Question

使用R我喜欢自动找到最佳觅食图形练习的收益递减曲线的切线。曲线如下所示：

const apiCall = function(searchInput) {
  const searchTerm = searchInput
  var apigClientFactory = require('aws-api-gateway-client').default;

  const config =  {
                apiKey: 'xxxxxx',
                invokeUrl:'https://xxxx.execute-api.us-east-2.amazonaws.com'
                }

  var apigClient = apigClientFactory.newClient(config);
  var params = {
        //This is where any header, path, or querystring request params go. The key is the parameter named as defined in the API
        //  userId: '1234',
        search_keyword: 'Ambassador'
      };
      // Template syntax follows url-template https://www.npmjs.com/package/url-template
  var pathTemplate = '/beta/testDB'
  var method = 'GET';
  var additionalParams = {
      //If there are any unmodeled query parameters or headers that need to be sent with the request you can add them here
      headers: {

            },
      queryParams: {
        search_keyword: 'Ambassador'
              }
    }
debugger
    apigClient.invokeApi(params, pathTemplate, method, additionalParams)
             .then(function(result)
                {
                    //This is where you would put a success callback
                    return JSON.parse(result.data)

                }).catch( function(result){
                    //This is where you would put an error callback
                })
}

切线从点（-40,40）开始，理想情况下，我会在切线所触及的曲线上找到该点。

通过试用＆amp;它看起来应该是错误的：

ins <- 40
t <- 30
avg <- 30
curve(ins/2^(x/10+1-1),
    0,2*t,xlim=c(-2*t,2*t),ylim=c(ins,0),
    xlab="time",ylab="food",type="l",lty=1,col=4,lwd=3,axes=FALSE)
axis(1,pos=ins); axis(2,pos=0)

Answer 1

设定：

library(sjPlot)
library(sjmisc)

set_theme(base = theme_classic(),  axis.title.size = 0,  geom.label.size = 4.5, 
                  axis.textsize.x = 1.1, axis.textsize.y = 1.1 )

sjp.frq(base$x, type = c("bar"), sort.frq = c("desc"), geom.colors = "grey")

ins <- 40 t <- 30 avg <- 30返回一个函数，该函数将函数的值作为主要结果，并将渐变作为属性。

deriv(...,function.arg=TRUE)

我们需要求解等式FUN <- deriv(~ins/2^(x/10+1-1),"x",function.arg=TRUE) curve(FUN(x), 0,2*t,xlim=c(-2*t,2*t), ylim=c(ins,0), xlab="time",ylab="food",type="l", lty=1,col=4,lwd=3,axes=FALSE) axis(1,pos=ins); axis(2,pos=0)（其中((40+x)*D(x)+40=f(x))是渐变，D(x)是函数值）：

将该等式转换为当等式为真时将返回0的函数：

f(x)

交叉点的衍生物：

rfun <- function(x) {
  f <- FUN(x)
  (40+x)*attr(f,"gradient")+40-f
} 
u1 <- uniroot(rfun,c(-40,60))

绘制细分：

d1 <- attr(FUN(u1$root),"gradient")

找到收益递减曲线的切线

1 个答案: