Array replacement on relationship property in Neo4j - neo4j

So, let's say I have relationship r, with property r.myarray:
[1,2,3,4,5,6,7]
and I need to write a query which will replace the items in the array - up to an including an arbitrary member guaranteed to be in the array (let's say 3 in this case) - with another array - let's say:
[6,12,13]
to get result:
[6,12,13,4,5,6,7]
I got as far as seeing that you can use RANGE or subset notation for the array (e.g. r.myarray[0..x]) to specify part of the array, and could theoretically do SET to replace the array with the first array plus the second subset (r.myarray[x..r.myarray.length], or something like that). I am about half a mile from a complete answer here, though.
edit: Final, interpolat-able query:
START r=relationship(726)
SET r.myarray = [1,2,3,4] + filter(y in r.ancestors where NOT (y IN [718]));

Range probably isn't what you want. Range produces a collection of numbers. It's good for looping, like if you want to go through all the numbers from 1-10, but it's not that useful with other array indexes. You probably want a combination of the + operator on collections, index operations, with possibly a dash of extract and filter. Combining those will let you do basically whatever you like. Here are some examples of the things you can do. I'm using a WITH clause just to show a data sample, you could of course do this on any node property:
/* Return only the first three items */
with [1,2,3,4,5,6,7] as arr return arr[0..3];
/* Cut out the 4th item, otherwise return everything */
with [1,2,3,4,5,6,7] as arr return arr[0..3] + arr[4..];
/* Return only the even numbers */
with [1,2,3,4,5,6,7] as arr
return filter(y in
extract(x in arr | case when (x % 2 = 0) then x end) where y > 0);

Related

Lua: Sort table of numbers with multiple dots

I have a table of strings like this:
{
"1",
"1.5",
"3.13",
"1.2.5.7",
"2.5",
"1.3.5",
"2.2.5.7.10",
"1.17",
"1.10.5",
"2.3.14.9",
"3.5.21.9.3",
"4"
}
And would like to sort that like this:
{
"1",
"1.2.5.7",
"1.3.5",
"1.5",
"1.10.5",
"1.17",
"2.2.5.7.10",
"2.3.14.9",
"2.5",
"3.5.21.9.3",
"3.13",
"4"
}
How do I sort this in Lua? I know that table.sort() will be used, I just don't know the function (second parameter) to use for comparison.
Given your requirements, you probably want something like natural sort order. I described several possible solution as well as their impact on the results in a blog post.
The simplest solution may look like this (below), but there are 5 different solutions listed with different complexity and the results:
function alphanumsort(o)
local function padnum(d) return ("%03d%s"):format(#d, d) end
table.sort(o, function(a,b)
return tostring(a):gsub("%d+",padnum) < tostring(b):gsub("%d+",padnum) end)
return o
end
table.sort sorts ascending by default. You don't have to provide a second parameter then. As you're sorting strings Lua will compare the strings character by character. Hence you must implement a sorting function that tells Lua which comes first.
I just don't know the function (second parameter) to use for
comparison.
That's why people wrote the Lua Reference Manual
table.sort (list [, comp])
Sorts the list elements in a given order, in-place, from list1 to
list[#list]. If comp is given, then it must be a function that
receives two list elements and returns true when the first element
must come before the second in the final order, so that, after the
sort, i <= j implies not comp(list[j],list[i]). If comp is not given,
then the standard Lua operator < is used instead.
The comp function must define a consistent order; more formally, the
function must define a strict weak order. (A weak order is similar to
a total order, but it can equate different elements for comparison
purposes.)
The sort algorithm is not stable: Different elements considered equal
by the given order may have their relative positions changed by the
sort.
Think about how you would do it with pen an paper. You would compare each number segment. As soon as a segment is smaller than the other you know this number comes first.
So a solution would probably require you to get those segments for the strings, convert them to numbers so you can compare their values...

Is it possible to create a variable and make its assignment based on certain conditions in a cypher query?

