[英]identify series of values preceded and followed by a certain value
I have a sequence of integers. 我有一系列整数。 What I would like to do is identify all sequences of 3's that is preceded AND followed by a 5. For example:
我想要做的是识别前面的3个序列,然后是5后面的序列。例如:
c(5,3,3,5,5,4,3,3,5)
The desired output would be: 期望的输出是:
c(F,T,T,F,F,F,F,F,F)
Explanation: The first sequence of 3's is preceded and followed by a 5. Hence True
. 说明:3的第一个序列之前和之后是5.因此为
True
。 The second sequence is preceded by a 4, hence False
. 第二个序列前面是4,因此是
False
。
Couldn't come up with a smarter solution so here is a for
loop 无法提出更智能的解决方案,所以这里是一个
for
循环
x <- c(5,3,3,5,5,4,3,3,5) #Initial vector
current_inds <- numeric() #Variable to hold indices which need to be changed
saw_3 <- FALSE #If 3 was seen before
output <- rep(FALSE, length(x)) #output vector
num_to_check <- 5 #Value to compare
last_val <- 0 #Last non-3 value
for (i in seq_along(x)) {
#If the current value is not equal to 3
if (x[i] != 3 ) {
#Check if we previously saw 3 and if previous non_3 value was 5
# and the next value is 5
if(saw_3 & x[i + 1] == num_to_check & last_val == num_to_check) {
#Change the group of 3 indices to TRUE
output[current_inds] <- TRUE
#Make the saw_3 flag as FALSE
saw_3 <- FALSE
}
#Update the last seen non_3 value to current value
last_val = x[i]
}
else {
#If it is a 3 then append the indices in current_inds
current_inds <- c(current_inds, i)
#Make saw_3 flag TRUE
saw_3 = TRUE
}
}
output
#[1] FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE
I have a very lengthy & ugly solution, but it works :p I hope someone can find a cleaner one :) I first create a matrix that contains 1 column which is every number in a non-repeated way (not unique, but without consecutives), and then 1 column with the number of times this number is repeated. 我有一个非常冗长和丑陋的解决方案,但它的工作原理:p我希望有人可以找到一个更清洁的:)我首先创建一个包含1列的矩阵,这是一个非重复的方式的每个数字(不是唯一的,但没有连续的),然后1列,重复该数字的次数。 Then I apply a logical function to see if a 3 is surrounded by 5s and in a final step, I unravel the vector back to its original length using the rep() function...
然后我应用逻辑函数来查看3是否被5s包围,在最后一步中,我使用rep()函数将向量解开回原始长度...
x <- c(5,3,3,5,5,4,3,3,5)
x_reduced <- x[x!=c(x[-1], FALSE)]
x_mat <- matrix(0, ncol = 3, nrow = length(x_reduced))
x_mat[ , 1] <- x_reduced
ctr = 1
x_ctr = 1
while (ctr < length(x)) {
x_mat[x_ctr ,1] = x[ctr]
x_mat[x_ctr, 2] = x_mat[x_ctr, 2] + 1
if(x[ctr+1] == x[ctr]){
ctr = ctr + 1
} else {
x_ctr = x_ctr + 1
ctr = ctr + 1
}
}
x_mat[nrow(x_mat), 1] <- x[length(x)]
x_mat[nrow(x_mat), 2] <- x_mat[nrow(x_mat), 2] + 1
check_element <- function(pos) {
if(pos == 1 | pos == nrow(x_mat)) return(FALSE)
if(x_mat[pos+1, 1] == 5 & x_mat[pos-1, 1] == 5){
return(TRUE)
} else {
return(FALSE)
}
}
x_mat[,3] <- sapply(1:nrow(x_mat), check_element)
rep(x_mat[,3], x_mat[,2])
There's room for optimization, but it's certainly possible with dplyr
and rle()
. 有优化的空间,但
dplyr
和dplyr
rle()
肯定是可能的。
> df_result
# A tibble: 9 x 1
result
<lgl>
1 FALSE
2 TRUE
3 TRUE
4 FALSE
5 FALSE
6 FALSE
7 FALSE
8 FALSE
9 FALSE
df_result <- df %>%
group_by(seq = {seq = rle(value); rep(seq_along(seq$lengths), seq$lengths)}) %>%
ungroup() %>%
mutate(last_3 = case_when(lag(seq) != seq ~ as.numeric(lag(value) == 5),
TRUE ~ NA_real_),
next_5 = case_when(lead(seq) != seq ~ as.numeric(lead(value) == 5),
TRUE ~ NA_real_)) %>%
group_by(seq, value) %>%
mutate(result = case_when(sum(last_3, na.rm = TRUE) + sum(next_5, na.rm = TRUE) == 2 ~ TRUE,
TRUE ~ FALSE)) %>%
ungroup() %>%
select(result)
library(dplyr)
df <- data.frame(value = c(5,3,3,5,5,4,3,3,5))
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.