在r

时间:2019-06-19 10:51:49

标签: r mean missing-data knn imputation

我想用最接近的邻居的平均值来估算缺失值,但是当我尝试kNN时,会给出错误消息。

因此向量是“股票价格”,这意味着我在周末不适用。我想用凹函数替换NA值(星期六,星期日):(星期五值+星期一值)/ 2。我认为k = 2的kNN函数将是适当的,但是我收到一条错误消息。

> Oriental_Stock$Stock
 [1] 42.80 43.05 43.00 43.00 42.20    NA    NA 42.50 40.00 40.25 40.55 
 41.50    NA    NA 40.85
> kNN(Oriental_Stock, variable = colnames("Stock"), k = 2)
Error in `[.data.table`(data, indexNA2s[, variable[i]], `:=`(imp_vars[i],  
 : i is invalid type (matrix). Perhaps in future a 2 column matrix could 
  return a list of elements of DT (in the spirit of A[B] in FAQ 2.14). 
  Please report to data.table issue tracker if you'd like this, or add 
  your comments to FR #657.

请让我知道是否可以执行此操作,也许还有比kNN更简单的选项。我不是数据科学家,而是学生,所以对此我不太了解。预先感谢您的任何建议!

1 个答案:

答案 0 :(得分:0)

Knn将在data.frame上工作,在其中它根据行之间的距离选择邻居。它不适用于矢量。

for循环可能是一个合理的解决方案:

#this finds the locations of the first NA of each couple of NAs
#the TRUE / FALSE part below picks only the first NA from each couple
idx <- which(is.na(stock))[c(TRUE, FALSE)]

#this iterates over the above indexes and calculates the mean and updates the NAs
for (x in idx) {
  stock[x] <- stock[x+1] <- (stock[x-1] + stock[x+2]) / 2
}

结果:

> stock
 [1] 42.800 43.050 43.000 43.000 42.200 42.350 42.350 42.500 40.000
[10] 40.250 40.550 41.500 41.175 41.175 40.850

我使用stock作为数据:

stock <- c(42.80,43.05, 43.00, 43.00, 42.20,    NA,    NA, 42.50, 40.00, 40.25, 40.55, 
           41.50,    NA,    NA, 40.85)