coding-efficiency

问题 Here is the code below (ans to a CP qs) The time limit for execution is 6 seconds but my code is slower than. How can I make it more memory and time efficient ? Input: the input begins with the number t of test cases in a single line ( t <= 10 ). In each of the next t lines there are two numbers m and n ( 1 <= m <= n <= 1000000000 , n-m <= 100000 ) separated by a space. Output : for every test case print all prime numbers p such that m <= p <= n , one number per line, test cases separated by

问题 I would like to construct a network in Caffe in which the incoming data is split up initially, passes separately through the same set of layers, and is finally recombined using an eltwise layer. After this, all the parts will move as a single blob. The layer configuration of the part of the network for which the data moves parallely will be identical, except for the learned parameters. Is there a way to define this network in Caffe without redefining the layers through which the different

问题 I want to compute a "distance" matrix, similarly to scipy.spatial.distance.cdist, but using the intersection over union (IoU) between "bounding boxes" (4-dimensional vectors), instead of a typical distance metric (like the Euclidean distance). For example, let's suppose that we have two collections of bounding boxes, such as import numpy as np A_bboxes = np.array([[0, 0, 10, 10], [5, 5, 15, 15]]) array([[ 0, 0, 10, 10], [ 5, 5, 15, 15]]) B_bboxes = np.array([[1, 1, 11, 11], [4, 4, 13, 13], [9

问题 I want to compute a "distance" matrix, similarly to scipy.spatial.distance.cdist, but using the intersection over union (IoU) between "bounding boxes" (4-dimensional vectors), instead of a typical distance metric (like the Euclidean distance). For example, let's suppose that we have two collections of bounding boxes, such as import numpy as np A_bboxes = np.array([[0, 0, 10, 10], [5, 5, 15, 15]]) array([[ 0, 0, 10, 10], [ 5, 5, 15, 15]]) B_bboxes = np.array([[1, 1, 11, 11], [4, 4, 13, 13], [9

问题 In R, the default method when using a factor variable in regression is to use contrasts. I.E we set a reference class, and then the results are reported as (factor) vs. reference. For example, if we had 3 groups, and set 1 to be the reference group, then the results will be (2 vs 1) and (3 vs 1). But we don't get to see 3 vs 2. I know that you can get this by rerunning the regression and re-leveling with 2 as the reference class. But is there any way to get all comparisons in one line of code

问题 In R, the default method when using a factor variable in regression is to use contrasts. I.E we set a reference class, and then the results are reported as (factor) vs. reference. For example, if we had 3 groups, and set 1 to be the reference group, then the results will be (2 vs 1) and (3 vs 1). But we don't get to see 3 vs 2. I know that you can get this by rerunning the regression and re-leveling with 2 as the reference class. But is there any way to get all comparisons in one line of code

问题 Yesterday I asked a question about improving efficiency in my code. Today I have another question in the same spirit of trying to write less lines of code to accomplish repetitive tasks. I have the following code: function myIntroductionText() { introPos.style.display = 'block'; posOne.style.display = 'none'; posTwo.style.display = 'none'; posThree.style.display = 'none'; posFour.style.display = 'none'; posFive.style.display = 'none'; posSix.style.display = 'none'; posSeven.style.display =

问题 I've asked this on math overflow and used comments to clarify/overstate my question. I hope it has the intended effect and doesn't come off as jarring. I'm attempting to find what subset of numbers reach a known average. I have a list of known values, negative and possible decimals. They look something like this {-.32,-.64,-.12,.08,-.54,-.43, ...} It's around 50 numbers in some cases, though this problem would be tested for other cases as well. The set mostly contains negative decimal numbers

问题 This code works, but reading posts on here I get the impression it is probably not a very "Pythonic" solution. Is there a better more efficient way to solve this specific problem: What this code does: it counts instances of one string found in another and return the count. It raises an error in case the user tries to pass in an empty string. The version of the code I came up with but wondering if there is a better more efficient more "Pythonic" way to do this: def count_string(raw_string,

问题 I want to create a sparse matrix consisting mainly of -1, but also includes some 0 and 1. This is part of a larger project, so it is important that I do not switch -1 with 0. As default, sparse(A) in Matlab keeps track of only non-zero elements. Is there a way to keep track of only non-(minus one) elements? For example, if A = -1 -1 -1 0 1 -1 -1 -1 Then new_sparse(A) = (1,4) = 0 (2,1) = 1 Thanks! 回答1: No, there is no way to override sparse to use different values. What you can do, though time