ddply {plyr}无法识别自定义函数，它告诉我我的函数不是一个函数 [英] Custom Function not recognized by ddply {plyr}, it tells me that my function is not a function

问题描述

我有一个名为(b2)的矩阵，其中包含3565行和125列，并且只有二分值(0和1)

I have a matrix called (b2) that contains 3565 rows and 125 columns with only dichotomous values (0 and 1)

我设计了一个函数来比较行i和行i+1，并将差异数量存储在新向量中.

I designed a function to compare row i and row i+1 and store the number of differences in a new vector.

loopPhudcf <- function(x){
  ## create a vector to store the results of your for loop
  output <- as.vector(rep(0, length(x[,1])))
  for (i in 1:(nrow(x))-1)  {
    output[i]<-as.vector(table(x[i,]==x[i+1,]))[1]
  }
  a<-nrow(x)
  b<-nrow(x)-1
  output<-t(as.matrix(output[c(a,1:b)]))
  output[output==ncol(x)]<-0
  return(output)
}

phudcfily123<-loopPhudcf(b2)

该函数运行良好，但是我还有一个ID变量，该变量使用以下命令添加到我的原始矩阵中:b2<-transform(b2,id=a$id)，然后导致3565 x 126是最后一个id变量

The function works fine, but I also have an ID variable which I added to my original matrix using: b2<-transform(b2,id=a$id), then resulting in a 3565 by 126 being the last one the id variable

我想使用ddply {plyr}应用我的函数，但是要做到这一点，我只需要对我的原始矩阵进行子集操作，而不需要ID变量(as.matrix(b2[,1:(ncol(b2)-1)]))，但是我一直在说我的函数不是一个函数:(

I wanted to apply my function using ddply {plyr} but to do this i need to subset only my original matrix without the ID variable (as.matrix(b2[,1:(ncol(b2)-1)])) but it keeps saying that my function is not a function :(

x <- ddply(.data = b2, .var = c("id"), .fun = loopPhudcf(as.matrix(b2[,1:(ncol(b2)-1)])))

Error in llply(.data = .data, .fun = .fun, ..., .progress = .progress,  : 
  .fun is not a function.

有人可以帮助我克服这个问题吗?

can anyone help me overcome this issue?

推荐答案

谢谢， 使用包reshape，我能够使用Brian的方法获得相同的结果，这是代码:

Thank you, using the package reshape i was able to get the same result reached using Brian's method, this is the code:

x<-sparseby(as.matrix(b2[,1:125]),list(group = b2[,126]), function(subset) loopPhudcf(as.matrix(b2[,1:125])))

令我有些奇怪的是，使用这种方法以及Brian I提出的方法，我获得了一个新的矩阵，而不是我想要的矢量

Something a little bit strange to me was that using this approach and the approach kindly suggested by Brian I obtained a new matrix instead of my desired vector

dim(x)
[1]  155 3566

因此，由于行包含相同的信息，因此我仅需对第一行进行子集即可获得向量.我的向量的长度为3565，是使用以下方法获得的:

So, I only had to subset the first row to get the vector since the rows contained the same information. My vector with a length of 3565 was obtained using:

x1<-x[1,2:ncol(x)]

我从2开始，因为第一列说明b2中的id变量. 再次谢谢你！

I started with 2 given that The first column accounts for the id variable in b2. Thank you again!

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