使用R查找非零元素组

I have the following problem and I can't find any easy solution.

我有以下问题,我找不到任何简单的解决方案。

I would like to find groups of nonzero elements in a vector (separated by at least one zero) and to assign an id to each group (subsequent integer). For example, in vector value <- c(1,1,2,3,4,3,0,0,0,1,2,3,9,8,0,0,3,2) there should be three groups:[1,1,2,3,4,3],[1,2,3,9,8],[3,2], so I would like to obtain such a data frame:

我想在向量中找到非零元素组(由至少一个零分隔)并为每个组分配一个id(后续整数)。例如,在向量值< - c(1,1,2,3,4,3,0,0,0,1,2,3,9,8,0,0,3,2)中应该有三个组:[1,1,2,3,4,3],[1,2,3,9,8],[3,2],所以我想获得这样一个数据框:

   value id
1      1  1
2      1  1
3      2  1
4      3  1
5      4  1
6      3  1
7      0 NA
8      0 NA
9      0 NA
10     1  2
11     2  2
12     3  2
13     9  2
14     8  2
15     0 NA
16     0 NA
17     3  3
18     2  3

5 个解决方案

#1

Using rle(). First create a new vector replacing the zeros with NA.

使用rle()。首先创建一个用NA替换零的新向量。

x <- match(value != 0, TRUE)
with(rle(!is.na(x)), {
    lv <- lengths[values]
    replace(x, !is.na(x), rep(seq_along(lv), lv))
})
# [1]  1  1  1  1  1  1 NA NA NA  2  2  2  2  2 NA NA  3  3

#2

You can try:

你可以试试:

as.integer(factor(cumsum(value==0)*NA^(value==0)))
#[1]  1  1  1  1  1  1 NA NA NA  2  2  2  2  2 NA NA  3  3

#3

Another possibility:

ifelse(value != 0, 
       cumsum(value != 0 & dplyr::lag(value) %in% c(0, NA)), 
       NA)

# [1]  1  1  1  1  1  1 NA NA NA  2  2  2  2  2 NA NA  3  3

#4

You could also do this:

你也可以这样做:

id <- (value>0)^NA
x <- rle(value>0)$lengths[c(TRUE, FALSE)]
id[!is.na(id)] <- rep(seq_along(x), times=x)

#[1]  1  1  1  1  1  1 NA NA NA  2  2  2  2  2 NA NA  3  3

#5

-2

You need to define a vector of vectors so in v[0] you will find all values of the first group, and in v[1] you will find all values of the second group and so on,

你需要定义一个向量向量,所以在v [0]中你会找到第一组的所有值,在v [1]中你会找到第二组的所有值,依此类推,

You need to loop on all values when you find a zero value. Continue until you find anon zero. Increment the vector with one and add the value and so on,

找到零值时,需要循环所有值。继续,直到你找到零。使用一个增加向量并添加值,依此类推,

I wish this answer be helpful.

我希望这个答案有所帮助。

#1