如何在R中查找和标记唯一序列的起点/终点

我在时间向量旁边有一个1和0的序列。我想找到所有1序列的开始和结束时间点，并给每个序列一个唯一的ID。这里是一些示例数据和我迄今为止的尝试

创建虚拟数据 这是我的努力

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我相信有更好的方法…

这里有一种方法可以找到上述向量的开始和停止位置：

# get positions of the 1s
onePos <- which(x == 1)
# get the ending positions
stopPos <- onePos[c(which(diff(onePos) != 1), length(onePos))]
# get the starting positions
startPos <- onePos[c(1, which(diff(onePos) != 1) + 1)]

最后，要添加一个id：

df <- data.frame(id=seq_along(startPos), on_t=t[startPos], off_t=t[stopPos])

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以下是一种查找上述向量的开始和停止位置的方法：

# get positions of the 1s
onePos <- which(x == 1)
# get the ending positions
stopPos <- onePos[c(which(diff(onePos) != 1), length(onePos))]
# get the starting positions
startPos <- onePos[c(1, which(diff(onePos) != 1) + 1)]

最后，要添加一个id：

df <- data.frame(id=seq_along(startPos), on_t=t[startPos], off_t=t[stopPos])

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你也可以这样做

x = c(0,0,0,0,1,1,1,1,0,0,0,1,1,1,1,1,1,0,0,0,1,1,1,1,0)
# Create the time vector
t = 10:34
xy <- data.frame(x, t)

mr <- rle(xy$x)$lengths
xy$group <- rep(letters[1:length(mr)], times = mr)

onesies <- xy[xy$x == 1, ]
out <- by(onesies, INDICES = onesies$group, 
             FUN = function(x) {
               data.frame(on_t = x$t[1], off_t = x$t[nrow(x)], ID = unique(x$group))
               })

do.call("rbind", out)

  on_t off_t ID
b   14    17  b
d   21    26  d
f   30    33  f

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你也可以这样做

x = c(0,0,0,0,1,1,1,1,0,0,0,1,1,1,1,1,1,0,0,0,1,1,1,1,0)
# Create the time vector
t = 10:34
xy <- data.frame(x, t)

mr <- rle(xy$x)$lengths
xy$group <- rep(letters[1:length(mr)], times = mr)

onesies <- xy[xy$x == 1, ]
out <- by(onesies, INDICES = onesies$group, 
             FUN = function(x) {
               data.frame(on_t = x$t[1], off_t = x$t[nrow(x)], ID = unique(x$group))
               })

do.call("rbind", out)

  on_t off_t ID
b   14    17  b
d   21    26  d
f   30    33  f

将两个向量放入data.table中，然后执行典型的分组、筛选和变异转换是另一个选项：

library(data.table)
dt = data.table(seq = x, time = t)
dt[, .(on_t = min(time), off_t = max(time), lab = unique(seq)), .(ID = rleid(seq))]
  # Use rleid to create a unique ID for each sequence as a group by variable, find the start 
  # and end point for each sequence as well as a label for each sequence;
  [lab == 1]
  # filter label so that the result only contains time for sequence of 1
  [, `:=`(lab = NULL, ID = seq_along(ID))][]
  # Remove label and recreate the ID

#    ID on_t off_t
# 1:  1   14    17
# 2:  2   21    26
# 3:  3   30    33

遵循OP的逻辑，哪种可能是更好的方法：

d = diff(c(0, x, 0))
# prepend and append a 0 at the beginning and ending of x to make sure this always work 
# if the sequence starts or ends with 1.
results = data.frame(on_t = t[d == 1], off_t = t[(d == -1)[-1]])
# pick up the time where 1 sequence starts as on time, and 0 starts as off time. Here d is 
# one element longer than t and x but since the last element for d == 1 will always be false, it won't affect the result.
results$ID = 1:nrow(results)
# create an ID

results
#   on_t off_t ID
# 1   14    17  1
# 2   21    26  2
# 3   30    33  3

将两个向量放入data.table中，然后执行典型的分组、筛选和变异转换是另一个选项：

library(data.table)
dt = data.table(seq = x, time = t)
dt[, .(on_t = min(time), off_t = max(time), lab = unique(seq)), .(ID = rleid(seq))]
  # Use rleid to create a unique ID for each sequence as a group by variable, find the start 
  # and end point for each sequence as well as a label for each sequence;
  [lab == 1]
  # filter label so that the result only contains time for sequence of 1
  [, `:=`(lab = NULL, ID = seq_along(ID))][]
  # Remove label and recreate the ID

#    ID on_t off_t
# 1:  1   14    17
# 2:  2   21    26
# 3:  3   30    33

遵循OP的逻辑，哪种可能是更好的方法：

d = diff(c(0, x, 0))
# prepend and append a 0 at the beginning and ending of x to make sure this always work 
# if the sequence starts or ends with 1.
results = data.frame(on_t = t[d == 1], off_t = t[(d == -1)[-1]])
# pick up the time where 1 sequence starts as on time, and 0 starts as off time. Here d is 
# one element longer than t and x but since the last element for d == 1 will always be false, it won't affect the result.
results$ID = 1:nrow(results)
# create an ID

results
#   on_t off_t ID
# 1   14    17  1
# 2   21    26  2
# 3   30    33  3

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第二种方法符合我的需要，非常简洁，谢谢！第二种方法符合我的需要，非常简洁，谢谢！