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std.algorithm.mutation.remove - multiple declarations

Function remove

Eliminates elements at given offsets from range and returns the shortened range.

Range remove(SwapStrategy s = SwapStrategy.stable, Range, Offset...) (
  Range range,
  Offset offset
)
if (Offset.length >= 1 && allSatisfy!(isValidIntegralTuple, Offset));

Range remove(SwapStrategy s = SwapStrategy.stable, Range, Offset...) (
  Range range,
  Offset offset
)
if (Offset.length >= 1 && !allSatisfy!(isValidIntegralTuple, Offset));

For example, here is how to remove a single element from an array:

string[] a = [ "a", "b", "c", "d" ];
a = a.remove(1); // remove element at offset 1
assert(a == [ "a", "c", "d"]);

Note that remove does not change the length of the original range directly; instead, it returns the shortened range. If its return value is not assigned to the original range, the original range will retain its original length, though its contents will have changed:

int[] a = [ 3, 5, 7, 8 ];
assert(remove(a, 1) == [ 3, 7, 8 ]);
assert(a == [ 3, 7, 8, 8 ]);

The element at offset 1 has been removed and the rest of the elements have shifted up to fill its place, however, the original array remains of the same length. This is because all functions in std.algorithm only change content, not topology. The value 8 is repeated because move was invoked to rearrange elements, and on integers move simply copies the source to the destination. To replace a with the effect of the removal, simply assign the slice returned by remove to it, as shown in the first example.

Multiple indices can be passed into remove. In that case, elements at the respective indices are all removed. The indices must be passed in increasing order, otherwise an exception occurs.

int[] a = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ];
assert(remove(a, 1, 3, 5) ==
    [ 0, 2, 4, 6, 7, 8, 9, 10 ]);

(Note that all indices refer to slots in the original array, not in the array as it is being progressively shortened.)

Tuples of two integral offsets can be used to remove an indices range:

int[] a = [ 3, 4, 5, 6, 7];
assert(remove(a, 1, tuple(1, 3), 9) == [ 3, 6, 7 ]);

The tuple passes in a range closed to the left and open to the right (consistent with built-in slices), e.g. tuple(1, 3) means indices 1 and 2 but not 3.

Finally, any combination of integral offsets and tuples composed of two integral offsets can be passed in:

int[] a = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ];
assert(remove(a, 1, tuple(3, 5), 9) == [ 0, 2, 5, 6, 7, 8, 10 ]);

In this case, the slots at positions 1, 3, 4, and 9 are removed from the array.

If the need is to remove some elements in the range but the order of the remaining elements does not have to be preserved, you may want to pass SwapStrategy.unstable to remove.

int[] a = [ 0, 1, 2, 3 ];
assert(remove!(SwapStrategy.unstable)(a, 1) == [ 0, 3, 2 ]);

In the case above, the element at slot 1 is removed, but replaced with the last element of the range. Taking advantage of the relaxation of the stability requirement, remove moved elements from the end of the array over the slots to be removed. This way there is less data movement to be done which improves the execution time of the function.

The function remove works on bidirectional ranges that have assignable lvalue elements. The moving strategy is (listed from fastest to slowest):

  • If s == SwapStrategy.unstable && isRandomAccessRange!Range && hasLength!Range && hasLvalueElements!Range, then elements are moved from the end of the range into the slots to be filled. In this case, the absolute minimum of moves is performed.
  • Otherwise, if s == SwapStrategy.unstable && isBidirectionalRange!Range && hasLength!Range && hasLvalueElements!Range, then elements are still moved from the end of the range, but time is spent on advancing between slots by repeated calls to range.popFront.
  • Otherwise, elements are moved incrementally towards the front of range; a given element is never moved several times, but more elements are moved than in the previous cases.

Parameters

NameDescription
s a SwapStrategy to determine if the original order needs to be preserved
range a bidirectional range with a length member
offset which element(s) to remove

Returns

A range containing all of the elements of range with offset removed.

Example

import std.typecons : tuple;

auto a = [ 0, 1, 2, 3, 4, 5 ];
writeln(remove!(SwapStrategy.stable)(a, 1)); // [0, 2, 3, 4, 5]
a = [ 0, 1, 2, 3, 4, 5 ];
writeln(remove!(SwapStrategy.stable)(a, 1, 3)); // [0, 2, 4, 5]
a = [ 0, 1, 2, 3, 4, 5 ];
writeln(remove!(SwapStrategy.stable)(a, 1, tuple(3, 6))); // [0, 2]

a = [ 0, 1, 2, 3, 4, 5 ];
writeln(remove!(SwapStrategy.unstable)(a, 1)); // [0, 5, 2, 3, 4]
a = [ 0, 1, 2, 3, 4, 5 ];
writeln(remove!(SwapStrategy.unstable)(a, tuple(1, 4))); // [0, 5, 4]

Example

import std.typecons : tuple;

// Delete an index
writeln([4, 5, 6].remove(1)); // [4, 6]

// Delete multiple indices
writeln([4, 5, 6, 7, 8].remove(1, 3)); // [4, 6, 8]

// Use an indices range
writeln([4, 5, 6, 7, 8].remove(tuple(1, 3))); // [4, 7, 8]

// Use an indices range and individual indices
writeln([4, 5, 6, 7, 8].remove(0, tuple(1, 3), 4)); // [7]

Example

SwapStrategy.unstable is faster, but doesn't guarantee the same order of the original array

writeln([5, 6, 7, 8].remove!(SwapStrategy.stable)(1)); // [5, 7, 8]
writeln([5, 6, 7, 8].remove!(SwapStrategy.unstable)(1)); // [5, 8, 7]

Function remove

Reduces the length of the bidirectional range range by removing elements that satisfy pred. If s = SwapStrategy.unstable, elements are moved from the right end of the range over the elements to eliminate. If s = SwapStrategy.stable (the default), elements are moved progressively to front such that their relative order is preserved. Returns the filtered range.

Range remove(alias pred, SwapStrategy s = SwapStrategy.stable, Range) (
  Range range
);

Parameters

NameDescription
range a bidirectional ranges with lvalue elements or mutable character arrays

Returns

the range with all of the elements where pred is true removed

Example

static immutable base = [1, 2, 3, 2, 4, 2, 5, 2];

int[] arr = base[].dup;

// using a string-based predicate
writeln(remove!("a == 2")(arr)); // [1, 3, 4, 5]

// The original array contents have been modified,
// so we need to reset it to its original state.
// The length is unmodified however.
arr[] = base[];

// using a lambda predicate
writeln(remove!(a => a == 2)(arr)); // [1, 3, 4, 5]

Authors

Andrei Alexandrescu

License

Boost License 1.0.