如何根据列中的值从data.table中删除列

Question

Background

I have some financial data (1.5 years SP500 stocks) that I have manipulated into a wide format using the data.table package. After following the whole data.table course on Datacamp, I'm starting to get a hang of the basics, but after searching for hours I'm at a loss on how to do this.

The Problem

The data contains columns with financial data for each stock. I need to delete columns that contain two consecutive NA s.

My guess is I have to use rle() , lapply() , to find consecutive values and DT[,x:= NULL] ) to delete the columns.

I read that rle() doesn't work on NA s, so I changed them to Inf instead. I just don't know how to combine the functions so that I can efficiently remove a few columns among the 460 that I have.

An answer using data.table would be great, but anything that works well is very much appreciated.

Alternatively I would love to know how to remove columns containing at least 1 NA

Example data

> test[1:5,1:5,with=FALSE]
         date     10104     10107     10138     10145
1: 2012-07-02  0.003199       Inf  0.001112 -0.012178
2: 2012-07-03  0.005873  0.006545  0.001428       Inf
3: 2012-07-05       Inf -0.001951 -0.011090       Inf
4: 2012-07-06       Inf -0.016775 -0.009612       Inf
5: 2012-07-09 -0.002742 -0.006129 -0.001294  0.005830
> dim(test)
[1] 377 461

Desired outcome

         date     10107     10138
1: 2012-07-02       Inf  0.001112
2: 2012-07-03  0.006545  0.001428
3: 2012-07-05 -0.001951 -0.011090
4: 2012-07-06 -0.016775 -0.009612
5: 2012-07-09 -0.006129 -0.001294

PS. This is my first question, I have tried to adhere to the rules, if I need to change anything please let me know.

Answer 1

Here's an rle version:

dt[, sapply(dt, function(x)
       setDT(rle(is.na(x)))[, sum(lengths > 1 & values) == 0]), with = F]

Or replace the is.na with is.infinite if you like.

Answer 2

To detect and delete columns containing atleast one NA, you can try the following

data = data.frame(A=c(1,2,3,4,5), B=c(2,3,4,NA,6), C=c(3,4,5,6,7), D=c(4,5,NA,NA,8))

colsToDelete = lapply(data, FUN = function(x){ sum(is.na(x)) >= 1 })

data.formatted = data[,c(!unlist(colsToDelete))]

Answer 3

Obviously the issue is finding consecutive missing. First, create a matrix TRUE/FALSE based on missing NA . Use that matrix to compare each row to next. Keep columns in original matrix where colSums == 0

Try this:

Missing.Mat <- apply(test, 2, is.na)
Consecutive.Mat <- Missing.Mat[-nrow(Missing.Mat),] * Missing.Mat[-1,]
Keep.Cols <- colSums(Consecutive.Mat) == 0

test[,Keep.Cols]

Answer 4

This is what I came up with. It calls rle on a vector y that is 1:length(column) unless a corresponding element of the column is Inf , in which case the corresponding value in y is zero. Then it checks if any of the runs are greater than 1.

keep <- c(date = T, apply(dat[, -1], 2,
              function(x) {
                y <- 1:length(x)
                y[!is.finite(x)] <- 0
                return(!any(rle(y)$lengths > 1))
              }))

dat2 <- dat[, keep]
dat2
#         date    X10107    X10138
# 1 2012-07-02       Inf  0.001112
# 2 2012-07-03  0.006545  0.001428
# 3 2012-07-05 -0.001951 -0.011090
# 4 2012-07-06 -0.016775 -0.009612
# 5 2012-07-09 -0.006129 -0.001294

Note that the column names are prepended with an "X" by read.table .

Now, the dput of the data:

dat <- structure(list(date = c("2012-07-02", "2012-07-03", "2012-07-05", 
"2012-07-06", "2012-07-09"), X10104 = c(0.003199, 0.005873, Inf, 
Inf, -0.002742), X10107 = c(Inf, 0.006545, -0.001951, -0.016775, 
-0.006129), X10138 = c(0.001112, 0.001428, -0.01109, -0.009612, 
-0.001294), X10145 = c(-0.012178, Inf, Inf, Inf, 0.00583)), .Names = c("date", 
"X10104", "X10107", "X10138", "X10145"), class = "data.frame", row.names = c(NA, 
-5L))