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Function std.algorithm.setops.largestPartialIntersection
Given a range of sorted forward ranges
ror
, copies to tgt
the elements that are common to most ranges, along with their number
of occurrences. All ranges in ror
are assumed to be sorted by less
. Only the most frequent tgt
elements are returned.
void largestPartialIntersection(alias less, RangeOfRanges, Range)
(
RangeOfRanges ror,
Range tgt,
SortOutput sorted = No .sortOutput
);
Parameters
Name | Description |
---|---|
less | The predicate the ranges are sorted by. |
ror | A range of forward ranges sorted by less . |
tgt | The target range to copy common elements to. |
sorted | Whether the elements copied should be in sorted order.
The function largestPartialIntersection is useful for
e.g. searching an inverted index for the documents most
likely to contain some terms of interest. The complexity of the search
is Ο(n * log(tgt ), where n is the sum of lengths of
all input ranges. This approach is faster than keeping an associative
array of the occurrences and then selecting its top items, and also
requires less memory (largestPartialIntersection builds its
result directly in tgt and requires no extra memory).
If at least one of the ranges is a multiset, then all occurences
of a duplicate element are taken into account. The result is
equivalent to merging all ranges and picking the most frequent
tgt elements. |
Warning
Because largestPartialIntersection
does not allocate
extra memory, it will leave ror
modified. Namely, largestPartialIntersection
assumes ownership of ror
and
discretionarily swaps and advances elements of it. If you want ror
to preserve its contents after the call, you may want to pass a
duplicate to largestPartialIntersection
(and perhaps cache the
duplicate in between calls).
Example
import std .typecons : tuple, Tuple;
// Figure which number can be found in most arrays of the set of
// arrays below.
double[][] a =
[
[ 1, 4, 7, 8 ],
[ 1, 7 ],
[ 1, 7, 8],
[ 4 ],
[ 7 ],
];
auto b = new Tuple!(double, uint)[1];
// it will modify the input range, hence we need to create a duplicate
largestPartialIntersection(a .dup, b);
// First member is the item, second is the occurrence count
writeln(b[0]); // tuple(7.0, 4u)
// 7.0 occurs in 4 out of 5 inputs, more than any other number
// If more of the top-frequent numbers are needed, just create a larger
// tgt range
auto c = new Tuple!(double, uint)[2];
largestPartialIntersection(a, c);
writeln(c[0]); // tuple(1.0, 3u)
// 1.0 occurs in 3 inputs
// multiset
double[][] x =
[
[1, 1, 1, 1, 4, 7, 8],
[1, 7],
[1, 7, 8],
[4, 7],
[7]
];
auto y = new Tuple!(double, uint)[2];
largestPartialIntersection(x .dup, y);
// 7.0 occurs 5 times
writeln(y[0]); // tuple(7.0, 5u)
// 1.0 occurs 6 times
writeln(y[1]); // tuple(1.0, 6u)
Authors
License
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