如果满足条件，如何对连续行进行子集化

Question

I am using R to analyze a number of time series (1951-2013) containing daily values of Max and Min temperatures. The data has the following structure:

YEAR MONTH  DAY     MAX    MIN
1985     1    1    22.8    9.4
1985     1    2    28.6   11.7
1985     1    3    24.7   12.2
1985     1    4    17.2    8.0
1985     1    5    17.9    7.6
1985     1    6    17.7    8.1

I need to find the frequency of heat waves based on this definition: A period of three or more consecutive days ‎with a daily maximum and minimum temperature exceeding the 90th percentile of the maximum ‎and minimum temperatures for all days in the studied period.

Basically, I want to subset those consecutive days (three or more) when the Max and Min temp exceed a threshold value. The output would be something like this:

YEAR MONTH   DAY     MAX     MIN
1989     7    18    45.0    23.5
1989     7    19    44.2    26.1
1989     7    20    44.7    24.4
1989     7    21    44.6    29.5
1989     7    24    44.4    31.6
1989     7    25    44.2    26.7
1989     7    26    44.5    25.0
1989     7    28    44.8    26.0
1989     7    29    44.8    24.6
1989     8    19    45.0    24.3
1989     8    20    44.8    26.0
1989     8    21    44.4    24.0
1989     8    22    45.2    25.0

I have tried the following to subset my full dataset to just the days that exceed the 90th percentile temperature:

HW<- subset(Mydata, Mydata$MAX >= (quantile(Mydata$MAX,.9)) &
                    Mydata$MIN >= (quantile(Mydata$MIN,.9)))

However, I got stuck in how I can subset only consecutive days that have met the condition.

Answer 1

An approach with data.table which is slightly different from @jlhoward's approach (using the same data):

library(data.table)

setDT(df)
df[, hotday := +(MAX>=44.5 & MIN>=24.5)
   ][, hw.length := with(rle(hotday), rep(lengths,lengths))
     ][hotday == 0, hw.length := 0]

this produces a datatable with a heat wave length variable ( hw.length ) instead of a TRUE / FALSE variable for a specific heat wave length:

> df
    YEAR MONTH DAY  MAX  MIN hotday hw.length
 1: 1989     7  18 45.0 23.5      0         0
 2: 1989     7  19 44.2 26.1      0         0
 3: 1989     7  20 44.7 24.4      0         0
 4: 1989     7  21 44.6 29.5      1         1
 5: 1989     7  22 44.4 31.6      0         0
 6: 1989     7  23 44.2 26.7      0         0
 7: 1989     7  24 44.5 25.0      1         3
 8: 1989     7  25 44.8 26.0      1         3
 9: 1989     7  26 44.8 24.6      1         3
10: 1989     7  27 45.0 24.3      0         0
11: 1989     7  28 44.8 26.0      1         1
12: 1989     7  29 44.4 24.0      0         0
13: 1989     7  30 45.2 25.0      1         1

Answer 2

I may be missing something here but I don't see the point of subsetting beforehand. If you have data for every day, in chronological order, you can use run length encoding (see the docs on the rle(...) function).

In this example we create an artificial data set and define "heat wave" as MAX >= 44.5 and MIN >= 24.5. Then:

# example data set
df <- data.frame(YEAR=1989, MONTH=7, DAY=18:30, 
                 MAX=c(45, 44.2, 44.7, 44.6, 44.4, 44.2, 44.5, 44.8, 44.8, 45, 44.8, 44.4, 45.2),
                 MIN=c(23.5, 26.1, 24.4, 29.5, 31.6, 26.7, 25, 26, 24.6, 24.3, 26, 24, 25))

r <- with(with(df, rle(MAX>=44.5 & MIN>=24.5)),rep(lengths,lengths))
df$heat.wave <- with(df,MAX>=44.5&MIN>=24.5) & (r>2)
df
#    YEAR MONTH DAY  MAX  MIN heat.wave
# 1  1989     7  18 45.0 23.5     FALSE
# 2  1989     7  19 44.2 26.1     FALSE
# 3  1989     7  20 44.7 24.4     FALSE
# 4  1989     7  21 44.6 29.5     FALSE
# 5  1989     7  22 44.4 31.6     FALSE
# 6  1989     7  23 44.2 26.7     FALSE
# 7  1989     7  24 44.5 25.0      TRUE
# 8  1989     7  25 44.8 26.0      TRUE
# 9  1989     7  26 44.8 24.6      TRUE
# 10 1989     7  27 45.0 24.3     FALSE
# 11 1989     7  28 44.8 26.0     FALSE
# 12 1989     7  29 44.4 24.0     FALSE
# 13 1989     7  30 45.2 25.0     FALSE

This creates a column, heat.wave which is TRUE if there was a heat wave on that day. If you need to extract only the hw days, use

df[df$heat.wave,]
#   YEAR MONTH DAY  MAX  MIN heat.wave
# 7 1989     7  24 44.5 25.0      TRUE
# 8 1989     7  25 44.8 26.0      TRUE
# 9 1989     7  26 44.8 24.6      TRUE

Answer 3

Your question really boils down to finding groupings of 3+ consecutive days in your subsetted dataset, removing all remaining data.

