fgeo.analyze provides functions to analyze ForestGEO data.
Install the latest stable version of fgeo.analyze from CRAN with:
Install the development version of fgeo.analyze with:
Or install all fgeo packages in one step.
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
library(fgeo.x)
library(fgeo.tool)
#>
#> Attaching package: 'fgeo.tool'
#> The following object is masked from 'package:stats':
#>
#> filter
library(fgeo.analyze)
Your data may have multiple stems per treeid and even multiple measures per stemid (if trees have buttresses).
# Trees with buttresses may have multiple measurements of a single stem.
# Main stems have highest `HOM`, then largest `DBH`.
vft <- tribble(
~CensusID, ~TreeID, ~StemID, ~DBH, ~HOM,
1, "1", "1.1", 88, 130,
1, "1", "1.1", 10, 160, # Main stem
1, "2", "2.1", 20, 130,
1, "2", "2.2", 30, 130, # Main stem
)
Fundamentally, abundance()
counts rows. All of these results are the same:
nrow(vft)
#> [1] 4
count(vft)
#> <print(tibble::tibble())>
summarize(vft, n = n())
#> <print(tibble::tibble())>
abundance(vft)
#> Warning: `treeid`: Duplicated values were detected. Do you need to pick main
#> stems?
#> <print(tibble::tibble())>
But that result is likely not what you expect. Instead, you likely expect this:
As shown above, you can get a correct result by combining summarize()
and n_distinct()
(from the dplyr package). But abundance()
includes some useful additional features (see ?abundance()
). This code conveys your intention more clearly, i.e. to calculate tree abundance by counting the number of main stems:
(main_stems <- pick_main_stem(vft))
#> Warning: The `add` argument of `group_by()` is deprecated as of dplyr 1.0.0.
#> Please use the `.add` argument instead.
#> This warning is displayed once every 8 hours.
#> Call `lifecycle::last_warnings()` to see where this warning was generated.
#> <print(tibble::tibble())>
abundance(main_stems)
#> <print(tibble::tibble())>
If you have data from multiple censuses, then you can compute by census (or any other group).
vft2 <- tribble(
~CensusID, ~TreeID, ~StemID, ~DBH, ~HOM,
1, "1", "1.1", 10, 130,
1, "1", "1.2", 20, 130, # Main stem
2, "1", "1.1", 12, 130,
2, "1", "1.2", 22, 130 # Main stem
)
by_census <- group_by(vft2, CensusID)
(main_stems_by_census <- pick_main_stem(by_census))
#> <print(tibble::tibble())>
abundance(main_stems_by_census)
#> <print(tibble::tibble())>
Often you will need to first subset data (e.g. by status
or DBH
) and then count.
If trees have buttresses, then you may need to pick the main stemid of each stem so you do not count the same stem more than once.
vft3 <- tribble(
~CensusID, ~TreeID, ~StemID, ~DBH, ~HOM,
1, "1", "1.1", 88, 130,
1, "1", "1.1", 10, 160, # Main stem
1, "2", "2.1", 20, 130,
1, "2", "2.2", 30, 130, # Main stem
2, "1", "1.1", 98, 130,
2, "1", "1.1", 20, 160, # Main stem
2, "2", "2.1", 30, 130,
2, "2", "2.2", 40, 130, # Main stem
)
(main_stemids <- pick_main_stemid(vft3))
#> <print(tibble::tibble())>
main_stemids
#> <print(tibble::tibble())>
basal_area(main_stemids)
#> Warning: `stemid`: Duplicated values were detected. Do you need to pick largest
#> `hom` values?
#> Warning: `censusid`: Multiple values were detected. Do you need to group by
#> censusid?
#> <print(tibble::tibble())>
basal_area()
also allows you to compute by groups.
But if you want to compute on a subset of data, then you need to pick the data first.
ten_to_twenty <- filter(by_census, DBH >= 10, DBH <= 20)
basal_area(ten_to_twenty)
#> <print(tibble::tibble())>
Example data.
vft <- tibble(
PlotName = c("luq", "luq", "luq", "luq", "luq", "luq", "luq", "luq"),
CensusID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L),
TreeID = c(1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L),
StemID = c(1.1, 1.2, 2.1, 2.2, 1.1, 1.2, 2.1, 2.2),
Status = c("alive", "dead", "alive", "alive", "alive", "gone",
"dead", "dead"),
DBH = c(10L, NA, 20L, 30L, 20L, NA, NA, NA),
Genus = c("Gn", "Gn", "Gn", "Gn", "Gn", "Gn", "Gn", "Gn"),
SpeciesName = c("spp", "spp", "spp", "spp", "spp", "spp", "spp", "spp"),
ExactDate = c("2001-01-01", "2001-01-01", "2001-01-01", "2001-01-01",
"2002-01-01", "2002-01-01", "2002-01-01",
"2002-01-01"),
PlotCensusNumber = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L),
Family = c("f", "f", "f", "f", "f", "f", "f", "f"),
Tag = c(1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L),
HOM = c(130L, 130L, 130L, 130L, 130L, 130L, 130L, 130L)
)
vft
#> <print(tibble::tibble())>
Abundance by year.
abundance_byyr(vft, DBH >= 10, DBH < 20)
#> <print(tibble::tibble())>
abundance_byyr(vft, DBH >= 10)
#> <print(tibble::tibble())>
Basal area by year.
Demography functions output a list that you can convert to a more convenient dataframe with as_tibble()
.
recruitment_ctfs(census1, census2)
#> Detected dbh ranges:
#> * `census1` = 10.9-323.
#> * `census2` = 10.5-347.
#> Using dbh `mindbh = 0` and above.
