std::erase_if (std::unordered_map)
From cppreference.com
< cpp | container | unordered map
Defined in header <unordered_map>
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template< class Key, class T, class Hash, class KeyEqual, class Alloc, class Pred > typename std::unordered_map<Key,T,Hash,KeyEqual,Alloc>::size_type |
(since C++20) | |
Erases all elements that satisfy the predicate pred
from the container. Equivalent to
auto old_size = c.size(); for (auto i = c.begin(), last = c.end(); i != last; ) { if (pred(*i)) { i = c.erase(i); } else { ++i; } } return old_size - c.size();
Parameters
c | - | container from which to erase |
pred | - | predicate that returns true if the element should be erased |
Return value
The number of erased elements.
Complexity
Linear.
Example
Run this code
#include <unordered_map> #include <iostream> template<typename Os, typename Container> inline Os& operator<<(Os& os, Container const& cont) { os << "{"; for (const auto& item : cont) { os << "{" << item.first << ", " << item.second << "}"; } return os << "}"; } int main() { std::unordered_map<int, char> data {{1, 'a'},{2, 'b'},{3, 'c'},{4, 'd'}, {5, 'e'},{4, 'f'},{5, 'g'},{5, 'g'}}; std::cout << "Original:\n" << data << '\n'; const auto count = std::erase_if(data, [](const auto& item) { auto const& [key, value] = item; return (key & 1) == 1; }); std::cout << "Erase items with odd keys:\n" << data << '\n' << count << " items removed.\n"; }
Possible output:
Original: {{5, e}{4, d}{3, c}{2, b}{1, a}} Erase items with odd keys: {{4, d}{2, b}} 3 items removed.
See also
removes elements satisfying specific criteria (function template) |