在R[复制]中需要TRUE/FALSE的值

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这个问题已经有了答案:

Error in if/while (condition) {: missing Value where TRUE/FALSE needed 2 answers
if/while(条件){:缺失值，TRUE/FALSE需要2个答案。

When I run the following code without commenting gr.ascent(MMSE, 0.5, verbose=TRUE) I receive this error Error in b1 * x : 'b1' is missing but when I comment that line I receive the following error when testing MMSE with these arguments MMSE(2,1,farmland$farm,farmland$area). Do you know where my problem is lying?

当我运行以下代码时，没有注释gr.ascent(MMSE, 0.5, verbose=TRUE)，我在b1 * x中收到这个错误错误:“b1”丢失了，但是当我注释这一行时，我在测试MMSE时收到以下错误，MMSE(2,1，农田$farm，农田$area)。你知道我的问题在哪里吗?

Error in if (abs(t[i]) <= k) { : missing value where TRUE/FALSE needed

Here's my code:

这是我的代码:

farmland <- read.csv("FarmLandArea.csv")
str(farmland)
fit=lm(farm~land,data=farmland)
mean.squared.residuals <- sum((lm(farm~land,data=farmland)$residuals)^2)/(length(farmland$farm)-2)

#gradient descent

#things I should possibly use: solve(t(X)%*%X, t(X)%*%y)
gr.ascent<- function(df, x0, alpha=0.2, eps=0.001, max.it = 50, verbose = FALSE){
  X1 <- x0
  cond <- TRUE
  iteration <- 0
  if(verbose) cat("X0 =",X1,"\n")
  while(cond){
    iteration <- iteration + 1
    X0 <- X1
    X1 <- X0 + alpha * df(X0)
    cond <- sum((X1 - X0)^2) > eps & iteration < max.it
    if(verbose) cat(paste(sep="","X",iteration," ="), X1, "\n")
  }
  return(X1)
}


k=19000

#rho <- function(t, k=19000){
#  for (i in seq(1,length(t))){
#    if (abs(t[i]) <= k)
#      return(t[i]^2)
#    else 
#      return(2*k*abs(t[i])-k^2)

#  }

#}

#nicer implementation of rho. ifelse works on vector
rho<-function(t,k) ifelse(abs(t)<=k,t^2,(2*k*abs(t))-k^2)
rho.prime <- function(t, k=19000){
  out <- rep(NA, length(t))
  for (i in seq(1,length(t))){
    if (abs(t[i]) <= k)
    { print(2*t[i])
      out[i] <- 2*t[i] 
    }
    else 
    {
      print(2*k*sign(t[i]))
      out[i] <- 2*k*sign(t[i])
    }
  }
  return(out)
}
MMSE <- function(b0, b1, y=farmland$farm, x=farmland$land){
   # Calls rho.prime() here with argument y-b0-b1*x


   #Why should we call rho.prime? in the html page you have used rho!?
  n = length(y)
  total = 0
  for (i in seq(1,n)) {
    #total = total + rho(t,k)*(y[i]-b0-b1*x[i])
    total = total + rho.prime(y-b0-b1*x,k)*(y[i]-b0-b1*x[i])
  }
  return(total/n)
}

gr.ascent(MMSE(1,2), 0.5, verbose=TRUE)

In which FarmLand csv data is like the following:

其中农田csv数据如下:

state,land,farm
Alabama,50744,14062
Alaska,567400,1375
Arizona,113635,40781
Arkansas,52068,21406
California,155959,39688
Colorado,103718,48750
Connecticut,4845,625
Delaware,1954,766
Florida,53927,14453
Georgia,57906,16094
Hawaii,6423,1734
Idaho,82747,17812
Illinois,55584,41719
Indiana,35867,23125
Iowa,55869,48125
Kansas,81815,72188
Kentucky,39728,21875
Louisiana,43562,12578
Maine,30862,2109
Maryland,9774,3203
Massachusetts,7840,812
Michigan,58110,15625
Minnesota,79610,42031
Mississippi,46907,17422
...

Here's the result of dput(farmland):

这是dput(农田)的结果:

