Mongdb combined limit and sort when using find function - mongodb

I have db mongdb example with document a and document b
a_id type
1 1
2 2
3 3
4 4
Now. I want to extract the last N (1,2,3,4,5,....) values in table b in the same order as in the example above. But if I use skip function :
b.find().skip(M)
if M > N then result empty => wrong. I want dynamic M.
If I use sort and limit then it does not give the correct order.
b.find().sort({$natural:-1}).limit(M)
result:
4 4
3 3
I want a solution!


You can use the same skip() to access the last N documents in the collection.
N = Last N documents to be accessed
So the query is
b.find().skip(b.count() - N).pretty()
or you can play with the mongo shell just as javascript like
var totalCount = b.count()
db.find().skip(totalCount - N).pretty()

Related

Extracting kdb list values based on some condition

Say we have a kdb list
L1:(1 2 3 4 5)
Apply condition
L1 < 3
And how can I retrieve result in another list (1 2)

You can use the where keyword for this:
q)l1 where l1<3
1 2
Applying l1<3 will return a list of booleans 11000b. Using where on this list will return the index of every 1b
q)where 11000b
0 1
Then indexing back into the original list will return the result in another list.

read up a table and analyze the elements matlab

I am trying to realize my idea in matlab.
I consider two column A and B.
A=data(:,1)
B=data(:,5)
the data look like:
A B
1 1
2 1
3 1
... ...
100 20
... ...
150 30
151 1
... ...
The values in column A are timepoints.
I start with the first element in column A. It schould be A(1,1) and look on the first element in the column B B(1,1). If B(1,1)==1its true,if not its false. Then I increase consider the second raw of the column A and second raw of the column B and so on until the last raw of A and B.
How can I construck this loop??

You can just consider B likes the following:
result = (B == 1);
The result would be the same size of B such as you want. Nowm you can get the value of A on result likes the following:
valid_times = A(result);

q kdb+ finding index of match elements using a list

Say I have a list
b:1 1 2 3 4
and I want to find the location of the element in list b using another list
a:1 2
When I type in b in\ a, I got
11000b
00000b
where it should be
11000b
00100b
What is going on and how to get the desired answer?
Thanks in advance!

You need to use each-right /:
q)b in/:a
11000b
00100b
With b in\a the first output is getting passed back in as b. Effectively:
q)1 1 2 3 4 in 1
11000b
q)11000b in 2
00000b

You can also use each-both ':
q)in[b]'[a]
11000b
00100b

Cannot mix aggregate and non-aggregate comparison with COUNT

I know this is very commom question. But I havenot still known why in my case as follows. Give me your idea about this issue:
Question: I want to count the number of user who appear in list < 3.
- First I created the "calculated Field"
- Here is my function:
If COUNT([User]) < 3 then [User] END
Finally, I count this Meseasure again to gain the final result.
Here's my example:
User
a
a
a
a
b
b
c
b
The result expected: 1 (only c)
Thanks all

Place your IF statement inside the COUNT().

Using SUM and UNIQUE to count occurrences of value within subset of a matrix

So, presume a matrix like so:
20 2
20 2
30 2
30 1
40 1
40 1
I want to count the number of times 1 occurs for each unique value of column 1. I could do this the long way by [sum(x(1:2,2)==1)] for each value, but I think this would be the perfect use for the UNIQUE function. How could I fix it so that I could get an output like this:
20 0
30 1
40 2
Sorry if the solution seems obvious, my grasp of loops is very poor.

Indeed unique is a good option:
u=unique(x(:,1))
res=arrayfun(#(y)length(x(x(:,1)==y & x(:,2)==1)),u)
Taking apart that last line:
arrayfun(fun,array) applies fun to each element in the array, and puts it in a new array, which it returns.
This function is the function #(y)length(x(x(:,1)==y & x(:,2)==1)) which finds the length of the portion of x where the condition x(:,1)==y & x(:,2)==1) holds (called logical indexing). So for each of the unique elements, it finds the row in X where the first is the unique element, and the second is one.

Try this (as specified in this answer):
>>> [c,~,d] = unique(a(a(:,2)==1))
c =
30
40
d =
1
3
>>> counts = accumarray(d(:),1,[],#sum)
counts =
1
2
>>> res = [c,counts]

Consider you have an array of various integers in 'array'
the tabulate function will sort the unique values and count the occurances.
table = tabulate(array)
look for your unique counts in col 2 of table.