evaluation

问题 I would like if possible how to make this Threshold for Evaluation and validation of created R-CNN object Detector, i tried to make it in the attached scripts but it does not work, I want to make Threshold for score that like below 0.58 that score and bboxes should not be appeared Herein the code:- load('gTruth.mat') output = selectLabels(gTruth,'signal'); if ~isfolder(fullfile('EvaluationData')) mkdir EvaluationData addpath('EvaluationData'); evaluationData = objectDetectorTrainingData

问题 The R package mice comes with following example: library("mice") imp <- mice(nhanes) fit <- with(data=imp,exp=lm(bmi~hyp+chl)) I want a flexible call of with() like: model_formula <- bmi~hyp+chl fit <- with(data=imp,exp=lm(model_formula)) But this throws Error in eval(predvars, data, env) : object 'bmi' not found . I searched for similar problems. The closet problem I found was Help understand the error in a function I defined in R. My impression is, that writing exp=lm(model_formula) the

问题 In my years of using Eclipse, I've never seen this and can't figure out what's wrong. For some reason, the Expressions View (as well as the Display View) will no longer evaluate even simple expressions (like "5", "Hello", "myVariable", "1+2"). What I get is, "Evaluations must contain either an expression or a block of well-formed statements". Weirder still, if I hover over a variable, the tooltip window shows the correct value and let's me drill into the object. I've tried restarting,

问题 In my years of using Eclipse, I've never seen this and can't figure out what's wrong. For some reason, the Expressions View (as well as the Display View) will no longer evaluate even simple expressions (like "5", "Hello", "myVariable", "1+2"). What I get is, "Evaluations must contain either an expression or a block of well-formed statements". Weirder still, if I hover over a variable, the tooltip window shows the correct value and let's me drill into the object. I've tried restarting,

问题 All this time my thinking of short circuit evaluations seems to be wrong. In javascript: var a = false, b = true, c=true; a && b || c; // Evaluates to true Compared to var a = false, b = true, c=true; a && (b || c); // Evaluates to true Why doesn't the VM stop when it sees that a is false? More explicit example: function a(){ console.log("I'm A"); return false; } function b(){ console.log("I'm B"); return true; } function c(){ console.log("I'm C"); return true; } a() && b() || c(); The output

问题 All this time my thinking of short circuit evaluations seems to be wrong. In javascript: var a = false, b = true, c=true; a && b || c; // Evaluates to true Compared to var a = false, b = true, c=true; a && (b || c); // Evaluates to true Why doesn't the VM stop when it sees that a is false? More explicit example: function a(){ console.log("I'm A"); return false; } function b(){ console.log("I'm B"); return true; } function c(){ console.log("I'm C"); return true; } a() && b() || c(); The output

问题 Given a R expression which represents a sum of terms like expr <- expression(a + b * c + d + e * f) I would like to retrieve the set of all the terms of the sum as a list of names or expressions. So in the example, the elements would be: a , b * c , d and e * f . The following comes from a comment. The tems could themselves contain a + operator as in expression(a + b * c + d + e * (f + g)) so we need some understanding of the R language. Is there a simple way to proceed e.g., using call_tree

问题 Using the rlang package, I wonder what is the difference between sym() and parse_expr() . Consider for example the following expressions: ex1 = sym('a') ex2 = parse_expr('a') They both return a identical(ex1, ex2) [1] TRUE Suppose now I need a quosure: ex3 = quo(!!sym('a')) ex4 = quo(!!parse_expr('a')) In both case, the result is: <quosure> expr: ^a env: global identical(ex3, ex4) [1] TRUE However, the two following are not the same for some reasons. ex5 = quo(!!sym('a - b')) ex6 = quo(!

问题 I use racket and I got the result 4 for following simple code: (let/cc done ((let/cc esc (done (+ 1 (let/cc k (esc k))))) 3)) and I was going to execute this code step-by-step. First, I changed the first let/cc into the form of call/cc like below: (call/cc (λ (done) ((let/cc esc (done (+ 1 (let/cc k (esc k))))) 3))) Of course, this produces 4 also. Second, since I found the mechanism of call/cc in the internet which says call/cc do following 4 steps: Captures the current continuation.

问题 According to ISO C11 - 6.5.16.3, it says that An assignment operator stores a value in the object designated by the left operand. An assignment expression has the value of the left operand after the assignment, but is not an lvalue. The type of an assignment expression is the type the left operand would have after lvalue conversion. The side effect of updating the stored value of the left operand is sequenced after the value computations of the left and right operands. The evaluations of the