Starting from .NET 6 MaxBy LINQ method is available. var result = items.MaxBy(i => i.Date); Prior to .NET 6: O(n log n): var result = items.OrderByDescending(i => i.Date).First(); O(n) – but iterates over the sequence twice: var max = items.Max(i => i.Date); var result = items.First(i => i.Date == max); Or you can use MoreLINQ … Read more
With this setup: >>> A = np.array([0,1,2]) >>> B = np.array([1,0,3]) >>> C = np.array([3,0,4]) You can either do: >>> np.maximum.reduce([A,B,C]) array([3, 1, 4]) Or: >>> np.vstack([A,B,C]).max(axis=0) array([3, 1, 4]) I would go with the first option.
How about getting the column name from the metadata: val selectedColumnName = df.columns(q) //pull the (q + 1)th column from the columns array df.agg(min(selectedColumnName), max(selectedColumnName))
Of course it would, because the start number should be Infinity for Math.min. All number that are lower than positive infinity should be the smallest from a list, if there are no smaller. And for Math.max it’s the same; all numbers that are larger than negative infinity should be the biggest if there are no … Read more
heapq.nlargest: >>> import heapq, random >>> heapq.nlargest(3, (random.gauss(0, 1) for _ in xrange(100))) [1.9730767232998481, 1.9326532289091407, 1.7762926716966254]
Use argmax() idxmax() to get the index of the max value. Then you can use loc df.loc[df[‘favcount’].idxmax(), ‘sn’] Edit: argmax() is now deprecated, switching for idxmax()
Before voting on this answer, please test (and verify) this on your machine and comment/add the results. Note that I used a vector size of 1000*1000*1000 for my tests. Currently, this answer has 19 upvotes but only one posted results, and these results did not show the effect described below (though obtained with a different … Read more
You need to include the header file <limits>, which is where std::numeric_limits is defined. Your Mac compiler was helping you out by automatically including that header file; however, you should not rely on that behavior and explicitly include any header files you need.
If possible, I recommend using C++11 or newer which allows you to compute the desired result w/out implementing your own function (std::min). As already pointed out in one of the comments, you can do T minimum(std::min({x, y, z})); or T minimum = std::min({x, y, z}); which stores the minimum of the variables x, y and … Read more