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Zip lists in jq's objects construction by {} instead of multiplying them like default


Zip lists in jq's objects construction by {} instead of multiplying them like default

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Zip lists in jq's objects construction by {} instead of multiplying them like default
Tag : json , By : nseibert
Date : November 28 2020, 12:01 PM

To fix this issue transpose/0 can be used to effectively zip the values together. And the nice thing about the way assignments work is that it can be assigned simultaneously over multiple variables.
([.titles,.years]|transpose[]) as [$title,$year] | {user,$title,$year}
def transpose:
  if . == [] then []
  else . as $in
  | (map(length) | max) as $max
  | length as $length
  | reduce range(0; $max) as $j
      ([]; . + [reduce range(0;$length) as $i ([]; . + [ $in[$i][$j] ] )] )
            end;
def transpose2:
    length as $cols
      | (map(length) | max) as $rows
      | [range(0;$rows) as $r | [.[range(0;$cols)][$r]]];
([.titles,.years]|transpose[]) as $p | {user,title:$p[0],year:$p[1]}

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Scala the default way to create objects that require a complex construction process


Tag : scala , By : Nickolas
Date : March 29 2020, 07:55 AM
With these it helps In general, you usually provide complex construction like this through methods on the companion object of your (base) class. It provides a nice clean separation of the initialisation code and the post-construction usage code.
Also, having immutable data may help design smaller, more focused concerns. For instance you have a box (left, right, top, bottom) that you could separate into its own class (or just a tuple if you like).

Multiplying lists and list of lists in Prolog


Tag : list , By : Ravenal
Date : March 29 2020, 07:55 AM
I hope this helps . You can do the zeroes by introducing indices that indicate row and column you are at and check for a match:
main(A, O) :-
    second(A, A, 0, O).

second([], _, _, []).
second([A|As], B, R, [O|Os]) :- %creates the list of lists.
    third(A, B, 0, R, O),
    R1 is R + 1,
    second(As, B, R1, Os).

third(_, [], _, _, []).
third(A, [B|Bs], C, R, [O|Os]) :-
    fourth(A, B, C, R, O),
    C1 is C + 1,
    third(A, Bs, C1, R, Os). %multiplies single digit by list.

fourth(_, _, X, X, 0).
fourth(A, B, C, R, O) :- C \== R, O is A * B.
| ?-  main([1,2,2,1], L).

L = [[0,2,2,1],[2,0,4,2],[2,4,0,2],[1,2,2,0]] ? ;

no
maplist_with_index(Pred, L, M) :-
    maplist_with_index_(Pred, 0, L, M).
maplist_with_index_(Pred, I, [H|T], [M|Ms]) :-
    Pred =.. [P|Pt],
    append([P,I|Pt], [H], NewPred),
    Call =.. NewPred,
    call(Call, M),
    I1 is I + 1,
    maplist_with_index_(Pred, I1, T, Ms).
maplist_with_index_(_, _, [], []).
main(A, O) :-
    second(A, A, O).

second(M, A, O) :-
    maplist_with_index(third(A), M, O).

third(R, A, E, O) :-
    maplist_with_index(fourth(R, E), A, O).

fourth(X, X, _, _, 0).
fourth(C, R, A, B, O) :- C \== R, O is A * B.

Multiplying two different lists objects in R


Tag : r , By : pad
Date : March 29 2020, 07:55 AM
To fix this issue I have read Multiplying Combinations of a list of lists in R. But i can't still apply it on my case. , You can use map as shown below:
Map('*',x,y)
> Map('*',x,y)
  [[1]]
      [,1] [,2]
[1,]    1    0
[2,]    0    4

 [[2]]
      [,1] [,2]
[1,]    0    0
[2,]    6    8
Listxy <- list(unlist(x)*unlist(y))
Listxy
[[1]]
[1] 1 0 0 4 0 6 0 8

multiplying lists of lists with different lengths


Tag : python , By : jazzyfox
Date : March 29 2020, 07:55 AM
I hope this helps you . That kind of combination is called the Cartesian product. I'd use itertools.product for this, which can easily cope with more than 2 lists, if you want.
Firstly, here's a short demo that shows how to get all of the pairs and how to use tuple assignment to grab the individual elements of the pairs of sublists.
from itertools import product

a = [[1,2],[3,4],[7,10]]
b = [[8,6],[1,9],[2,1],[8,8]]

for (u0, u1), (v0, v1) in product(a, b):
    print(u0, u1, v0, v1)
1 2 8 6
1 2 1 9
1 2 2 1
1 2 8 8
3 4 8 6
3 4 1 9
3 4 2 1
3 4 8 8
7 10 8 6
7 10 1 9
7 10 2 1
7 10 8 8
total = sum(u0 * v0 + u1 * v1 for (u0, u1), (v0, v1) in product(a, b))
print(total)
593
a = [[1,2],[3,4],[7,10]]
b = [[8,6],[1,9],[2,1],[8,8]]

print(sum([u*v for u,v in zip(*[[sum(t) for t in zip(*u)] for u in (a, b)])]))
593
(1 + 3 + 7) * (8 + 1 + 2 + 8) + (6 + 9 + 1 + 8) * (2 + 4 + 10)
# Use zip to transpose each of the a & b lists.
for u in (a, b):
    for t in zip(*u):
        print(t)
(1, 3, 7)
(2, 4, 10)
(8, 1, 2, 8)
(6, 9, 1, 8)
# Use zip to transpose each of the a & b lists and compute the partial sums.
partial_sums = []
for u in (a, b):
    c = []
    for t in zip(*u):
        c.append(sum(t))
    partial_sums.append(c)
print(partial_sums)        
[[11, 16], [19, 24]]
total = 0
for u, v in zip(*partial_sums):
    print(u, v)
    total += u * v
print(total)        
11 19
16 24
593

Creating a list of lists by multiplying a list unexpectedly leaves inner lists connected


Tag : python , By : Adil
Date : March 29 2020, 07:55 AM
This might help you Let's say I want to create a list of lists to be filled later, that looks like that:
B=[A[:] for x in range(rows)]
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