I have to give list of values into in clause of SQL query but while retrieving the values [ ] also come along with data which is not readable by query language.
For example I have list as:
def val = new ArrayList<Integer>(Arrays.asList(1,2,3,4,5,6,7,8))
while doing println(val) output come as [1, 2, 3, 4, 5, 6, 7, 8] but in the query it is needed as: 1, 2, 3, 4, 5, 6, 7, 8
In java this one works as System.out.println(val.toString().replaceAll("[\\[\\]]", "")) but not in groovy. Can't we use collection to remove like this?
Instead of:
def val = new ArrayList(Arrays.asList(1,2,3,4,5,6,7,8))
use:
def val = new ArrayList(Arrays.asList(1,2,3,4,5,6,7,8)).join(', ')
or simply:
def val = [1,2,3,4,5,6,7,8].join(', ')
Try using g-strings and the minus operator:
println "${val}" - '[' - ']'
Related
Let's say I have a vector, the values are from 1 to 10. I want that if you find 5 and 5 next to each other, remove them together with the next elements.
input
[1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10]
expected output
[1, 2, 3, 4]
This was my attempt. I'm finding index to remove, but borrowing rules are making me stuck.
let mut element = vec![1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10];
for (index, val) in element.iter().enumerate() {
if *val == 5 {
if let Some(next_val) = element.get(index + 1) {
if *next_val == 5 {
//element.drain(index..);
}
}
}
}
Rust is saving you from iterator invalidation (a common source of bugs in other languages). This is an error that usually happens when you try to modify a data structure while concurrently iterating over it. You cannot move on to the (now-deleted) next element after calling element.drain(index..). So you need to add a break after that point to avoid memory unsafety.
In this case just adding break; is sufficient to make the code compile. However, for a more concise, linear solution, take full advantage of the iterators and methods provided by the standard library:
if let Some(index) = element.windows(2).position(|pair| pair[0] == pair[1]) {
element.truncate(index);
}
windows(2) on a slice gives an iterator over subslices of length 2, and the position call returns the index of the first element of that iterator for which the two elements of the slice are equal. (If no such pair exists, position returns None.)
I find that the position closure becomes more obvious with the (currently unstable) array_windows feature:
if let Some(index) = element.array_windows().position(|[x, y]| x == y) {
element.truncate(index);
}
Playground
Related
is it possible to filter on a vector in-place?
You can't do what you want to do because you want to remove some elements from a vector while you are iterating it. And this is a big mistake. Note that removing any elements from a vector invalidates the iterators, hence you will access unexpected locations so rust doesn't allow UBs
You can use something like the following
let mut elements = vec![1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10];
let mut first_repeated_five_index_op = None;
for index in 0..elements.len()-1{
if elements[index] == 5 && elements[index + 1] == 5{
first_repeated_five_index_op = Some(index);
break;
}
}
if let Some(first_repeated_five_index) = first_repeated_five_index_op{
elements.truncate(first_repeated_five_index);
}
println!("{:?}", elements);
See a Demo
Is there a Qore operator/function to get sublist from a list without modifying source list, i.e. equivalent of substr(). extract operator removes items from original list.
list l = (1,2,3,4,5,6,7,8,9);
list l2 = extract l, 2, 4;
printf("l:%y\nl2:%y\n", l, l2);
l:[1, 2, 7, 8, 9]
l2:[3, 4, 5, 6]
select operator supports in condition argument $# macro expanded as index.
list l = (1,2,3,4,5,6,7,8,9);
list l2 = select l, $# >= 2 && $# <2+4;
printf("l:%y\nl2:%y\n", l, l2);
l:[1, 2, 3, 4, 5, 6, 7, 8, 9]
l2:[3, 4, 5, 6]
The select operator is the best solution as you stated in your answer to your own question.
The splice and extract operators both will modify the list operand, which is not what you want.
Note that there is an outstanding feature issue for this in Qore (1781) - not yet targeted to a release, but it could go in the next major release (0.8.13) if there is any interest.
I am new to Prolog and came across this practice excercise. The question asks to define a predicate
zipper([[List1,List2]], Zippered). //this is two lists within one list.
This predicate should interleave elements of List1 with elements of List2.
