How to find the indices where there are n consecut

问题:

Suppose I have this data:

  x = c(14,14, 6,  7 ,14 , 0 ,0  ,0 , 0,  0,  0 , 0 , 0,  0 , 0 , 0 , 0,  9  ,1 , 3  ,8  ,9 ,15,  9 , 8, 13,  8,  4 , 6 , 7 ,10 ,13,  3,
 0 , 0 , 0 , 0 , 0 , 0,  0,  0 , 0 , 0 , 0,  0,  0,  0,  0  ,0,  0 , 0 , 0,  0,  0,  0,  0 , 0,  0, 4 , 7  ,4,  5 ,16 , 5  ,5 , 9 , 4  ,4,  9 , 8,  2,  0  ,0  ,0  ,0  ,0,  0,  0,  0  ,0 , 0,  0,  0,  0,  0,  0,  0,  0,0)

x
 [1] 14 14  6  7 14  0  0  0  0  0  0  0  0  0  0  0  0  9  1  3  8  9 15  9  8
[26] 13  8  4  6  7 10 13  3  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
[51]  0  0  0  0  0  0  0  0  4  7  4  5 16  5  5  9  4  4  9  8  2  0  0  0  0
[76]  0  0  0  0  0  0  0  0  0  0  0  0  0  0  

I want to recover the indices beginning where there are more than 3 zeroes in a row and terminating with the last 0 before a nonzero.

For example,

I would get

6, 17 for the first rash of zeroes, etc.

回答1:

Here are two base R approaches:

1) rle First run rle and then compute ok to pick out the sequences of zeros that are more than 3 long. We then compute the starts and ends of all repeated sequences subsetting to the ok ones at the end.

with(rle(x), {
  ok <- values == 0 & lengths > 3
  ends <- cumsum(lengths)
  starts <- ends - lengths + 1
  data.frame(starts, ends)[ok, ]
})

giving:

  starts ends
1      6   17
2     34   58
3     72   89

2) gregexpr Take the sign of each number -- that will be 0 or 1 and then concatenate those into a long string. Then use gregexpr to find the location of at least 4 zeros. The result gives the starts and the ends can be computed from that plus the match.length attribute minus 1.

s <- paste(sign(x), collapse = "")
g <- gregexpr("0{4,}", s)[[1]]
data.frame(starts = 0, ends = attr(g, "match.length") - 1) + g

giving:

  starts ends
1      6   17
2     34   58
3     72   89

回答2:

Starts = which(diff(x == 0) == 1) + 1
Ends   = which(diff(x == 0) == -1)
if(length(Ends) < length(Starts)) {
    Ends = c(Ends, length(x)) }

Starts
[1]  6 34 72
Ends
[1] 17 58 89

This works for your test data, but allows any sequence of zeros, including short ones. To insure that you get sequences of length at least n, add:

n=3
Long = which((Ends - Starts) >= n)
Starts = Starts[Long]
Ends = Ends[Long]

回答3:

By using dplyr , get the diff then if the diff not equal to 0 , they are not belong to same group , after cumsum we get the grouid

library(dplyr)
df=data.frame('x'=x,rownumber=seq(length(x)))
df$Groupid=cumsum(c(0,diff(df$x==0))!=0)
df%>%group_by(Groupid)%>%summarize(start=first(rownumber),end=last(rownumber),number=first(x),size=n())%>%filter(number==0&size>=3)
# A tibble: 3 x 5
  Groupid start   end number  size
    <int> <int> <int>  <dbl> <int>
1       1     6    17      0    12
2       3    34    58      0    25
3       5    72    89      0    18

回答4:

If x happens to be a column of a data.table you can do

library(data.table)
dt <- data.table(x = x)

dt[, if(.N > 3 & all(x == 0)) .(starts = first(.I), ends = last(.I))
   , by = rleid(x)]

#    rleid starts ends
# 1:     5      6   17
# 2:    22     34   58
# 3:    34     72   89

Explanation:

• rleid(x) gives an ID (integer) for each element in x indicating which "run" the element is a member of, where "run" means a sequence of adjacent equal values.
• dt[, <code>, by = rle(x)] partitions dt according to rleid(x) and computes <code> for each subset of dt's rows. The results are stacked together in a single data.table.
• .N is the number of elements in the given subset
• .I is the vector of row numbers corresponding to the subset
• first and last give the first and last element of a vector

• .(<stuff>) is the same as list(<stuff>)

  The rleid function, by grouping within the brackets, .N and .I symbols, first and last functions are part of the data.table package.