assign

问题 I've got this code: function margeOptions(options, passedOptions) { options = Object.assign(options, passedOptions); } let passedOpts = {a: true}; let opts = {a: false}; margeOptions(opts, passedOpts); console.log(opts); // as expected returns {a: true} but when I change function a little bit, like this: function margeOptions(options, passedOptions) { options = Object.assign({}, options, passedOptions); } let passedOpts = {a: true}; let opts = {a: false}; margeOptions(opts, passedOpts);

问题 In my code I have to create an object assigning some values, something like this: assign(paste("a","bis",sep="."),rep(NA,5)) then I have to replace some of them, like this: get(paste("a","bis",sep="."))[1:2] <- 7:8 But I get the following error: "Error in get(paste("a", "bis", sep = "."))[1:2] <- 7:8 : target of assignment expands to non-language object". Of course the code above is a simplified version of the real one. What I'm trying to do is to build a loop which allows me to replace in a

问题 I am looking for a way to use a list of filter arguments to produce different objects. I have a data set for which I want to make several graphs. However, I want all these graphs based on subsets of the dataset. For illustrative purposes I have made the following data. df <- data.frame(type = c("b1", "b2", "b1", "b2"), yield = c("15", "10", "5", "0"), temperature = c("2", "21", "26", "13"), Season = c("Winter", "Summer", "Summer", "Autumn"), profit = c(TRUE, TRUE, FALSE, FALSE)) Also, I have

问题 I wrote a test program, and now it seems that if I don't use $/ in a method signature because I have to use .match inside the method, I can no long make anything. What did I do wrong? A further question is that if .match sets $/ , and $/ is read-only, then I cannot have $/ in the signature of a method that contains a .match statement, and I cannot have more than one .match inside the method because each .match will try to set the read-only $/ . This will be very awkward to program. Here is

问题 I wrote a test program, and now it seems that if I don't use $/ in a method signature because I have to use .match inside the method, I can no long make anything. What did I do wrong? A further question is that if .match sets $/ , and $/ is read-only, then I cannot have $/ in the signature of a method that contains a .match statement, and I cannot have more than one .match inside the method because each .match will try to set the read-only $/ . This will be very awkward to program. Here is

问题 I'd implementing a simple linked list. This is the (working) code I had so far: pub struct LinkedList<T> { start: Option<Box<Link<T>>>, } impl<T> LinkedList<T> { pub fn new() -> LinkedList<T> { return LinkedList { start: None }; } } struct Link<T> { value: Box<T>, next: Option<Box<Link<T>>>, } impl<T> Link<T> { fn new_end(value: T) -> Link<T> { return Link::new(value, None); } fn new(value: T, next: Option<Box<Link<T>>>) -> Link<T> { return Link { value: Box::new(value), next, }; } } Next on

问题 I'd implementing a simple linked list. This is the (working) code I had so far: pub struct LinkedList<T> { start: Option<Box<Link<T>>>, } impl<T> LinkedList<T> { pub fn new() -> LinkedList<T> { return LinkedList { start: None }; } } struct Link<T> { value: Box<T>, next: Option<Box<Link<T>>>, } impl<T> Link<T> { fn new_end(value: T) -> Link<T> { return Link::new(value, None); } fn new(value: T, next: Option<Box<Link<T>>>) -> Link<T> { return Link { value: Box::new(value), next, }; } } Next on

问题 I'd implementing a simple linked list. This is the (working) code I had so far: pub struct LinkedList<T> { start: Option<Box<Link<T>>>, } impl<T> LinkedList<T> { pub fn new() -> LinkedList<T> { return LinkedList { start: None }; } } struct Link<T> { value: Box<T>, next: Option<Box<Link<T>>>, } impl<T> Link<T> { fn new_end(value: T) -> Link<T> { return Link::new(value, None); } fn new(value: T, next: Option<Box<Link<T>>>) -> Link<T> { return Link { value: Box::new(value), next, }; } } Next on

问题 This question already has answers here : Why wouldn't I use Child.prototype = Parent.Prototype rather than Child.prototype = new Parent(); for Javascript inheritance? (3 answers) Benefits of using `Object.create` for inheritance (4 answers) Closed 1 year ago . So I'm trying to understand the difference between o1.prototype = Object.create(o2.prototype) and o1.prototype = o2.prototype . According to the answer to this question, the former sets obj2.prototype to by the prototype for obj1

问题 Why does b have a value? I think b should be null, because there is no return in function f. f <- function(){ a <- 10 } b <- f() b # [1] 10 回答1: <- operator returns assignement invisibly, which allows b <- a <- 1 b a > b [1] 1 > a [1] 1 来源： https://stackoverflow.com/questions/63950864/why-does-b-have-a-value