计算一组连续值的最大长度

Question

我有一个随机的 10x10 DF（实际上它有几百万行）：

df <- replicate(10, sample(0:5, 10, rep=T))

我需要在我的 df 末尾计算一个列，该列是连续值的最大长度等于或超过一个设定值的计数，例如3 个或更多。

因此，包含以下值的单行：2,4,3,3,4,5,1,0,5,1 将返回值 5，作为值 4,3,3 的集合， 4,5都是3个或更多，并且是连续的。

虽然 5 确实在高于 3 的行中再次出现，但其连续出现少于 5 个连续数字，超过 3 的行中较早。

任何帮助表示赞赏。

Answer 1

# condition is that x should be larger or equal to 3
condition <- function(x) x >= 3

# example row
row = c(2,4,3,3,4,5,1,0,5,1)

# we can use condition on row:
condition(row)

# and we can emplay rle on that:
rle(condition(row))

# we need to filter those rle results for TRUE:
r <- rle(condition(row))
r$length[r$values == TRUE]

# The answer is the max of the latter
max(r$length[r$values])

或用于您的数据框示例

# condition is that x should be larger or equal to 3
condition <- \(x) x >= 3

 
number <- function(row, condition){
  r <- row |>
         condition() |>
         rle()
  max(r$length[r$values])
}

df <- replicate(10, sample(0:5, 10, rep=T))
apply(df, 1, number, condition)

Answer 2

在此处使用rle 进行行程编码。

vec <- c(2,4,3,3,4,5,1,0,5,1)
r <- rle(vec >= 3)
r
# Run Length Encoding
#   lengths: int [1:5] 1 5 2 1 1
#   values : logi [1:5] FALSE TRUE FALSE TRUE FALSE
ind <- head(which(r$values), 1)
ind
# [1] 2
r$lengths[ind]
# [1] 5

### to see what those five values are ...
r$values[-ind] <- FALSE
vec[inverse.rle(r)]
# [1] 4 3 3 4 5

这使我们获得了该行中最长的长度。要将此逐行应用于帧，

set.seed(42)
df <- replicate(10, sample(0:5, 10, rep=T))
df
#       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#  [1,]    0    0    3    1    3    5    1    3    5     3
#  [2,]    4    4    4    5    4    5    4    0    1     1
#  [3,]    0    5    4    2    3    1    0    2    1     1
#  [4,]    0    3    4    5    1    3    0    2    0     2
#  [5,]    1    1    3    1    1    2    3    4    1     4
#  [6,]    3    1    1    3    2    5    4    4    4     4
#  [7,]    1    2    3    3    0    4    1    3    5     5
#  [8,]    1    0    2    5    4    1    0    5    4     2
#  [9,]    0    0    1    1    1    5    4    4    3     5
# [10,]    3    2    0    4    1    1    3    3    0     3

func <- function(x, lim = 3) {
  r <- rle(x >= lim)
  ind <- head(which(r$values), 1)
  if (length(ind) == 1 && !anyNA(ind)) r$lengths[ind] else 0
}
apply(df, 1, func)
#  [1] 1 7 2 3 1 1 2 2 5 1