I have this code for a class where I'm supposed to use the reduce() method to find the min and max values in an array. However, we are required to use only a single call to reduce. The return array should be of size 2, but I know that the reduce() method always returns an array of size 1.
Reduce to the min
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I'm able to obtain the minimum value using the code below, however I don't know how to obtain the max value in that same call. I assume that once I do obtain the max value that I just push it to the array after the reduce() method finishes.
function minMax(items) var minMaxArray = items.reduce(function (r, n) r[0] = (!r[0])? n : Math.min(r[0], n); r[1] = (!r[1])? n : Math.max(r[1], n); return r; , []); return minMaxArray;console.log(minMax([4, 1, 2, 7, 6]));
let arr = [8978, 'lol', -78, 989, NaN, null, undefined, 6, 9, 55, 989];let minMax = arr.reduce(([min, max], v) => [ Math.min(min, v) min, Math.max(max, v) max], [Infinity, -Infinity]);console.log(minMax);
We can accomplish this by declaring an empty array as the accumulator value for the reduce function, then carrying out a different set of operations on the last iteration of the reduce method. We do this by passing all four parameters to the reduce method (total, item, index, array) and using a comparison of the index to the array length to do something different on that last iteration.
if (index === pricesArray.length-1) means that on the last iteration of the reduce method through the prices array, a different set of operations occurs. Up to that point, we are merely recreating the prices array, which is trivial. But on the last iteration, after fully recreating the prices array, we do something different. We create another empty array, the one we intend to return. We then sort through the 'accumulatorArray' variable - which is the prices array, recreated, sorting it from lowest to highest. We now take the lowest price and highest price and store them in variables. Having sorted the array in ascending order, we know that the lowest is at index 0 and the highest, at index array.length - 1. We then push those variables into our previously declared return array. And instead of returning the accumulator variable itself, we return our own specially declared return array. The result is an array with the lowest price and then the highest price.
Computes the mean of elements across dimensions of a tensor.Reduces input_tensor along the dimensions given in axis.Unless keepdims is true, the rank of the tensor is reduced by 1 foreach entry in axis. If keepdims is true, the reduced dimensions are retainedwith length 1.
pruning away boxes that have high intersection-over-union (IOU) overlapwith previously selected boxes. Bounding boxes with score less thanscore_threshold are removed. Bounding boxes are supplied as[y1, x1, y2, x2], where (y1, x1) and (y2, x2) are the coordinates of anydiagonal pair of box corners and the coordinates can be provided as normalized(i.e., lying in the interval [0, 1]) or absolute. Note that this algorithmis agnostic to where the origin is in the coordinate system and moregenerally is invariant to orthogonal transformations and translationsof the coordinate system; thus translating or reflections of the coordinatesystem result in the same boxes being selected by the algorithm.The output of this operation is a set of integers indexing into the inputcollection of bounding boxes representing the selected boxes. The boundingbox coordinates corresponding to the selected indices can then be obtainedusing the tf.gather operation. For example: selected_indices = tf.image.non_max_suppression_v2( boxes, scores, max_output_size, iou_threshold, score_threshold) selected_boxes = tf.gather(boxes, selected_indices)This op also supports a Soft-NMS (with Gaussian weighting) mode (c.f.Bodla et al, ) where boxes reduce the scoreof other overlapping boxes instead of directly causing them to be pruned.To enable this Soft-NMS mode, set the soft_nms_sigma parameter to belarger than 0.
Reduces input along the dimensions given in axis. Unlesskeep_dims is true, the rank of the tensor is reduced by 1 for each entry inaxis. If keep_dims is true, the reduced dimensions areretained with length 1.
Over the years, new features such as list comprehensions, generator expressions, and built-in functions like sum(), min(), max(), all(), and any() were viewed as Pythonic replacements for map(), filter(), and reduce(). Guido planned to remove map(), filter(), reduce(), and even lambda from the language in Python 3.
