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Element Access

Method
MapContainer
Access and throw exception if it doesn't exist
mapped_type &at(const key_type &k);
const mapped_type &at(const key_type &k) const;
Access and create new element if it doesn't exist
mapped_type &operator[] (const key_type &k);
mapped_type &operator[] (key_type &&k);
template <typename... Targs> mapped_type &operator()(const dimension_type &x1, const Targs &...xs);

Parameters

  • k - the key of the element to find
  • x1 - the value of the element to find in the first dimension
  • xs - the value of the element to find in other dimensions

Return value

A reference to the element associated with that key.

Exceptions

std::out_of_range if the container does not have an element with the specified key

Complexity

\[ O(m \log n) \]

Notes

Unlike in a pareto::spatial_map, the insert operation for fronts is allowed to fail when the new element is already dominated by the front. In this case, the operator[] will return a reference to a placeholder that is not ultimately inserted in the front.

Info

See the section on spatial containers / element access for more information.

Example

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front<double, 3, unsigned> pf({min, max, min});
// Set some values
pf(-2.57664, -1.52034, 0.600798) = 17;
pf(-2.14255, -0.518684, -2.92346) = 32;
pf(-1.63295, 0.912108, -2.12953) = 36;
pf(-0.653036, 0.927688, -0.813932) = 13;
pf(-0.508188, 0.871096, -2.25287) = 32;
pf(-2.55905, -0.271349, 0.898137) = 6;
pf(-2.31613, -0.219302, 0) = 8;
pf(-0.639149, 1.89515, 0.858653) = 10;
pf(-0.401531, 2.30172, 0.58125) = 39;
pf(0.0728106, 1.91877, 0.399664) = 25;
pf(-1.09756, 1.33135, 0.569513) = 20;
pf(-0.894115, 1.01387, 0.462008) = 11;
pf(-1.45049, 1.35763, 0.606019) = 17;
pf(0.152711, 1.99514, -0.112665) = 13;
pf(-2.3912, 0.395611, 2.78224) = 11;
pf(-0.00292544, 1.29632, -0.578346) = 20;
pf(0.157424, 2.30954, -1.23614) = 6;
pf(0.453686, 1.02632, -2.24833) = 30;
pf(0.693712, 1.12267, -1.37375) = 12;
pf(1.49101, 3.24052, 0.724771) = 24;

// Access value
if (pf.contains({1.49101, 3.24052, 0.724771})) {
    std::cout << "Element access: " << pf(1.49101, 3.24052, 0.724771) << std::endl;
} else {
    std::cout << "{1.49101, 3.24052, 0.724771} was dominated" << std::endl;
}

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pf = pareto.front()
# Set some values
pf[-2.57664, -1.52034, 0.600798] = 17
pf[-2.14255, -0.518684, -2.92346] = 32
pf[-1.63295, 0.912108, -2.12953] = 36
pf[-0.653036, 0.927688, -0.813932] = 13
pf[-0.508188, 0.871096, -2.25287] = 32
pf[-2.55905, -0.271349, 0.898137] = 6
pf[-2.31613, -0.219302, 0] = 8
pf[-0.639149, 1.89515, 0.858653] = 10
pf[-0.401531, 2.30172, 0.58125] = 39
pf[0.0728106, 1.91877, 0.399664] = 25
pf[-1.09756, 1.33135, 0.569513] = 20
pf[-0.894115, 1.01387, 0.462008] = 11
pf[-1.45049, 1.35763, 0.606019] = 17
pf[0.152711, 1.99514, -0.112665] = 13
pf[-2.3912, 0.395611, 2.78224] = 11
pf[-0.00292544, 1.29632, -0.578346] = 20
pf[0.157424, 2.30954, -1.23614] = 6
pf[0.453686, 1.02632, -2.24833] = 30
pf[0.693712, 1.12267, -1.37375] = 12
pf[1.49101, 3.24052, 0.724771] = 24

# Access value
if [1.49101, 3.24052, 0.724771] in pf:
    print('Element access:', pf[1.49101, 3.24052, 0.724771])
else:
    print("[1.49101, 3.24052, 0.724771] was dominated")

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Element access: 24