5

我有如下数据：

library(dplyr)
ex <- data.frame(bool = c(rep(FALSE, 2), rep(TRUE, 3), rep(FALSE, 2), rep(TRUE, 5), 
                      FALSE, FALSE, rep(TRUE, 6), FALSE, FALSE, FALSE)) %>%
  mutate(seq = data.table::rleid(bool)) %>% 
  group_by(seq) %>% 
  mutate(n = n()) %>% 
  ungroup() %>% 
  mutate(expected_output = c(4, 4, NA, NA, NA, 4, 4, rep(NA,5), 4, 4, rep(NA, 6), rep(6, 3)))

对于每一个FALSE我需要找到TRUE长度至少为4. 但是如果之前没有这样的序列（比如 rows1:2或6:7），我们应该向前检查，即找到观察后出现的第一个长度为 4 或更大的序列。

的最后一列ex包含预期输出。我怎么能做到这一点（最好是tidyverse）？

编辑

tidyverse仍然非常感谢使用的解决方案。

4

3 回答 3

1

你可以做：

定义函数：（健壮且具有错误处理功能）

fun1<-
function(vec, min_rep = 4) {

    stopifnot(length(vec)>0, all(vec %in% 0:1))

    runL <- do.call(rbind,rle(vec))
    lngth<- ncol(runL)
    runL <- rbind(runL, seq = 1:lngth, seq2 = NA^runL[2,])

    runL[3,] <- ifelse(!runL[2,]|runL[1,]<min_rep, NA, runL[3,]) 

    cases <- na.omit(runL[3,])

    if(length(cases)>0) {

        for(i in rev(cases)) {
            runL[4,1:i][!is.na(runL[4,1:i])] <- i
        }

        for(i in cases) {
            runL[4,i:lngth][!is.na(runL[4,i:lngth])] <- i
        }
    } else { runL[4,] <- NA }

    return(rep(runL[4,],runL[1,]))
}

调用函数：

vec = c(rep(FALSE, 2), rep(TRUE, 3), rep(FALSE, 2), rep(TRUE, 5), 
        FALSE, FALSE, rep(TRUE, 6), FALSE, FALSE, FALSE)
cbind(vec,fun1(vec))

vec = rep(T,5)
cbind(vec,fun1(vec))

vec = rep(F,5)
cbind(vec,fun1(vec))

vec = c(rep(F,5),T)
cbind(vec,fun1(vec))

vec = c()
cbind(vec,fun1(vec))

vec = 1:3
cbind(vec,fun1(vec))
于 2018-09-12T14:31:34.587 回答
1

以下应该使用base R.

function(col,min_seq =4)
{
    end = c(which(c(col[-1],NA)!=col),length(col))   
    num = diff(c(0,end))     
    start = end-num+1 
    seq_n = seq_along(start) 
    v=col[end]

    accept = num >= min_seq & v
    st = start[accept]
    sn = seq_n[accept]
    en = end[accept]
    en_ = en
    en_[1]=1
    place = rep(sn, diff(c(en_,length(col) + 1 )))      # If row with start of sequence is wanted instead of sequence number sn can be replaced with st
    place[col]=NA

    return(place)
}
于 2018-09-12T13:57:26.807 回答
0

如果 OP 严格不想要data.table解决方案，我可以删除这篇文章。

这是一种可能的data.table方法：

#aggregate the dataset by bool and rleid
agg <- DT[, .(rn=.GRP, N=.N), by=.(bool, seq=rleid(bool))]

#extract all the TRUE sequences with length >= 4
true4s <- agg[(bool) & N >= 4L]

    #for rows that are FALSE
agg[(!bool), expOut := {
        prev <- NA

        #find the previous sequence of TRUEs by using data.table non-equi join 
        #(a rolling join will work too here)
        #in addition, do the match in reverse so that we can fill NA with prev value
        ans <- true4s[.SD[order(-rn)], {
            if (.N > 0L) {
                prev <- seq[.N] 
            } 
            prev
              #for each row in i (see ?data.table for i argument and also ?.EACHI)
                          #non equi join where earlier row in x to be join with later row in i 
        }, by=.EACHI, on=.(rn<rn)]$V1

        #for the rolling version
        #}, by=.EACHI, on=.(rn), roll=Inf]$V1

        rev(ans)
    }]

#add expected output to original dataset
DT[, expected_output := inverse.rle(list(values=agg$expOut, lengths=agg$N))]

输出：

     bool expected_output
 1: FALSE               4
 2: FALSE               4
 3:  TRUE              NA
 4:  TRUE              NA
 5:  TRUE              NA
 6: FALSE               4
 7: FALSE               4
 8:  TRUE              NA
 9:  TRUE              NA
10:  TRUE              NA
11:  TRUE              NA
12:  TRUE              NA
13: FALSE               4
14: FALSE               4
15:  TRUE              NA
16:  TRUE              NA
17:  TRUE              NA
18:  TRUE              NA
19:  TRUE              NA
20:  TRUE              NA
21: FALSE               6
22: FALSE               6
23: FALSE               6
     bool expected_output

数据：

library(data.table)
DT <- data.table(bool = c(rep(FALSE, 2), rep(TRUE, 3), rep(FALSE, 2), rep(TRUE, 5), 
    FALSE, FALSE, rep(TRUE, 6), FALSE, FALSE, FALSE))
于 2018-09-13T02:13:55.550 回答