n <- 1:30
f <- c(1,1)
for (i in n) f <- c(f, f[length(f)-1]+f[length(f)])
for (i in 1:(length(f)-1)) print(f[i]/f[i+1])
从结果可以看到,貌似收敛到 0.618
Q2
1) answer <- c(3)
在for循环里:
#每次都从answer中取最后一项,并把计算结果存到answer中(作为最后一项);
2) j =1时: answer <- c(answer, ( 7* answer[ 1 ] ) %% 31) ==> answer = c(3, 21)
3) j =2时:answer <- c(answer, ( 7* answer[ 2 ] ) %% 31)=>answer = c(3,21, 21*7%%31)
--->23<-----
4) ...
16)j=15时: answer中有16个元素;
$curl = curl_init()curl_setopt($curl,CURLOPT_URL,$_GET['url'])
curl_setopt($curl,CURLOPT_RETURNTRANSFER,true)
curl_setopt($curl,CURLOPT_USERAGENT,"Mozilla/4.0
(compatibleMSIE 6.0Windows NT 5.1SV1.NET CLR 1.1.4322.NET CLR 2.0.50727)")
$output = curl_exec($curl)
curl_close($curl)
print_r($output)
k <- list()
for(i in 1:1000)
{
k[[i]] <- nn2()
}
newdata=c() #1
for(i in 1:1000)
{
#方法一:三次样条法
library(splines)
m1 <- lm(h~bs(a,df=3),data=k[[i]])
#预测百分位数值
new <- data.frame(a=7:20)
cs.p <- predict(m1, new)
#均方差
mse.cs <- sum( (st$p50-cs.p)^2 )/14
#最大范数误差
mne.cs <- max(abs(st$p50-cs.p))
newdata<-rbind(newdata,mse.cs) #2
print(newdata) #3
}
aa<-mean(newdata) #4
新建newdata来保存循环的结果,以便对循环的结果进行后续操作比如求均值并保存在aa中