等效于带有键/Schwartzian转换的Python列表排序 [英] Equivalent of Python's list sort with key / Schwartzian transform
问题描述
在Python中,给定一个列表,我可以按一个键函数对其进行排序,例如:
In Python, given a list, I can sort it by a key function, e.g.:
>>> def get_value(k):
... print "heavy computation for", k
... return {"a": 100, "b": 30, "c": 50, "d": 0}[k]
...
>>> items = ['a', 'b', 'c', 'd']
>>> items.sort(key=get_value)
heavy computation for a
heavy computation for b
heavy computation for c
heavy computation for d
>>> items
['d', 'b', 'c', 'a']
如您所见,列表不是按字母数字排序的,而是按返回值get_value()
排序的.
As you see, the list was sorted not alphanumerically but by the return value of get_value()
.
C ++是否有等效功能? std::sort()
仅允许我提供一个自定义比较器(等效于Python的items.sort(cmp=...)
),而不是键函数.如果没有,我是否可以在我的代码中找到经过测试的,有效的,可公开获得的等效方法?
Is there an equivalent in C++? std::sort()
only allows me to provide a custom comparator (equivalent of Python's items.sort(cmp=...)
), not a key function. If not, is there any well-tested, efficient, publicly available implementation of the equivalent I can drop into my code?
请注意,Python版本仅每个元素调用一次key
函数,而不是每个比较调用两次.
Note that the Python version only calls the key
function once per element, not twice per comparison.
推荐答案
您可以自己滚动:
template <typename RandomIt, typename KeyFunc>
void sort_by_key(RandomIt first, RandomIt last, KeyFunc func)
{
using Value = decltype(*first);
std::sort(first, last, [=](const ValueType& a, const ValueType& b) {
return func(a) < func(b);
});
}
如果KeyFunc
太昂贵,则必须使用值创建一个单独的向量.
If KeyFunc
is too expensive, you'll have to create a separate vector with the values.
我们甚至可以一起破解一个类,使我们仍然可以使用std::sort
:
We can even hack together a class that will allow us to still use std::sort
:
template <typename RandomIter, typename KeyFunc>
void sort_by_key(RandomIter first, RandomIter last, KeyFunc func)
{
using KeyT = decltype(func(*first));
using ValueT = typename std::remove_reference<decltype(*first)>::type;
struct Pair {
KeyT key;
RandomIter iter;
boost::optional<ValueT> value;
Pair(const KeyT& key, const RandomIter& iter)
: key(key), iter(iter)
{ }
Pair(Pair&& rhs)
: key(std::move(rhs.key))
, iter(rhs.iter)
, value(std::move(*(rhs.iter)))
{ }
Pair& operator=(Pair&& rhs) {
key = std::move(rhs.key);
*iter = std::move(rhs.value ? *rhs.value : *rhs.iter);
value = boost::none;
return *this;
}
bool operator<(const Pair& rhs) const {
return key < rhs.key;
}
};
std::vector<Pair> ordering;
ordering.reserve(last - first);
for (; first != last; ++first) {
ordering.emplace_back(func(*first), first);
}
std::sort(ordering.begin(), ordering.end());
}
或者,如果这太过分了,这就是我原来的解决方案,它要求我们编写自己的sort
Or, if that's too hacky, here's my original solution, which requires us to write our own sort
template <typename RandomIt, typename KeyFunc>
void sort_by_key_2(RandomIt first, RandomIt last, KeyFunc func)
{
using KeyT = decltype(func(*first));
std::vector<std::pair<KeyT, RandomIt> > ordering;
ordering.reserve(last - first);
for (; first != last; ++first) {
ordering.emplace_back(func(*first), first);
}
// now sort this vector by the ordering - we're going
// to sort ordering, but each swap has to do iter_swap too
quicksort_with_benefits(ordering, 0, ordering.size());
}
尽管现在我们必须重新实现quicksort:
Although now we have to reimplement quicksort:
template <typename Key, typename Iter>
void quicksort_with_benefits(std::vector<std::pair<Key,Iter>>& A, size_t p, size_t q) {
if (p < q) {
size_t r = partition_with_benefits(A, p, q);
quicksort_with_benefits(A, p, r);
quicksort_with_benefits(A, r+1, q);
}
}
template <typename Key, typename Iter>
size_t partition_with_benefits(std::vector<std::pair<Key,Iter>>& A, size_t p, size_t q) {
auto key = A[p].first;
size_t i = p;
for (size_t j = p+1; j < q; ++j) {
if (A[j].first < key) {
++i;
std::swap(A[i].first, A[j].first);
std::iter_swap(A[i].second, A[j].second);
}
}
if (i != p) {
std::swap(A[i].first, A[p].first);
std::iter_swap(A[i].second, A[p].second);
}
return i;
}
下面给出一个简单的示例:
Which, given a simple example:
int main()
{
std::vector<int> v = {-2, 10, 4, 12, -1, -25};
std::sort(v.begin(), v.end());
print(v); // -25 -2 -1 4 10 12
sort_by_key_2(v.begin(), v.end(), [](int i) { return i*i; });
print(v); // -1 -2 4 10 12 -25
}
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