Following this question about sorting a list by another list, I tried to do the same thing - but from some reason it doesn\'t work for me. What am I missing?
You are sorting the numbers by their position in the order
list, but none of the numbers occur in the order list. In this case, indexOf will return -1 for everything, meaning everything is equal to everything else. In such a case, the resulting sort order is unspecified - though you may realistically assume that it would not change.
You can make a list of pairs :
[3.0, 5.0]
[1.0, 0.9]
[2.0, 10.4]
Then sort this list of pairs by the first value of each array :
[1.0, 0.9]
[2.0, 10.4]
[3.0, 5.0]
Here is the code :
List<Double> nums = Arrays.asList(5.0, 0.9, 10.4);
List<Double> order = Arrays.asList(3.0, 1.0, 2.0);
List<Double[]> pairs = new ArrayList<>();
for (int i = 0; i < nums.size(); i++) {
pairs.add(new Double[] {order.get(i), nums.get(i)});
}
pairs.sort(Comparator.comparing(pair -> pair[0]));
for (Double[] pair : pairs) {
System.out.print(pair[1] + " ");
}
Output :
0.9 10.4 5.0
The Comparator you are supplying calls indexOf
for every num
passed.
The returned values are -1
on all calls, so the order is preserverd as-is.
You need to sort natural.
Sorting by another list of Double
should be possible, but unnecessarily complicated, it would be simpler to provide a custom Object which sorts as desired.
Update
List<Double> nums = Arrays.asList(5.0, 0.9, 10.4);
List<Double> order = Arrays.asList(3.0, 1.0, 2.0);
Map<Double,Double> numToOrder = new HashMap<>();
for (int i = 0; i < nums.size(); ++i) {
numToOrder.put(nums.get(i), order.get(i));
}
nums.sort(Comparator.comparing(num -> numToOrder.get(num)));
System.out.println(nums);
Original (wrong) answer
(nums is modified in place, and the lambda returning key returns wrong results)
List<Double> nums = Arrays.asList(5.0, 0.9, 10.4);
List<Double> order = Arrays.asList(3.0, 1.0, 2.0);
nums.sort(Comparator.comparing(num -> order.get(nums.indexOf(num))));
System.out.println(nums);