For example,
zipper([[1,3,5,7], [2,4,6,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8].
zipper([[1,3,5], [2,4,6,7,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8].
So far I have a solution for two different list:
zipper ([],[],Z).
zipper([X],[],[X]).
zipper([],[Y],[Y]).
zipper([X|List1],[Y|List2],[X,Y|List]) :- zipper(List1,List2,List).
I am not sure how I can translate this solution for one list. Any suggestion on where I can start would be greatly helpful!
Firstly you should change zipper ([],[],Z). to zipper ([],[],[]).. Then to make it work for one list you could do what mat recommended in the comment or you could change it a little. So my version is:
zipper([],[],[]).
zipper([X,[]],X).
zipper([[],Y],Y).
zipper([[X|List1],[Y|List2]],[X,Y|List]) :- zipper([List1,List2],List).
And for your examples:
?- zipper([[1,3,5,7], [2,4,6,8]], Zippered).
Zippered = [1, 2, 3, 4, 5, 6, 7, 8] ;
Zippered = [1, 2, 3, 4, 5, 6, 7, 8] ;
false.
?- zipper([[1,3,5],[2,4,6,7,8]],Zippered).
Zippered = [1, 2, 3, 4, 5, 6, 7, 8] ;
false.
I've a query in Neo4j/Cypher which returns a collection like that:
MATCH z = (a:X)-[b:Y*2..]->(a)
RETURN relationships(z)
It returns (the paths) respectively the names of the edges:
[name 1, name 2, name 3, name 5],
[name 2, name 3, name 5, name 1],
[name 3, name 5, name 1, name 2],
[name 5, name 1, name 2, name 3],
[name 6, name 7],
[name 7, name 6],
[name 4, name 8],
[name 8, name 4]
In the result above the first 4 rows are equal, then the next two are equal and the last two rows are equal. I want to have no permutations like above i want something like that:
[name 1, name 2, name 3, name 5],
[name 6, name 7],
[name 4, name 8]
Does someone know how I would solve the problem?
Thanks in advance ;)
Cypher does not have a direct way to sort a collection. But there's a way to do it: use UNWIND followed by a WITH ... ORDER BY for sorting on a row level and use collect() to aggregate again.
Your case is even more complex since you want to partially order things on a pre-path base. First you need to build a kind of hashsum for each path (we're here concating relationship's ids) then do the unwind followed by order by and a collect based on the hash calculated previously.
I assume you want to use id of your relationship for ordering.
MATCH z = (a:X)-[b:Y*2..]->(a)
WITH relationships(z) as rels,
reduce(s="", x in relationships(z) | s+x+"_") as hash
UNWIND rels as r
WITH hash, id(r) as r_id order by hash, r_id
WITH hash, collect(r_id) as sorted_r_id
RETURN distinct sorted_r_id
Suppose I have a tuple of (1, 2, 3) and want to index a multidimensional array with it such as:
index = (1, 2, 3)
table[index] = 42 # behaves like table[1][2][3]
index has an unknown number of dimensions, so I can't do:
table[index[0]][index[1]][index[2]]
I know I could do something like this:
functools.reduce(lambda x, y: x[y], index, table)
but it's utterly ugly (and maybe also inefficient), so I wonder if there's a better, more Pythonic choice.
EDIT: Maybe a simple loop is best choice:
elem = table
for i in index:
elem = elem[i]
EDIT2: Actually, there's a problem with both solutions: I can't assign a value to the indexed array :-(, back to ugly:
elem = table
for i in index[:-1]:
elem = elem[i]
elem[index[-1]] = 42
The question is very interesting and also your suggested solution looks good (havn't checked it, but this kind of problem requires a recursive treatment and you just did it in one line).
However, the pythonic way I use in my programs is to use dictionaries of tuples. The syntax is array-like, the performance - of a dictionary, and there was no problem in it for me.
For example:
a = {(1, 2, 3): 'A', (3, 4, 5): 'B', (5, 6, 7, 8): 'C'}
print a[1, 2, 3]
print a[5, 6, 7, 8]
Will output:
A
B
And assigning to an index is super easy:
a[1, 4, 5] = 42. (But you might want to first check that (1, 4, 5) is within the dict, or else it will be created by the assignment)