I am trying to create an array of values that will be assigned based on the outcome of a case test. This test will be inside a query that I already know works with a preset value in the query.
The query I am trying to embed in the case test is something like this:
WITH SPLIT (('07/28/2015'), '/' AS cd
MATCH (nodeA: NodeTypeA)-(r:ARelation)->(nodeB: NodeTypeB)
WITH cd, SPLIT (nodeA.ADate, '/') AS dd, nodeA, nodeB, r
WHERE
(TOINT(cd[2])> TOINT(dd[2])) OR (TOINT(cd[2]= TOINT(dd[2]) AND ((TOINT(cd[0])> TOINT(dd[0])) OR (TOINT(cd[0])= TOINT(dd[0]) AND (TOINT(cd[1])>= TOINT(dd[1])))))
RETURN nodeA, nodeB, r
I want to replace the current date with whatever date will be 6 months from the current date, and I came up with something like this, though I am not sure where I would put it in my query or if it would even work (do I initialize the new variable for instance somehow?):
WHEN ((TOINT(cd[0])> 6))
THEN
TOINT(fd[2])=TOINT(cd[2])+1, TOINT(fd[0])=TOINT(cd[0])-6, TOINT(fd[1])=TOINT(cd[1])
ELSE
TOINT(fd[2])=TOINT(cd[2]), TOINT(fd[0])=TOINT(cd[0])+6, TOINT(fd[1])=TOINT(cd[1])
fd would then replace the cd in the original query's WHERE segment. Where would my case test go, is it correctly written (and if not, what is wrong), and would I need something else added to make it all work?
Just use a WITH block to do a computation and bind it to a new variable, like this:
WITH 2 + 2 as y RETURN y;
That basically assigns the value 4 to y.
In your query, you already have a big WITH block. Just put your computations in those, bound to new variables, and you can then refer to those variables in subsequent expressions.
Don't try to modify these variables, just create new ones (with new WITH blocks) as needed. If you need variables that can actually change, then...well hey you're working with a database, the ultimate way to store and update information. Create a new node, and then update it as you see fit. :)
This is my proposed solution
Explanation: I have declared four variables in my query i.e. name1, name2, ken and lana and I am using these variables for creating MATCH pattern (in the MATCH clause) and filtering those in the Where clause.
WITH "Lau" AS name1,
"L" AS name2,
"Keanu Reeves" AS ken,
"Lana Wachowski" AS lana
MATCH(x:Person{ name: ken})-[:ACTED_IN]->(m:Movie)<-[:ACTED_IN]-(y:Person),
(x1:Person{name: lana})-[:DIRECTED]->(m)<-[:DIRECTED]-(y1:Person)
WHERE y.name CONTAINS name1 OR
y.name CONTAINS name2 OR
(y.name CONTAINS name1 AND y.name CONTAINS name2)
RETURN x, m, y, x1;

.indexOf() equivalent in Neo4j Cypher

No matter how I swing it, I need some kind of function to find the index of a item in an array supplied as a parameter.
I am trying to simply update items in a collection based on the index of one of their properties in an array, and have been poring through Cypher docs for nearly 2 hours...
It would also be acceptable to order the items by that array, and then run a foreach on the ordered list...
Following #stefan-armbruster answer and great blog post, a slow but simple index_of can be done with:
reduce(x=[-1,0], i IN [1,2,7,5,21,5,1,435] |
CASE WHEN i = 21 THEN [x[1], x[1]+1] ELSE [x[0], x[1]+1] END
)[0]
Here reduce function works with a two elements array: the position and the current index. If an element in your array matches the given condition, the first element of the reduced array will be replaced with the current index.
I put an example on neo4j console http://console.neo4j.org/?id=34byv
I've blogged about that recently. You can use the reduce function with an three element array as state variable containing
the index of the highest occupation so far
the current index (aka iteration number)
the value of highest occupation so far
As an example to find the index of max element in an array:
RETURN reduce(x=[0,0,0], i IN [1,2,2,5,2,1] |
CASE WHEN i>x[2] THEN [x[1],x[1]+1,o] ELSE [x[0], x[1]+1,x[2]] END
)[0]

Is it possible to make a nested FOREACH without COGROUP in PigLatin?

