evaluation

问题 My question is related to Latent Dirichlet Allocation . Suppose we apply LDA on our dataset, then apply fit transform on that. the output is a matrix that is a collection of five documents. Each document consists of three topics. othe output is below: [[ 0.0922935 0.09218227 0.81552423] [ 0.81396651 0.09409428 0.09193921] [ 0.05265482 0.05240119 0.89494398] [ 0.05278187 0.89455775 0.05266038] [ 0.85209554 0.07338382 0.07452064]] So, this is the matrix that will be sent to a classification

问题 #include <iostream> #include <sstream> #include <stack> #include <limits> #include <string> using namespace std; int main() { string input; cout << "Enter a postfix expression: " << endl; getline(cin, input); int operand1, operand2, result,number; stack<char>operation; stringstream temp; int i=0; while (i < input.length()) { if (isdigit(input[i])) { operation.push(input[i]); } else { operand2 = operation.top(); temp << operation.top(); operation.pop(); operand1 = operation.top(); temp <<

问题 If I give Mathematica the input TreeForm[Unevaluated[4^5]] I expect to see three boxes -- power, 4, and 5. Instead I see a single box with 1024. Can anyone explain? 回答1: Compare TreeForm@Unevaluated[4^5] with TreeForm@Hold[4^5] From the help: Unevaluated[expr] represents the unevaluated form of expr when it appears as the argument to a function. and Hold[expr] maintains expr in an unevaluated form. so, as Unevaluated[4^5] gets to TreeForm ... it gets evaluated ... It works like this: f[x_+y_]

问题 If I give Mathematica the input TreeForm[Unevaluated[4^5]] I expect to see three boxes -- power, 4, and 5. Instead I see a single box with 1024. Can anyone explain? 回答1: Compare TreeForm@Unevaluated[4^5] with TreeForm@Hold[4^5] From the help: Unevaluated[expr] represents the unevaluated form of expr when it appears as the argument to a function. and Hold[expr] maintains expr in an unevaluated form. so, as Unevaluated[4^5] gets to TreeForm ... it gets evaluated ... It works like this: f[x_+y_]

问题 Could someone please help me with the following: I need to change my xgboost training model with caret package to an undefault metric RMSLE. By default caret and xgboost train and measure in RMSE. Here are the lines of code: create custom summary function in caret format custom_summary = function(data, lev = NULL, model = NULL){ out = rmsle(data[, "obs"], data[, "pred"]) names(out) = c("rmsle") out } create control object control = trainControl(method = "cv", number = 2, summaryFunction =

问题 I thought that i-- is a shorthand for i = i - 1 , but I discovered that both evaluate different: var i = 1; while (i = i - 1) {…} In this case, i is 0 , which evaluates to false . This works as expected. var i = 1; while (i--) {…} i should be 0 and evaluate to false , but it does not. It evaluates to true . Is this a bug, or is there a reason for it? 回答1: The i-- will be evaluated only after the loop condition is evaluated but before the statements within the loop. This is the decrement

问题 From a data.table d such as for example require(data.table) d = data.table(a = 1:4, b = 11:14, c = 21:24, group = c(1,1,2,2)) I would like to sum all variables which names are stored in the vector varsToSum by unique values of group . varsToSum = c("a", "b") For the above d and varsToSum , the expected outcome is d[,list(a = sum(a), b = sum(b)),list(group)] group a b 1: 1 3 23 2: 2 7 27 Related posts: Select / assign to data.table variables which names are stored in a character vector How to

问题 i'm trying to evaluate SIFT and SURF Detectors by Repeatability criteria. i find out that below method can find Repeatability ,Correspondence of SIFT and SURF cv::evaluateFeatureDetector(img_1c, img_2c, h12, &key_points_1, &key_points_2, repeatability, corrCounter); some of the result are listed below: Number Repeatibility Correspond Keypoint 1st Keypoint 2th 1to2 0.7777778 140 224 180 1to3 0.7125 114 224 161 1to4 0.704918 86 224 123 1to5 0.6853933 61 224 89 1to6 0.6521739 45 224 69 for first

问题 suppose i a have a multiplicative expression with lots of multiplicands (small expressions) expression = a*b*c*d*....*w where for example c is (x-1), d is (y**2-16), k is (x y-60)..... x,y are numbers and i know that c,d,k,j maybe zero Does the order i write the expression matters for faster evaluation? Is it better to write c d k j....*w or python will evaluate all expression no matter the order i write? 回答1: Python v2.6.5 does not check for zero values. def foo(): a = 1 b = 2 c = 0 return a

问题 I have question about how to evaluate the information retrieve result is good or not such as calculate the relevant document rank, recall, precision ,AP, MAP..... currently, the system is able to retrieve the document from the database once the users enter the query. The problem is I do not know how to do the evaluation. I got some public data set such as "Cranfield collection" dataset link it contains 1.document 2.query 3.relevance assesments DOCS QRYS SIZE* Cranfield 1,400 225 1.6 May I