calculate mean daylength across a varying range of dates

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I'm using the daylength function in the geosphere package to calculate day length at a location between two points. Depending on the specific individual and location, in the number of days over which I'm averaging day length varies.

While my function works when I hard code variables (i.e., provide a specific value for lat and date), it does not work when I supply a vector of values and get the following error and warning messages:

Error in mutate_impl(.data, dots) : Evaluation error: NA/NaN argument.
In addition: Warning messages:
1: In doy.prev:doy :
  numerical expression has 379 elements: only the first used
2: In doy.prev:doy :
  numerical expression has 379 elements: only the first used

I know this error pertains to my day length calculations because the other part of the code runs fine when it is omitted.

My code and a subset of data (i.e., first 25 observations):

df %>%
  mutate( = if_else((ID == lag(ID) & site != lag(site)),
                          (lat + lag(lat))/2, NA_real_),
         doy.prev = if_else((ID == lag(ID) & 
                                site != lag(site)),
                             NA_real_), = if_else((ID == lag(ID) & 
                                site != lag(site) &
                                yday(ts) != yday(lag(ts)) & 
                                ! & 
                             mean(daylength(, doy.prev:doy)), 
structure(list(ID = structure(c(1L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 4L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 7L, 8L, 
8L), .Label = c("NB2014.12", "NB2014.13", "NB2014.14", "NB2014.15", 
"NB2014.16", "NB2014.42", "NB2014.43", "NB2014.44", "NB2014.45", 
"NB2014.47", "NB2014.48", "NB2014.49", "NB2014.70", "NB2014.71", 
"NB2014.72", "NB2014.73", "NB2014.74", "NB2014.75", "NB2014.76", 
"NB2014.77", "NB2014.78", "NB2014.79", "NB2014.80", "NB2014.81", 
"NB2015.156", "NB2015.157", "NB2015.158", "NB2015.159", "NB2015.160", 
"NB2015.312", "NB2015.313", "NB2015.314", "NB2015.315", "NB2015.316", 
"NB2015.317", "NB2015.318", "NB2015.320", "NB2015.321", "NB2015.322", 
"NB2015.323", "NB2015.324", "NB2015.325", "NB2015.326", "NB2015.327", 
"NB2015.328", "NB2015.329", "NB2015.330", "NB2015.331", "NB2015.332", 
"NB2015.333", "NB2015.334", "NB2015.335", "NB2015.336", "NB2015.337", 
"NB2015.338", "NB2015.339", "NB2015.340", "NB2015.341", "NB2015.342", 
"NB2015.343", "NB2015.344", "NB2015.345", "NB2015.346", "NB2015.347", 
"NB2015.348", "NB2015.349", "NB2015.350", "NB2015.351", "NB2018.10", 
"NB2018.11", "NB2018.12", "NB2018.13", "NB2018.14", "NB2018.15", 
"NB2018.16", "NB2018.17", "NB2018.18", "NB2018.19", "NB2018.20", 
"NB2018.21", "NB2018.22", "NB2018.23", "NB2018.24", "NB2018.25", 
"NB2018.26", "NB2018.27", "NB2018.28", "NB2018.29", "NB2018.30", 
"NB2018.31", "NB2018.32", "NB2018.33", "NB2018.34", "NB2018.35", 
"NB2018.37", "NB2018.38", "NB2018.39", "NB2018.40", "NB2018.41", 
"NB2018.42", "NB2018.43", "NB2018.44", "NB2018.45", "NB2018.46", 
"NB2018.47", "NB2018.48", "NB2018.49", "NB2018.5", "NB2018.50", 
"NB2018.51", "NB2018.52", "NB2018.53", "NB2018.54", "NB2018.55", 
"NB2018.56", "NB2018.57", "NB2018.58", "NB2018.59", "NB2018.6", 
"NB2018.60", "NB2018.61", "NB2018.62", "NB2018.63", "NB2018.64", 
"NB2018.7", "NB2018.8", "NB2018.9"), class = "factor"), site = c("Tantramar", 
"Tantramar", "HPWLR", "Tantramar", "Beaubassin", "Marsh Landings", 
"Eddie rd. ", "Marsh Landings", "Marsh Landings", "Marsh Landings", 
"Eddie rd. ", "Beaubassin", "AMHRST", "HPWLR", "Tantramar", "Tantramar", 
"Fork Field Farms", "WNERR", "GB_ferryway", "GB_thomas", "Tantramar", 
"HPWLR", "Tantramar", "Tantramar", "Marsh Landings"), lat = c(45.900303030303, 
45.900303030303, 45.83, 45.900303030303, 45.85, 45.85, 45.85, 
45.85, 45.85, 45.85, 45.85, 45.85, 45.79, 45.83, 45.900303030303, 
45.900303030303, 45.94, 43.34, 43.09, 43.08, 45.900303030303, 
45.83, 45.900303030303, 45.900303030303, 45.85), doy = c(213, 
206, 206, 217, 217, 217, 217, 217, 218, 218, 218, 218, 218, 218, 
194, 206, 207, 211, 211, 211, 220, 220, 207, 210, 210), ts = structure(c(1406899801.4133, 
1406297348.1112, 1406299522.4141, 1407276094.4158, 1407277417.7616, 
1407279028.1764, 1407279972.1813, 1407281880.08955, 1407285413.4387, 
1407314856.6032, 1407315906.52065, 1407316678.29125, 1407316887.28, 
1407319828.1424, 1405278154.7126, 1406330632.0613, 1406364501.8284, 
1406713079.0338, 1406716251.3933, 1406716449.5783, 1407490305.4993, 
1407491817.085, 1406370738.3239, 1406655731.0996, 1406673688.1819
), class = c("POSIXct", "POSIXt"), tzone = "UTC"), timeS = c(NA, 
NA, 2174.30289983749, NA, 1323.34579992294, 1610.41479992867, 
944.004900217056, 1907.90824985504, NA, NA, 1049.91744995117, 
771.77060008049, 208.988749980927, 2940.86240005493, NA, NA, 
33869.7670998573, 348577.20539999, 3172.35950016975, 198.18499994278, 
NA, 1511.5857000351, NA, NA, 17957.0822999477)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -25L))