When you call reduce(), passing my_add() and numbers as arguments, you get an output that shows all the operations that reduce() performs to come up with a final result of 10. In this case, the operations are equivalent to ((((0 + 1) + 2) + 3) + 4) = 10.
The call to reduce() in the above example applies my_add() to the first two items in numbers (0 and 1) and gets 1 as the result. Then reduce() calls my_add() using 1 and the next item in numbers (which is 2) as arguments, getting 3 as the result. The process is repeated until numbers runs out of items and reduce() returns a final result of 10.
This means that the first call to function will use the value of initializer and the first item of iterable to perform its first partial computation. After this, reduce() continues working with the subsequent items of iterable.
In this example, add() takes two arguments and returns their sum. So, you can use add() with reduce() to compute the sum of all the items of numbers. Since add() is written in C and optimized for efficiency, it may be your best choice when using reduce() for solving the sum use case. Note that the use of operator.add() is also more readable than using a lambda function.
The function my_prod() multiplies two numbers, a and b. The call to reduce() iterates over the items of numbers and computes their product by applying my_prod() to successive items. The final result is the product of all the items in numbers, which in this example is 24.
When you run reduce() with my_min_func() and my_max_func(), you get the minimum and maximum value in numbers, respectively. reduce() iterates over the items of numbers, compares them in cumulative pairs, and finally returns the minimum or maximum value.
Even though this solution takes only one line of code, it can still make your code unreadable or at least difficult to understand. Again, Python provides a tool to efficiently solve the any-true problem without using reduce(): the built-in function any().
A Python function called accumulate() lives in itertools and behaves similarly to reduce(). accumulate(iterable[, func]) accepts one required argument, iterable, which can be any Python iterable. The optional second argument, func, needs to be a function (or a callable object) that takes two arguments and returns a single value.
Avoid complex user-defined functions when using reduce(). These kinds of functions can make your code difficult to read and understand. You can use an explicit and readable for loop instead.
Your second-best option would be to use reduce() with operator.add(). The functions in operator are written in C and are highly optimized for performance. So, they should perform better than a user-defined function, a lambda function, or a for loop.
Over the years, reduce() has been replaced by more Pythonic tools like sum(), min(), max() all(), any(), among others. However, reduce() is still there and is still popular among functional programmers. If you have questions or thoughts about using reduce() or any of its Python alternatives, then be sure to post them in the comments below.
Apply function of two arguments cumulatively to the items of iterable, fromleft to right, so as to reduce the iterable to a single value. For example,reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) calculates ((((1+2)+3)+4)+5).The left argument, x, is the accumulated value and the right argument, y, isthe update value from the iterable. If the optional initializer is present,it is placed before the items of the iterable in the calculation, and serves asa default when the iterable is empty. If initializer is not given anditerable contains only one item, the first item is returned.Roughly equivalent to:
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Each listed input is operated on independently. For per-atom inputs,the group specified with this command means only atoms within thegroup contribute to the result. For per-atom inputs, if the computereduce/region command is used, the atoms must also currently be withinthe region. Note that an input that produces per-atom quantities maydefine its own group which affects the quantities it returns. Forexample, if a compute is used as an input which generates a per-atomvector, it will generate values of 0.0 for atoms that are not in thegroup specified for that compute.
Using a wildcard is the same as if the individual columns of the arrayhad been listed one by one. For example, the following two compute reducecommands are equivalent, since thecompute stress/atom command creates a per-atomarray with six columns:
The first two input values in the compute reduce command are vectorswith the IDs of the 2 atoms in each bond, using thecompute property/local command. The last inputvalue is bond distance, using thecompute bond/local command. Instead of taking themax of the two atom ID vectors, which does not yield usefulinformation in this context, the replace keywords will extract theatom IDs for the two atoms in the bond of maximum stretch. These atomIDs and the bond stretch will be printed with thermodynamic output. 2ff7e9595c
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