my TIME Enum contains Annual, Monthly, weekly, daily and Hourly.
Here I want to decide which is the minimum and want to return that.
How can I do this ? Here is the code I tried.
private Time DecideMinTime(IEnumerable<Time> g)
{
var minTime = Time.Hourly;
foreach (var element in g)
{
minTime = element;
}
return minTime;
}
Assuming that the numeric value of the enum elements decides what the minimum is:
private Time DecideMinTime(IEnumerable<Time> g)
{
if (g == null) { throw new ArgumentNullException("g"); }
return (Time)g.Cast<int>().Min();
}
If the numeric values indicate the opposite order then you would use .Max()
instead of .Min()
.
As indicated, the numeric order is not consistent. This can be worked around simply by using a mapping indicating the correct order:
static class TimeOrdering
{
private static readonly Dictionary<Time, int> timeOrderingMap;
static TimeOrdering()
{
timeOrderingMap = new Dictionary<Time, int>();
timeOrderingMap[Time.Hourly] = 1;
timeOrderingMap[Time.Daily] = 2;
timeOrderingMap[Time.Weekly] = 3;
timeOrderingMap[Time.Monthly] = 4;
timeOrderingMap[Time.Annual] = 5;
}
public Time DecideMinTime(IEnumerable<Time> g)
{
if (g == null) { throw new ArgumentNullException("g"); }
return g.MinBy(i => timeOrderingMap[i]);
}
public TSource MinBy<TSource, int>(
this IEnumerable<TSource> self,
Func<TSource, int> ordering)
{
if (self == null) { throw new ArgumentNullException("self"); }
if (ordering == null) { throw new ArgumentNullException("ordering"); }
using (var e = self.GetEnumerator()) {
if (!e.MoveNext()) {
throw new ArgumentException("Sequence is empty.", "self");
}
var minElement = e.Current;
var minOrder = ordering(minElement);
while (e.MoveNext()) {
var curOrder = ordering(e.Current);
if (curOrder < minOrder) {
minOrder = curOrder;
minElement = e.Current;
}
}
return minElement;
}
}
}
To make it easier you can assign int values to your enum:
enum Time : byte {Hourly=1, Daily=2, Weekly=3, Monthly=4, Annual=5};
and then
private static Time DecideMinTime(IEnumerable<Time> g)
{
return g.Min();
}
That way you avoid casting back and forth.
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