Let's consider an example where we would want to keep some rows and remove others:

dat <- data.frame(year = 1989, month=c(6, 7, 7, 7, 7, 7, 8, 8, 8, 10, 10), day=c(12, 11, 12, 13, 14, 21, 5, 6, 7, 12, 13))
dat
#    year month day
# 1  1989     6  12
# 2  1989     7  11
# 3  1989     7  12
# 4  1989     7  13
# 5  1989     7  14
# 6  1989     7  21
# 7  1989     8   5
# 8  1989     8   6
# 9  1989     8   7
# 10 1989    10  12
# 11 1989    10  13

I've excluded the temperature data, because I'm assuming we've already subsetted to just the days that exceed the 90th percentile using the code from your question.

In this dataset there is a 4-day heat wave in July and a three-day heat wave in August. The first step would be to convert the data to date objects and compute the number of days between consecutive observations (I assume the data is already ordered by day here):

dates <- as.Date(paste(dat$year, dat$month, dat$day, sep="-"))
(dd <- as.numeric(difftime(tail(dates, -1), head(dates, -1), units="days")))
# [1] 29  1  1  1  7 15  1  1 66  1

We're close, because now we can see the time periods where there were multiple date gaps of 1 day -- these are the ones we want to grab. We can use the rle function to analyze runs of the number 1, keeping only the runs of length 2 or more:

(valid.gap <- with(rle(dd == 1), rep(values & lengths >= 2, lengths)))
# [1] FALSE  TRUE  TRUE  TRUE FALSE FALSE  TRUE  TRUE FALSE FALSE

Finally, we can subset the dataset to just the days that were on either side of a 1-day date gap that is part of a heat wave:

dat[c(FALSE, valid.gap) | c(valid.gap, FALSE),]
#   year month day
# 2 1989     7  11
# 3 1989     7  12
# 4 1989     7  13
# 5 1989     7  14
# 7 1989     8   5
# 8 1989     8   6
# 9 1989     8   7

Answer 4

A simple approach, not full vectorized..

# play data
year <- c("1960")
month <- c(rep(1,30), rep(2,30), rep(3,30))
day <- rep(1:30,3)
maxT <- round(runif(90, 20, 22),1)
minT <- round(runif(90, 10, 12),1)

df <- data.frame(year, month, day, maxT, minT)

# target and tricky data...
df[1:3, 4] <- 30
df[1:4, 5] <- 14
df[10:13, 4] <- 30
df[10:11, 5] <- 14

# limits
df$maxTope <- df$maxT - quantile(df$maxT,0.9)
df$minTope <- df$minT - quantile(df$minT,0.9)

# define heat day
df$heat <- ifelse(df$maxTope > 0 & df$minTope >0, 1, 0)

# count heat day2
for(i in 2:dim(df)[1]){ 
    df$count[1] <- ifelse(df$heat[1] == 1, 1, 0)
    df$count[i] <- ifelse(df$heat[i] == 1, df$count[i-1]+1, 0)
}

# select last day of heat wave (and show the number of days in $count)
df[which(df$count >= 3),]

Answer 5

Here's a quick little solution:

is_High_Temp <- ((quantile(Mydata$MAX,.9)) &
                    Mydata$MIN >= (quantile(Mydata$MIN,.9)))
start_of_a_series <- c(T,is_High_Temp[-1] != is_High_Temp[-length(x)]) # this is the tricky part
series_number <- cumsum(start_of_a_series) 
series_length <- ave(series_number,series_number,FUN=length())
is_heat_wave  <-  series_length >= 3 & is_High_Temp 

Answer 6

A solution with dplyr , also using rle()

library(dplyr)

cond <- expr(MAX >= 44.5 & MIN >= 24.5)

df %>% 
  mutate(heatwave = 
           rep(rle(!!cond)$values & rle(!!cond)$lengths >= 3, 
               rle(!!cond)$lengths)) %>%
  filter(heatwave)

#>   YEAR MONTH DAY  MAX  MIN heatwave
#> 1 1989     7  24 44.5 25.0     TRUE
#> 2 1989     7  25 44.8 26.0     TRUE
#> 3 1989     7  26 44.8 24.6     TRUE

^{Created on 2020-05-16 by the reprex package (v0.3.0)}

data

#devtools::install_github("alistaire47/read.so")
df <- read.so::read.so("YEAR MONTH   DAY     MAX     MIN
1989     7    18    45.0    23.5
1989     7    19    44.2    26.1
1989     7    20    44.7    24.4
1989     7    21    44.6    29.5
1989     7    24    44.4    31.6
1989     7    25    44.2    26.7
1989     7    26    44.5    25.0
1989     7    28    44.8    26.0
1989     7    29    44.8    24.6
1989     8    19    45.0    24.3
1989     8    20    44.8    26.0
1989     8    21    44.4    24.0
1989     8    22    45.2    25.0")