Subset where there are at least five consecutive years in a data.frame column

问题

I have a data.frame / data.table in R as follows:

df <- data.frame(
  ID = c(rep("A", 20)),
  year = c(1968, 1971, 1972, 1973, 1974, 1976, 1978, 1980, 1982, 1984, 1985, 
           1986, 1987, 1988, 1990, 1991, 1992, 1993, 1994, 1995)
)

I'd like to subset the df in order to keep only those entries which have at least five consecutive years. In this example this is the case in two periods (1984:1988 and 1990:1995). How can I do this in R?

回答1:

A compact solution using diff and cumsum:

setDT(df)[, grp := cumsum(c(0, diff(year)) > 1), by = ID
          ][, if (.N > 4) .SD, by = .(ID, grp)][, grp := NULL][]

which gives the desired result:

    ID year
 1:  A 1984
 2:  A 1985
 3:  A 1986
 4:  A 1987
 5:  A 1988
 6:  A 1990
 7:  A 1991
 8:  A 1992
 9:  A 1993
10:  A 1994
11:  A 1995

Explanation:

With grp := cumsum(c(0, diff(year)) > 1), by = ID you create a (temporary) grouping variable for consecutive years for each ID.
With if (.N > 4) .SD, by = .(ID, grp) you select only groups with 5 or more consecutive years for each ID.
With grp := NULL you remove the (temporary) grouping variable.

A compareble approach in base R:

i <- with(df, ave(year, ID, FUN = function(x) { 
  r <- rle(cumsum(c(0, diff(year)) > 1));
  rep(r$lengths, r$lengths)
  } ))

df[i > 4,] # or df[which(i > 4),]

which will get you the same result.

回答2:

Here is another way:

df2 <- NULL 
   sapply(seq(nrow(df)), function(x)
             {
              ifelse((sum(diff(df[x:(x+4), "year"], 1)) == 4 &
                      sum(diff(df[x:(x+4), "year"], 1) == 1) == 4),
                      df2 <<- rbind(df2, df[x:(x+4),]),"")
             })
df2 <- unique(df2)

回答3:

We can try

i1 <- with(df, as.logical(ave(year, ID,  FUN = function(x) {
                        i1 <- (x[-1] - x[-length(x)]) ==1
                        i2 <- c(FALSE, i1)
                        i3 <- c(i1, FALSE)
                        rl <- rle(i2|i3)
                        rl$values[rl$values][rl$lengths[rl$values] <5] <- FALSE
                        rep(rl$values, rl$lengths)
                      })))

df[i1,]
#   ID year
#10  A 1984
#11  A 1985
#12  A 1986
#13  A 1987
#14  A 1988
#15  A 1990
#16  A 1991
#17  A 1992
#18  A 1993
#19  A 1994
#20  A 1995

Or use data.table

library(data.table)
i1 <- setDT(df)[, ind := (year - shift(year, fill= year[1L]))==1L , ID][, 
         {i1 <- .I[.N * ind > 3]
         .(v1 = head(i1,1)-1, v2 = tail(i1, 1))}, 
         .(ID, rl = rleid(ind))][, seq(v1, v2) , rl]$V1
df[, ind := NULL][i1]
#     ID year
# 1:  A 1984
# 2:  A 1985
# 3:  A 1986
# 4:  A 1987
# 5:  A 1988
# 6:  A 1990
# 7:  A 1991
# 8:  A 1992
# 9:  A 1993
#10:  A 1994
#11:  A 1995

Or a slightly compact option

i1 <- setDT(df)[, (shift(year, type="lead", fill = year[.N])-year)==1 |
       (year - shift(year, fill = year[1L]))==1, ID][, .I[.N>4 & V1] , .(rleid(V1), ID)]$V1
df[i1]

data

df <- data.frame(
   ID=c(rep("A", 20)),
   year=c(1968, 1971, 1972, 1973, 1974, 1976, 1978, 1980, 1982, 1984, 1985, 
   1986, 1987, 1988, 1990, 1991, 1992, 1993, 1994, 1995))

回答4:

Another option using rowid:

DT[, c("rl", "rw") := {
    iscons <- cumsum(c(0L, diff(year)!=1L))
    .(iscons, rowid(ID, iscons))
}]

DT[rl %in% DT[rw>=5L]$rl]

data:

#adding one more group
DT <- rbindlist(list(setDT(df), copy(df)[, ID := "B"]))

回答5:

I first sorted the elements:

sorted = sort(df$year, decreasing = F)
count = 0 ## count sequences
keep=c()  ## which to keep
keep_num = c()  ##counting the sequence length
keep_count=1
for(i in 2:length(sorted)){
    if((sorted[i]- sorted[i-1]) == 1){ ## if they are in a row
        count = count + 1
        if(count == 4){  ## if there 4+1 years involved in a row
            keep=c(keep, sorted[i]- 4)
        }
        if(count >= 4){ ## if length more than 5, update 
            keep_num[keep_count]=count
        }
    }
    else{ ##reset
        count =0
        keep_count = keep_count + 1
    }
}
keep_num = keep_num[!is.na(keep_num)]

Reconstructing the kept ones:

y = c()
for(i in 1:length(keep)){
    y = c(y, seq(keep[i], keep[i]+keep_num[i]))
}

Subsetting among the ones we wanted to keep:

selected = df[match(y, df$year, nomatch = 0),]

This will select the rows with the desired condition.

回答6:

step1. data in data table "d"

d
hdrY     mvanoyP
 1: 1981 -14.3520324
 2: 1982   0.4900168
 3: 1983   2.6518741
 4: 1984   5.2284595
 5: 1985  -6.2874634
 6: 1986  -1.3287914
 7: 1987  20.6385345
 8: 1988  24.2090114
 9: 1989  21.5302571
10: 1990   9.0267066
11: 1991  10.4148838
12: 1992  13.9189716
13: 1993   7.8816196
14: 1994   3.4650221
15: 1995   2.8722555
16: 1996  -4.1442363
17: 1997  -3.2359926
18: 1998  -5.7479137
19: 1999   2.3481127
20: 2000   0.8089402
21: 2001 -14.4741916
22: 2002 -22.9272540
23: 2003 -27.3105212
24: 2004 -13.9726022
25: 2005 -14.0055281
26: 2006 -15.8456991
27: 2007 -21.0369933
28: 2008 -13.1031347
29: 2009   4.1517341
30: 2010  20.3711446
31: 2011  27.4202037

step2. select nvanoyP <0 and find consecutive 6 year

d %<>% data.table()
    db <- d[mvanoyP < 0, ] %>%
        .[, tag := cumsum(c(0, diff(hdrY)) > 1)] %>%
        .[, if (.N > 6) .SD,.(tag)] #
    if (nrow(db) > 0){
        db[, start := min(hdrY), by = tag]
        db[,   end := max(hdrY), by = tag]
    }
    db

step3. output

db
   tag hdrY   mvanoyP start  end
1:   3 2001 -14.47419  2001 2008
2:   3 2002 -22.92725  2001 2008
3:   3 2003 -27.31052  2001 2008
4:   3 2004 -13.97260  2001 2008
5:   3 2005 -14.00553  2001 2008
6:   3 2006 -15.84570  2001 2008
7:   3 2007 -21.03699  2001 2008
8:   3 2008 -13.10313  2001 2008

来源：https://stackoverflow.com/questions/40831682/subset-where-there-are-at-least-five-consecutive-years-in-a-data-frame-column

标签

dataframe

data.table