#> $N2
#> [1] 29
#>
#> $R
#> [1] 3
#>
#> $rate
#> [1] 0.02413113
#>
#> $lower
#> [1] 0.0084585
#>
#> $upper
#> [1] 0.06812388
#>
#> $time
#> [1] 4.525246
#>
#> $date1
#> [1] 18937.96
#>
#> $date2
#> [1] 20600.72
as_tibble(
recruitment_ctfs(census1, census2, quiet = TRUE)
)
#> <print(tibble::tibble())>
Except if you use split2
: This argument creates a complex data structure that as_tibble()
cannot handle.
# Errs
as_tibble(
recruitment_ctfs(
census1, census2,
split1 = census1$sp,
split2 = census1$quadrat, # `as_tibble()` can't handle this
quiet = TRUE
)
)
#> Warning: `split2` is deprecated.
#> * Bad: `split1 = x1, split2 = x2`
#> * Good: `split1 = interaction(x1, x2)`
#> This warning is displayed once per session.
#> Error: Can't deal with data created with `split2` (deprecated).
#> * Bad: `split1 = x1, split2 = x2`
#> * Good: `split1 = interaction(x1, x2)`
Instead, pass the multiple grouping variables to split
via interaction()
. This approach allows you to use any number of grouping variables and the output always works with as_tibble()
.
# Recommended
by_sp_and_quadrat <- interaction(census1$sp, census1$quadrat)
as_tibble(
recruitment_ctfs(
census1, census2,
split1 = by_sp_and_quadrat,
quiet = TRUE
)
)
#> <print(tibble::tibble())>
The same applies for other demography functions.
as_tibble(
mortality_ctfs(
census1, census2,
split1 = by_sp_and_quadrat,
quiet = TRUE
)
)
#> <print(tibble::tibble())>
A simple way to separate the grouping variables is with tidyr::separate()
.
growth <- growth_ctfs(
census1, census2,
split1 = by_sp_and_quadrat,
quiet = TRUE
)
as_tibble(growth)
#> <print(tibble::tibble())>
as_tibble(growth) %>%
tidyr::separate(groups, into = c("species", "quadrats"))
#> <print(tibble::tibble())>
# Pick alive trees, of 10 mm or more
tree <- download_data("luquillo_tree5_random")
census <- filter(tree, status == "A", dbh >= 10)
# Pick sufficiently abundant species
pick <- filter(add_count(census, sp), n > 50)
# Use your habitat data or create it from elevation data
elevation <- download_data("luquillo_elevation")
habitat <- fgeo_habitat(elevation, gridsize = 20, n = 4)
tt_test_result <- tt_test(pick, habitat)
#> Using `plotdim = c(320, 500)`. To change this value see `?tt_test()`.
#> Using `gridsize = 20`. To change this value see `?tt_test()`.
#> Warning: Is `census` a tree table (not a stem table)? See `?tt_test()`.
# A list or matrices
tt_test_result
#> [[1]]
#> N.Hab.1 Gr.Hab.1 Ls.Hab.1 Eq.Hab.1 Rep.Agg.Neut.1 Obs.Quantile.1 N.Hab.2
#> CASARB 35 1313 282 5 0 0.820625 24
#> Gr.Hab.2 Ls.Hab.2 Eq.Hab.2 Rep.Agg.Neut.2 Obs.Quantile.2 N.Hab.3
#> CASARB 394 1204 2 0 0.24625 11
#> Gr.Hab.3 Ls.Hab.3 Eq.Hab.3 Rep.Agg.Neut.3 Obs.Quantile.3 N.Hab.4
#> CASARB 482 1114 4 0 0.30125 8
#> Gr.Hab.4 Ls.Hab.4 Eq.Hab.4 Rep.Agg.Neut.4 Obs.Quantile.4
#> CASARB 1217 377 6 0 0.760625
#>
#> [[2]]
#> N.Hab.1 Gr.Hab.1 Ls.Hab.1 Eq.Hab.1 Rep.Agg.Neut.1 Obs.Quantile.1 N.Hab.2
#> PREMON 94 1005 594 1 0 0.628125 97
#> Gr.Hab.2 Ls.Hab.2 Eq.Hab.2 Rep.Agg.Neut.2 Obs.Quantile.2 N.Hab.3
#> PREMON 1478 120 2 0 0.92375 39
#> Gr.Hab.3 Ls.Hab.3 Eq.Hab.3 Rep.Agg.Neut.3 Obs.Quantile.3 N.Hab.4
#> PREMON 230 1367 3 0 0.14375 15
#> Gr.Hab.4 Ls.Hab.4 Eq.Hab.4 Rep.Agg.Neut.4 Obs.Quantile.4
#> PREMON 130 1465 5 0 0.08125
#>
#> [[3]]
#> N.Hab.1 Gr.Hab.1 Ls.Hab.1 Eq.Hab.1 Rep.Agg.Neut.1 Obs.Quantile.1 N.Hab.2
#> SLOBER 21 270 1328 2 0 0.16875 25
#> Gr.Hab.2 Ls.Hab.2 Eq.Hab.2 Rep.Agg.Neut.2 Obs.Quantile.2 N.Hab.3
#> SLOBER 516 1082 2 0 0.3225 21
#> Gr.Hab.3 Ls.Hab.3 Eq.Hab.3 Rep.Agg.Neut.3 Obs.Quantile.3 N.Hab.4
#> SLOBER 1336 260 4 0 0.835 8
#> Gr.Hab.4 Ls.Hab.4 Eq.Hab.4 Rep.Agg.Neut.4 Obs.Quantile.4
#> SLOBER 1193 396 11 0 0.745625
# A dataframe
as_tibble(tt_test_result)
#> <print(tibble::tibble())>
# A simple summary to help you interpret the results
summary(tt_test_result)
#> <print(tibble::tibble())>