> dput(farmland)
structure(list(state = structure(1:50, .Label = c("Alabama", 
"Alaska", "Arizona", "Arkansas", "California", "Colorado", "Connecticut", 
"Delaware", "Florida", "Georgia", "Hawaii", "Idaho", "Illinois", 
"Indiana", "Iowa", "Kansas", "Kentucky", "Louisiana", "Maine", 
"Maryland", "Massachusetts", "Michigan", "Minnesota", "Mississippi", 
"Missouri", "Montana", "Nebraska", "Nevada", "New Hampshire", 
"New Jersey", "New Mexico", "New York", "North Carolina", "North Dakota", 
"Ohio", "Oklahoma", "Oregon", "Pennsylvania", "Rhode Island", 
"South Carolina", "South Dakota", "Tennessee", "Texas", "Utah", 
"Vermont", "Virginia", "Washington", "West Virginia", "Wisconsin", 
"Wyoming"), class = "factor"), land = c(50744L, 567400L, 113635L, 
52068L, 155959L, 103718L, 4845L, 1954L, 53927L, 57906L, 6423L, 
82747L, 55584L, 35867L, 55869L, 81815L, 39728L, 43562L, 30862L, 
9774L, 7840L, 58110L, 79610L, 46907L, 68886L, 145552L, 76872L, 
109826L, 8968L, 7417L, 121356L, 47214L, 48711L, 68976L, 40948L, 
68667L, 95997L, 44817L, 1045L, 30109L, 75885L, 41217L, 261797L, 
82144L, 9250L, 39594L, 66544L, 24230L, 54310L, 97105L), farm = c(14062L, 
1375L, 40781L, 21406L, 39688L, 48750L, 625L, 766L, 14453L, 16094L, 
1734L, 17812L, 41719L, 23125L, 48125L, 72188L, 21875L, 12578L, 
2109L, 3203L, 812L, 15625L, 42031L, 17422L, 45469L, 95000L, 71250L, 
9219L, 734L, 1141L, 67500L, 10938L, 13438L, 61875L, 21406L, 55000L, 
25625L, 12109L, 109L, 7656L, 68281L, 17031L, 203750L, 17344L, 
1906L, 12578L, 23125L, 5703L, 23750L, 47188L)), .Names = c("state", 
"land", "farm"), class = "data.frame", row.names = c(NA, -50L
))

1 个解决方案

#1

OK, by the numbers:

好的,通过数字:

In your call to gr.ascent(...) you pass a function, MMSE as the first argument. Inside gr.ascent(...) you refer to this function as df(...).
在你的呼叫中，提升(…)你传递一个函数，MMSE作为第一个参数。在gr.ascent(…)中，您将此函数称为df(…)。
The function MMSE(...) has 2 arguments, b0 and b1 for which there are no defaults - so these must be specified or there will be an error, but
函数MMSE(…)有两个参数b0和b1，其中没有默认值，因此必须指定这些参数，否则将出现错误。
When you call the function df(...) inside gr.ascent(...), in the line: X1 <- X0 + alpha * df(X0) you pass only 1 argument, which is b0.
当你调用函数df(…)在gr.ascent(…)中，在这条线上:X1 <- X0 + alpha * df(X0)你只传递1个参数，即b0。
So the second argument, b1 is missing, hence the error.
第二个参数是b1，所以是错误。

When you call MMSE(...) directly, as in:

当您直接调用MMSE(…)时，如:

MMSE(2,1,farmland$farm,farmland$area)

you pass farmland$area as the 4th argument. But there is no column area in the farmland data frame! So this gets passed as NA, which, when used in

您将农田$area作为第4个参数。但农田数据框中没有列区!所以这个就像NA一样传递，在这里。

total = total + rho.prime(y-b0-b1*x,k)*(y[i]-b0-b1*x[i])

coerces the t argument to rho.prime(...) to NA, hence the second error.

强制t参数到rho.prime(…)到NA，因此是第二个错误。

I can't suggest a solution because I have no clue what you are trying to accomplish here.

我不能提出一个解决方案，因为我不知道你在这里想要完成什么。

EDIT (Response to OP's comment).

编辑(响应OP的评论)。

Notwithstanding @thelatemail's comment, which I agree with wholeheartedly, your new error is rather obscure.

尽管有@thelatemail的评论，我完全同意，你的新错误是相当模糊的。

In your earlier version, you were passing a function, MSEE(...) to gr.ascent(...), and using it incorrectly. This time, you are passing a value to gr.ascent(...), that value being the return value when you call MSEE(1,2). So what happens when you try to treat this value as a function, as in:

在您的早期版本中，您正在传递一个函数，MSEE(…)到gr.ascent(…)，并错误地使用它。这一次，您将一个值传递给gr. rise(…)，该值是当您调用MSEE(1,2)时的返回值。当你把这个值当做函数来处理时，会发生什么呢?

X1 <- X0 + alpha * df(X0)

Well, normally this would throw an error telling you that df is not a function. In this case, it's just your bad luck that df is a function. It is the probability density function for the F distribution, which has a required argument df1, among others (type ?df to see the documentation). That's why you are getting the error.

通常情况下会抛出一个错误告诉你df不是一个函数。在这种情况下，df是一个函数是你的坏运气。它是F分布的概率密度函数，它有一个必需的参数df1，以及其他的(类型?df来查看文档)。这就是为什么你会得到错误。

To "fix" this you need to go back to passing the function, as in:

要“修复”，您需要返回传递函数，如:

gr.ascent(MSEE,...)

and then use it correctly inside gr.ascent(...), as in:

然后在gr. rise(…)中正确地使用它:

X1 <- X0 + alpha * df(X0, <some other argument>).

#1