I want to use the FOREACH like:
a:{a_attr:chararray}
b:{b_attr:int}
FOREACH a {
res = CROSS a, b;
-- some processing
GENERATE res;
}
By this I mean to make for each element of a a cross-product with all the elements of b, then perform some custom filtering and return tuples.
==EDIT==
Custom filetering = res_filtered = FILTER res BY ...;
GENERATE res_filtered.
==EDIT-2==
How to do it with a nested CROSS no more no less inside a FOR loop without prior GROUP or COGROUP?
Depending on the specifics of your filtering, you may be able to design a limited set of disjoint classes of elements in a and b, and then JOIN on those. For example:
If your filtering rules are
if a_attr starts with "Foo" and b is 4, accept
if a_attr starts with "Bar" and b is greater than 17, accept
if a_attr begins with a letter in [m-z] and b is less than 0, accept
otherwise, reject
Then you can write a UDF that will return 1 for items satisfying the first rule, 2 for the second, 3 for the third, and NULL otherwise. Your CROSS/FILTER then becomes
res = JOIN a BY myUDF(a), b BY myUDF(b);
Pig drops null values in JOINs, so only pairs satisfying your filtering criteria will be passed.
CROSS generates a cross-product of all the tuples in each relation. So there is no need to have a nested FOREACH. Just do the CROSS and then FILTER:
a: {a_attr: chararray}
b: {b_attr: int}
crossed = CROSS a, b;
crossed: {a::a_attr: chararray,b::b_attr: int}
res = FILTER crossed BY ... -- your custom filtering
If you have the FILTER immediately after the CROSS, you should not have (unnecessary) excessive IO trouble from the CROSS writing the entire cross-product to disk before filtering. Records that get filtered will never be written at all.

How do I collect and combine multiple arrays for calculation?

I am collecting the values for a specific column from a named_scope as follows:
a = survey_job.survey_responses.collect(&:base_pay)
This gives me a numeric array for example (1,2,3,4,5). I can then pass this array into various functions I have created to retrieve the mean, median, standard deviation of the number set. This all works fine however I now need to start combining multiple columns of data to carry out the same types of calculation.
I need to collect the details of perhaps three fields as follows:
survey_job.survey_responses.collect(&:base_pay)
survey_job.survey_responses.collect(&:bonus_pay)
survey_job.survey_responses.collect(&:overtime_pay)
This will give me 3 arrays. I then need to combine these into a single array by adding each of the matching values together - i.e. add the first result from each array, the second result from each array and so on so I have an array of the totals.
How do I create a method which will collect all of this data together and how do I call it from the view template?
Really appreciate any help on this one...
Thanks
Simon
s = survey_job.survey_responses
pay = s.collect(&:base_pay).zip(s.collect(&:bonus_pay), s.collect(&:overtime_pay))
pay.map{|i| i.compact.inject(&:+) }
Do that, but with meaningful variable names and I think it will work.
Define a normal method in app/helpers/_helper.rb and it will work in the view
Edit: now it works if they contain nil or are of different sizes (as long as the longest array is the one on which zip is called.
Here's a method that will combine an arbitrary number of arrays by taking the sum at each index. It'll allow each array to be of different length, too.
def combine(*arrays)
# Get the length of the largest array, that'll be the number of iterations needed
maxlen = arrays.map(&:length).max
out = []
maxlen.times do |i|
# Push the sum of all array elements at a given index to the result array
out.push( arrays.map{|a| a[i]}.inject(0) { |memo, value| memo += value.to_i } )
end
out
end
Then, in the controller, you could do
base_pay = survey_job.survey_responses.collect(&:base_pay)
bonus_pay = survey_job.survey_responses.collect(&:bonus_pay)
overtime_pay = survey_job.survey_responses.collect(&:overtime_pay)
#total_pay = combine(base_pay, bonus_pay, overtime_pay)
And then refer to #total_pay as needed in your view.

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