如何找到具有相同值的三个连续行

Question

I have a dataframe as follows:

chr     leftPos    Sample1  X.DD   3_samples    MyStuff
1        324         -1        1        1           1
1        4565        -1        0        0           0 
1        6887        -1        1        0           0
1        12098        1       -1        1           1
2        12          -1        1        0           1
2        43          -1        1        1           1
5        1           -1        1        1           0
5        43           0        1       -1           0
5        6554         1        1        1           1
5        7654        -1        0        0           0
5        8765         1        1        1           0
5        9833         1        1        1          -1
6        12           1        1        0           0
6        43           0        0        0           0
6        56           1        0        0           0
6        79           1        0       -1           0
6        767          1        0       -1           0
6        3233         1        0       -1           0

I would like to convert it according to the following rules For each chromosome:

a. If there are three or more 1's or -1's consecutively in a column then the value stays as it is.

b. If there are less than three 1's or -1s consecutively in a column then the value of the 1 or -1 changes to 0

The rows in a column have to have the same sign (+ or -ve) to be called consecutive.

The result of the dataframe above should be:

chr     leftPos    Sample1  X.DD   3_samples    MyStuff
    1        324         -1        0        0           0
    1        4565        -1        0        0           0 
    1        6887        -1        0        0           0
    1        12098        0        0        0           0
    2        12           0        0        0           0
    2        43           0        0        0           0
    5        1            0        1        0           0
    5        43           0        1        0           0
    5        6554         0        1        0           0
    5        7654         0        0        0           0
    5        8765         0        0        0           0
    5        9833         0        0        0           0
    6        12           0        0        0           0
    6        43           0        0        0           0
    6        56           1        0        0           0
    6        79           1        0       -1           0
    6        767          1        0       -1           0
    6        3233         1        0       -1           0

I have managed to do this for two consecutive rows but I'm not sure how to change this for three or more rows.

DAT_list2res <-cbind(DAT_list2[1:2],DAT_list2res)
colnames(DAT_list2res)[1:2]<-c("chr","leftPos")
DAT_list2res$chr<-as.numeric(gsub("chr","",DAT_list2res$chr))
DAT_list2res<-as.data.frame(DAT_list2res)
dx<-DAT_list2res
f0 <- function( colNr, dx)
{
  col <- dx[,colNr]
  n1 <- which(col == 1| col == -1)          # The `1`-rows.
  d0 <- which( diff(col) == 0)      # Consecutive rows in a column are equal.
  dc0 <- which( diff(dx[,1]) == 0)  # Same chromosome.
  m <- intersect( n1-1, intersect( d0, dc0 ) )
  return ( setdiff( 1:nrow(dx), union(m,m+1) ) )
}
g <- function( dx )
{
  for ( i in 3:ncol(dx) ) { dx[f0(i,dx),i] <- 0 }  
  return ( dx )
}
dx<-g(dx)

Answer 1

Here is one solution only using base R .

First define a function that will replace any repetitions which are less than 3 for zeros:

replace_f <- function(x){
  subs <- rle(x)
  subs$values[subs$lengths < 3] <- 0
  inverse.rle(subs)
}

Then split your data.frame by chr and then apply the function to all columns that you want to change (in this case columns 3 to 6):

df[,3:6] <- do.call("rbind", lapply(split(df[,3:6], df$chr), function(x) apply(x, 2, replace_f)))

Notice that we combine the results together with rbind before replacing the original data. This will give you the desired result:

   chr leftPos Sample1 X.DD X3_samples MyStuff
1    1     324      -1    0          0       0
2    1    4565      -1    0          0       0
3    1    6887      -1    0          0       0
4    1   12098       0    0          0       0
5    2      12       0    0          0       0
6    2      43       0    0          0       0
7    5       1       0    1          0       0
8    5      43       0    1          0       0
9    5    6554       0    1          0       0
10   5    7654       0    0          0       0
11   5    8765       0    0          0       0
12   5    9833       0    0          0       0
13   6      12       0    0          0       0
14   6      43       0    0          0       0
15   6      56       1    0          0       0
16   6      79       1    0         -1       0
17   6     767       1    0         -1       0
18   6    3233       1    0         -1       0

Answer 2

A data.table solution using rleid would be

require(data.table)
setDT(dat)
dat[,Sample1 := Sample1 * as.integer(.N>=3), by=.(chr, rleid(Sample1))]

This used the grouping by rleid(Sample1) and data.table 's helpful .N -variable.

Doing it for all columns you could use the eval(parse(text=...)) syntax as follows:

for(i in names(dat)[3:6]){
  by_string = paste0("list(chr, rleid(", i, "))")
  def_string = paste0(i, "* as.integer(.N>=3)")
  dat[,(i) := eval(parse(text=def_string)), by=eval(parse(text=by_string))]
}

So it results in:

> dat[]
    chr leftPos Sample1 X.DD X3_samples MyStuff
 1:   1     324      -1    0          0       0
 2:   1    4565      -1    0          0       0
 3:   1    6887      -1    0          0       0
 4:   1   12098       0    0          0       0
 5:   2      12       0    0          0       0
 6:   2      43       0    0          0       0
 7:   5       1       0    1          0       0
 8:   5      43       0    1          0       0
 9:   5    6554       0    1          0       0
10:   5    7654       0    0          0       0
11:   5    8765       0    0          0       0
12:   5    9833       0    0          0       0
13:   6      12       0    0          0       0
14:   6      43       0    0          0       0
15:   6      56       1    0          0       0
16:   6      79       1    0         -1       0
17:   6     767       1    0         -1       0
18:   6    3233       1    0         -1       0