#`k;1#`p;`v;::]
k v p
-----
1 1 a
1 2 b
1 3 c
2 4 a
2 5 b
2 6 c
This solution is probably more complicated than it needs to be for your use case as it tries to solve for the general case of unpivoting.
Just an update to this, I solved this problem a different way to the selected answer.
In the end, I:
Converted each row into a table with one row in it and all the columns I needed.
Joined all the tables together.

how to find max value from multiple columns in dataframe in spark [duplicate]

This question already has an answer here:
Scala/Spark dataframes: find the column name corresponding to the max
(1 answer)
Closed 3 years ago.
I have input spark dataframe as
sample A B C D
1 1 3 5 7
2 6 8 10 9
3 6 7 8 1
I need to find the max among A,B,C,D columns which are subject marks.
I need to create a new dataframe with max_marks as the new column.
sample A B C D max_marks
1 1 3 5 7 7
2 6 8 10 9 10
3 6 7 8 1 8
I have done this using scala as
val df = df.columns.toSeq
val df1=df.foldLeft(df){(df,colName)=> df.withColumn("max_sub",max((colName)))
df.show()
I am getting an error message
"main" org.apache.spark.sql.AnalysisException:grouping expression sequence is empty
this dataframe has about 100 columns so how to iterate over this dataframe
It would be helpful to iterate over the data frame as the columns where the mean has to be found out are about 10 out of 100 column dataframe with about 10000 records
I am looking to dynamically pass the columns without giving the column names manually which means to loop over the columns that i choose and perform any mathematical operation
There are many ways to accomplish this one of the ways would be using map.
Simple pseudo code to do what you want (It wont work in anyway but I think the idea is clear)
df = df.withColumn("max_sub", "A")
df.map({x=> {
max = "A"
maxVal = 0
for col in x{
if(col != "max_sub" && x.col > maxVal){
max = col
maxVal = x.col
}
}
x.max_sub = max
x
})

OpenRefine: Fill down with increasing counter

Is it possible in OpenRefine to fill down blank cells with a counter instead of copying the top non-blank value?
In this example image:
Or here the same example as typed text - image this as a column from top to bottom:
1
1
blank
1
blank
blank
blank
blank
blank
1
I would like to see the column filled as follows (again, imagine top to bottom):
1
1
2
1
2
3
4
5
6
1
Thanks, help is very much appreciated.
It's not really simple. You have to:
1 Replace the blanks with something else, such as an "x"
2 Create a unique record for the entire dataset
3 Use this Jython script:
import itertools
data = row['record']['cells']['YOUR COLUMN NAME']['value']
x = itertools.count(2)
liste = []
for i, el in enumerate(data):
if data[i] == "x":
liste.append(x.next())
else:
x = itertools.count(2)
liste.append(el)
return ",".join([str(x) for x in liste])
4 Use Blank down to clear duplicates
5 Split the first multivalued cell.
Here is a screencast of the operations described above.
If you know a little Python, you can also transform your file using pandas. I do not know what is the most elegant way to do it, but this script should work.
import itertools
import pandas as pd
x = itertools.count(2)
def set_x():
global x
x = itertools.count(2)
set_x()
def increase(value):
if not value:
return next(x)
else:
set_x()
return value
data = pd.read_csv("your_file.csv", na_values=['nan'], keep_default_na=False)
data['column 1'] = data['column 1'].apply(lambda row: increase(row))
print(data)
data.to_csv("final_file.csv")
Here are two simple solutions using GREL.
Use records
You could move the column to the beginning, telling OpenRefine to use the numbers as records. You might need to transform the column to text to really convince OpenRefine to use it as records.
Then either add a new column or transform the existing one with the following expression.
1 + row.index - row.record.fromRowIndex
Use record markers
In case you don't want to use records or don't have a static number, you can create a similar setup. Imagine you have an incomplete counter like in the following table and want to fill it.
Origin
Desired
1
1
2
1
1
2
2
3
1
1
To fill the missing cells first add a new column based on your orignal column using the following expression and name it record_row_index.
if(isNonBlank(value), row.index, "")
After that fill down the original column and the new column record_row_index.
Then create a new column based on the original filled column using the following expression.
value + row.index - cells["record_row_index"].value
Hint: the expression is expecting both columns to be of type number.
If one of them is of type text, you can either transform the column beforehand or use toNumber() in the expression.
The following table shows how these operations are working together.
Origin
Origin filled
row.index
record_row_index
Desired
1
1
0
0
1 + 0 - 0 = 1
1
1
0
1 + 1 - 0 = 2
1
1
2
2
1 + 2 - 2 = 1
2
2
3
3
2 + 3 - 3 = 2
2
4
3
2 + 4 - 3 = 3
1
1
5
5
1 + 5 - 5 = 1

create a new matrix from values obtained iterating through other matricies

In Matlab I have 4 matricies which are all 1(row) by 4(coloumns) (ABDC, EFGH, IJKL, MNOP)
Their names are also stored in a list
Stock_List2 = {'ABCD' 'EFGH' 'IJKL' 'MNOP'} and is a 1 by 4 cell.
I want to iterate through the list and create a new matrix called "display" which takes the values of the indvidual matricies and places them underneath each other)
I am trying something like
for e = 1:length(Stock_List2)
display(e) = eval(strcat(Stock_List2)(e))
end
Error: ()-indexing must appear last in an index expression.
However getting the following error expression which truthfully may well just be that I'm way off the mark.
As an example if the orginal matricies are as follows:
ABCD 1 2 3 4
DEFG 5 6 7 8
HIJK 9 8 7 6
LMNO 5 4 3 2
I would like the final output ie the 'display matrix to be a 4 by 4 matrix looking like
display
1 2 3 4
5 6 7 8
9 8 7 6
5 4 3 2
If I understood right you want to concatenate vertically the matrices ABDC, EFGH, IJKL and MNOP saving them in the matrix "display".
You could do:
display = [ABDC; EFGH; IJKL; MNOP]
or:
for i=1:length(Stock_List2)
display(i,:) = Stock_List2{i}
end
Apologies if what I wanted wasnt clear - I've got the following from a colleague which achieves the desired result
for e=1:length(Stock_List2)
eval(strcat('display_mat(e,:) = ',Stock_List2{e}));
end