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可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I want to fit lognormal distribution to my data, using python scipy.stats.lognormal.fit . According to the manual , fit returns shape, loc, scale parameters. But, lognormal distribution normally needs only two parameters : mean and standard deviation. How to interpret the results from scipy fit function? How to get mean and std.dev.? 回答1: The distributions in scipy are coded in a generic way wrt two parameter location and scale so that location is the parameter ( loc ) which shifts the distribution to the left or right, while scale is the

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have a 2D matrix and I want to take norm of each row. But when I use numpy.linalg.norm(X) directly, it takes the norm of the whole matrix. I can take norm of each row by using a for loop and then taking norm of each X[i] , but it takes a huge time since I have 30k rows. Any suggestions to find a quicker way? Or is it possible to apply np.linalg.norm to each row of a matrix? 回答1: Note that, as perimosocordiae shows , as of NumPy version 1.9, np.linalg.norm(x, axis=1) is the fastest way to compute the L2-norm. If you are computing an L2-norm

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I know that elasticsearch takes in account the length of a field when computing the score of the documents retrieved by a query. The shorter the field, the higher the weight (see The field-length norm ). I like this behaviour: when I search for iphone I am much more interested in iphone 6 than in Crappy accessories for: iphone 5 iphone 5s iphone 6 . Now, I would like to try to boost this stuff, let's say that I want to double its importance. I know that one can modify the score using the function score , and I guess that I can

Pandas数据处理基本操作 以下是本次实验使用的数据表，如有需要联系，请留言 import pandas as pd import numpy as np #读取文件 不同媒体或网站对电影的评分 fandango = pd . read_csv ( 'fandango_score_comparison.csv' ) fandango . head ( ) FILM RottenTomatoes RottenTomatoes_User Metacritic Metacritic_User IMDB Fandango_Stars Fandango_Ratingvalue RT_norm RT_user_norm Metacritic_norm Metacritic_user_nom IMDB_norm RT_norm_round RT_user_norm_round Metacritic_norm_round Metacritic_user_norm_round IMDB_norm_round Metacritic_user_vote_count IMDB_user_vote_count Fandango_votes Fandango_Difference 0 Avengers: Age of Ultron (2015) 74 86 66 7.1 7.8 5.0 4.5 3.70 4

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Given a 2-dimensional array in python, I would like to normalize each row with the following norms: Norm 1: L_1 Norm 2: L_2 Norm Inf: L_Inf I have started this code: from numpy import linalg as LA X = np.array([[1, 2, 3, 6], [4, 5, 6, 5], [1, 2, 5, 5], [4, 5,10,25], [5, 2,10,25]]) print X.shape x = np.array([LA.norm(v,ord=1) for v in X]) print x Output: (5, 4) # array dimension [12 20 13 44 42] # L1 on each Row How can I modify the code such that WITHOUT using LOOP, I can directly have the rows of the matrix normalized? (Given the norm

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I tried norm , but I think it gives the wrong result. (the norm of c(1, 2, 3) is sqrt(1*1+2*2+3*3) , but it returns 6 .. x1 <- 1:3 norm(x1) # Error in norm(x1) : 'A' must be a numeric matrix norm(as.matrix(x1)) # [1] 6 as.matrix(x1) # [,1] # [1,] 1 # [2,] 2 # [3,] 3 norm(as.matrix(x1)) # [1] 6 Does anyone know what's the function to calculate the norm of a vector in R? 回答1: This is a trivial function to write yourself: norm_vec <- function(x) sqrt(sum(x^2)) 回答2: norm(c(1,1), type="2") # 1.414214 norm(c(1, 1, 1), type="2") # 1.732051 回答3:

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: My knowledge of maths is limited which is why I am probably stuck. I have a spectra to which I am trying to fit two Gaussian peaks. I can fit to the largest peak, but I cannot fit to the smallest peak. I understand that I need to sum the Gaussian function for the two peaks but I do not know where I have gone wrong. An image of my current output is shown: The blue line is my data and the green line is my current fit. There is a shoulder to the left of the main peak in my data which I am currently trying to fit, using the following

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I want to simulate keydown events on a given textarea element in an html page. Since I am using chrome, I called initKeyboardEvent on my variable and I passed the keyCode I want to type into the textarea. Here is what I tried: var keyEvent = document . createEvent ( 'KeyboardEvent' ); keyEvent . initKeyboardEvent ( 'keydown' , true , false , null , 0 , false , 0 , false , 77 , 0 ); inputNode . dispatchEvent ( keyEvent ); In this code I'm typing the letter m however the textarea is only getting the keyCode 13 which is the Enter key.

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I create a Qt3D mesh like this: Qt3DCore::QEntity *newEntity = new Qt3DCore::QEntity(); Qt3DExtras::QConeMesh *mesh =new Qt3DExtras::QConeMesh(); mesh->setTopRadius(0.2); mesh->setBottomRadius(1.0); mesh->setLength(2.0); for(int i = 0; i < mesh->geometry()->attributes().size(); ++i) { mesh->geometry()->attributes().at(i)->buffer()->setSyncData(true); // To have access to data } newEntity->addComponent(mesh); The created mesh looks like this: Later in the code, I try to export the above mesh in STL binary format . To do so, I extract the

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I having trouble getting rid of this error in the code below. There are 3 functions; dev, norm and clcMatA. The first two functions are called in the third one. But they are not recognized as functions. I have defined them like I do for other functions but I didn't get such errors before. The errors: Error 1 error #6404: This name does not have a type, and must have an explicit type. [DEV] D:\Users\Vahid\Documents\Visual Studio 2008\Projects\Tst\Tst\Source1.for 66 Error 2 error #6404: This name does not have a type, and must have an explicit