2017-06-26 38 views
0

我有以下代码:简化我的功能

z7 <- function(data, k, e){ 
    require(zoo) 
    df = data 
    r = df$ROA 
    t = df$t 
    EA = df$EA 
    k = k 
    e = e 

    #Estimate rolling linear models 
    models = rollapply(df, width = k, FUN = function(z) 
    coef(lm(r~t, data = as.data.frame(z))), by.column = FALSE, align ="right") 

    #Extract residuals from the models 
    res = rollapply(df, width= k, FUN = function(x) 
    residuals(lm(r~t, data = as.data.frame(x))), by.column = FALSE, align ="right") 

    #Standard deviation and Mean of residuals, on a row basis 
    s = as.data.frame(apply(res, 1, sd)) 
    m = as.data.frame(apply(res, 1, mean)) #note that this is aproximately 0 due to detrending. 

    #Combine the data define n as number of rows in the dataset 
    dataset = cbind(models, res, m, s) 
    n = as.vector(nrow(dataset)) 
    n 
    dataset 

    #Compute predictions at k+1 
    for(i in n){ 
    x = k + 1 
    preds = dataset$`(Intercept)` + dataset$t*(x) 
    x = x + 1 
    } 

    #Compute coefficient of variation 
    for(j in n){ 
    n2 = k +1 
    tau = ((1 + 1/(4*(n2))) * (dataset$apply.res..1..sd./dataset$apply.res..1..mean.)) 
    } 

    dataset3 = cbind(dataset, tau) 
    dataset3 
    #Compute mean of chi distribution and the adjusted standard deviation 
    Mchi <- sqrt(2)*((gamma((k+1)/2))/gamma(k/2)) 
    S = s*Mchi*(k+1)/sqrt(k) 

    #Compute z7, checking whether the adjusted sd or cv should be used 
    for(i in nrow(dataset3)){ 
    if (abs(dataset3$tau*dataset3$preds) < e) { 
     z = -(dataset3$EA + dataset3$preds)/S 
    } else 
     z = -(dataset3$EA + dataset3$preds) /(dataset3$tau*dataset3$preds) 
    } 
} 

由于是明显的,我创建创建一个调整的标准化得分的功能。通常,Z分数被定义为(x-均值)/ sd。

在这种情况下,我们考虑到x是一个非平稳随机变量的事实。因此,必须在滚动基础上估算该度量,并在观察次数上进行迭代构建。

df是感兴趣的数据集,k是用于估计轧制线性模型的窗口长度,而e是简单地用于测试调整的标准偏差是否太小使用变异系数,而不是一个一个值替代标准差是针对异方差性进行调整的。

t = seq(0,15,1) 
r = (100+50*sin(0.8*t)) 
EA = rnorm(0:15) 
df = data.frame(t,r,EA) 

test = z7(df, 3, 0.00000000001) 

的错误是::

Error in data.frame(..., check.names = FALSE) : 
arguments imply differing number of rows: 14, 0 

回溯是:

5. 
stop(gettextf("arguments imply differing number of rows: %s", 
    paste(unique(nrows), collapse = ", ")), domain = NA) 
4. 
data.frame(..., check.names = FALSE) 
3. 
cbind(deparse.level, ...) 
2. 
cbind(dataset, tau) 
1. 
z7(df, 3, 1e-11) 

如何

运行我的功能与下面的测试措施,我得到一个错误我可以修复这个错误吗?另外,有没有办法简化我的代码?

谢谢。

回答

0

我认为发生错误的行

tau = ((1 + 1/(4*(n2))) * (dataset$apply.res..1..sd./dataset$apply.res..1..mean.)) 

我改成了

tau = ((1 + 1/(4*(n2))) * (dataset$`apply(res, 1, sd)`/dataset$`apply(res, 1, mean)`)) 

,在过去for环路我想有一个与dataset3$preds

>dataset3$preds 
NULL 

问题并且在开始时你宣布r = df$ROA但我认为这设置r等于NULL

希望是有用的!

问候

WW