In plain R:

d <- data.frame(
    meanLat = c(45.0, 44.6),
    doy = c(207,211),
    doy.prev = 206:207
# one row
sum(daylength(d$meanLat[1], d$doy.prev[1]:d$doy[1]))
#[1] 29.96547
# all rows
apply(d, 1, function(x) sum(daylength(x[1], x[3]:x[2])))
#[1] 29.96547 74.25768

# you could also first make a proper long matrix
x <-, apply(d, 1, function(x) cbind(x[1], x[3]:x[2])))
# followed by
tapply(daylength(x[,1], x[,2]), x[,1], sum)
#    44.6       45
#74.25768 29.96547, Check sunrise, sunset, dusk and dawn times for any location in the world! The time range during which the day becomes night or vice versa is called twilight. We can As you may have noticed, daylength or length of daytime varies over the year. The average day, a century ago, was 1.7 milliseconds shorter than today. Duration Between Two Dates – Calculates number of days. Time and Date Duration – Calculate duration, with both date and time included Birthday Calculator – Find when you are 1 billion seconds old

I added additional filter and it gave below warnings. Does it give any hint?

df %>%
  mutate( = if_else((ID == lag(ID) & site != lag(site)),
                            (lat + lag(lat))/2, NA_real_),
         doy.prev = if_else((ID == lag(ID) & 
                               site != lag(site)),
                            NA_real_)) %>%
  filter(! %>%
  mutate( = if_else(((ID == lag(ID) & 
                               site != lag(site) &
                               yday(ts) != yday(lag(ts))) & 
                               ! & 
                            mean(daylength(, doy.prev:doy)), 
12 NB2014.16 GB_ferryway       43.1   211 2014-07-30 10:30:51   3172.     43.2      211   3172. 
13 NB2014.16 GB_thomas         43.1   211 2014-07-30 10:34:09    198.     43.1      211    198. 
14 NB2014.42 HPWLR             45.8   220 2014-08-08 09:56:57   1512.     45.9      220   1512. 
15 NB2014.44 Marsh Landings    45.8   210 2014-07-29 22:41:28  17957.     45.9      210  17957. 
Warning messages:
1: In doy.prev:doy :
  numerical expression has 15 elements: only the first used
2: In doy.prev:doy :
  numerical expression has 15 elements: only the first used

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I was unable to find a way to get the daylength function to work within dplyr. However, here is a work-around:

df2 <- df %>%

df2$timeHday = apply((df2 %>% select(meanLat, doy.local, doy.prev1)), 
                   function(x) sum(daylength(x[1], x[3]:x[2])))

df <- df %>% left_join(df2, by = c("ID", ""))

Many thanks to Robert Hijmans for the assistance!

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  • What value do you expect doy.prev to have instead of NA, since daylength will give that error if doy contains NA?
  • @Sonny if doy = NA, then I don't need the daylength calculation. that's why I added ! to the if_else statement.
  • But you have used & condition , so it will ignore those lines where not all are NA.
  • Sorry @Sonny, I'm not following your point. To me ! means do not include any values where doy.prev = NA.
  • Unfortunately, no that doesn't make much sense to me. I don't want to include the filter because I need all the data for a later calculation. Also, without the previous rows, it's hard for me to confirm that the code is doing what I want it to. Based on my full spreadsheet the first two lines are definitely wrong because the individual was observed twice on the same day. I only need to calculate daylength when consecutive observations for the same individual (ID == lag(ID)), are for two different locations (site != lag(site)) on two different days (yday(ts) != lag(yday(ts)).
  • When I substitute in an innocuous value for the daylength function, like 100, I see that the only individuals that should get a day length calculation (in the first 25 rows) are rows 17 and 18 (NB2014.16 at WNERR and GB_ferryway).