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Programming-Idioms

# 280 Filter map
Remove all the elements from the map m that don't satisfy the predicate p.
Keep all the elements that do satisfy p.

Explain if the filtering happens in-place, i.e. if m is reused or if a new map is created.
New implementation

Be concise.

Be useful.

All contributions dictatorially edited by webmasters to match personal tastes.

Please do not paste any copyright violating material.

Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations 
for (auto it = m.begin(); it != m.end();) {
  if (!p(it->second)) {
    it = m.erase(it);
  } else {
    ++it;
  }
}
  real, dimension(:), allocatable :: a
  a = pack(a,p(a))
!
  elemental logical function p(x)
    real, intent(in) :: x
    p = x > 0.7
  end function p


for k, v := range m {
	if !p(v) {
		delete(m, k)
	}
}
import "maps"
maps.DeleteFunc(m, func(k K, v V) bool {
	return !p(v)
})
m.entrySet().removeIf(e -> !p.test(e));
import java.util.Iterator;
Iterator<?> i = m.entrySet().iterator();
while (i.hasNext()) 
    if (!p.test(i.next())) i.remove();
uses classes;
for i := m.count-1 downto 0 do
  if not p(m.items[i]) then m.delete(i);
$p = sub { $_[0] };

while ( ($k,$v) = each %m ) {
    $f{$k} = $v if $p->($v);
}

%m = %f;
$p = sub { $_[0] };

foreach $k (keys %m) {
    delete $m{$k} if not $p->( $m{$k} );            
}

m = {k:v for k, v in m.items() if p(v)}
for k in list(m):
    if p(m[k]): m.pop(k)

m = dict(filter(p, m.items()))
m.select{|k,v| p(v) }
m.retain(|_, &mut v| p(v));

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