Please email me a copy of your R script for this assignment by start of class on Friday February 2nd with your first and last name followed by assignment 2 as the file name (e.g. ‘marissa_dyck_assignment2.R’)
You should always be following best coding practices (see Intro to R module 1) but especially for assingment submissions. Please make sure each problem has its own header so that I can easily navigate to your answers and that your code is well organized with spaces as described in the best coding practices section and comments as needed.
Save the altered turtles data as a comma separated file to the data/processed folder in your working directory using the ‘readr’ package and name it ‘turtles_tidy’
write_csv(turtles.df,
'data/processed/turtles_tidy.csv')Download the brown bear damage data (bear_2008_2016.csv) and save it in the data folder.
bear_data <- read_csv('data/raw/bear_2008_2016.csv') %>%
# set column names to lowercase
set_names(
names(.) %>%
tolower()) %>%
# rename columns to include units
rename(m_to_town = dist_to_town,
m_to_forest = dist_to_forest)## Rows: 3024 Columns: 25
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (1): Targetspp
## dbl (24): Damage, Year, Month, POINT_X, POINT_Y, Bear_abund, Landcover_code,...
##
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.head(bear_data) ## # A tibble: 6 × 25
## damage year month targetspp point_x point_y bear_abund landcover_code
## <dbl> <dbl> <dbl> <chr> <dbl> <dbl> <dbl> <dbl>
## 1 0 2008 0 <NA> 542583. 532675. 38 311
## 2 0 2008 0 <NA> 540963. 528113. 40 311
## 3 0 2008 0 <NA> 542435. 510385. 38 211
## 4 0 2008 0 <NA> 541244. 507297. 38 231
## 5 0 2008 0 <NA> 542791. 497755. 32 211
## 6 0 2008 0 <NA> 544639. 610454. 32 313
## # ℹ 17 more variables: altitude <dbl>, human_population <dbl>,
## # m_to_forest <dbl>, m_to_town <dbl>, livestock_killed <dbl>,
## # numberdamageperplot <dbl>, shannondivindex <dbl>, prop_arable <dbl>,
## # prop_orchards <dbl>, prop_pasture <dbl>, prop_ag_mosaic <dbl>,
## # prop_seminatural <dbl>, prop_deciduous <dbl>, prop_coniferous <dbl>,
## # prop_mixedforest <dbl>, prop_grassland <dbl>, prop_for_regen <dbl>str(bear_data)## spc_tbl_ [3,024 × 25] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
## $ damage : num [1:3024] 0 0 0 0 0 0 0 0 0 0 ...
## $ year : num [1:3024] 2008 2008 2008 2008 2008 ...
## $ month : num [1:3024] 0 0 0 0 0 0 0 0 0 0 ...
## $ targetspp : chr [1:3024] NA NA NA NA ...
## $ point_x : num [1:3024] 542583 540963 542435 541244 542791 ...
## $ point_y : num [1:3024] 532675 528113 510385 507297 497755 ...
## $ bear_abund : num [1:3024] 38 40 38 38 32 32 37 45 45 36 ...
## $ landcover_code : num [1:3024] 311 311 211 231 211 313 313 312 324 312 ...
## $ altitude : num [1:3024] 800 666 504 553 470 ...
## $ human_population : num [1:3024] 2 0 2 0 0 0 0 0 0 0 ...
## $ m_to_forest : num [1:3024] 0 0 1829 398 942 ...
## $ m_to_town : num [1:3024] 2398 2441 571 1598 1068 ...
## $ livestock_killed : num [1:3024] 0 0 0 0 0 0 0 0 0 0 ...
## $ numberdamageperplot: num [1:3024] 0 0 0 0 0 0 0 0 0 0 ...
## $ shannondivindex : num [1:3024] 0.963 1.119 1.056 1.506 1.067 ...
## $ prop_arable : num [1:3024] 0 0 65.6 27.3 53.2 ...
## $ prop_orchards : num [1:3024] 0 0 0 0 0 0 0 0 0 0 ...
## $ prop_pasture : num [1:3024] 0 25.3 11.13 20 4.27 ...
## $ prop_ag_mosaic : num [1:3024] 0 0 8.55 0 0 ...
## $ prop_seminatural : num [1:3024] 16.3 30.6 0 19.8 0 ...
## $ prop_deciduous : num [1:3024] 57.796 43.095 0 0.769 32.454 ...
## $ prop_coniferous : num [1:3024] 0 0 0 0 0 ...
## $ prop_mixedforest : num [1:3024] 0 0 0 0 0 ...
## $ prop_grassland : num [1:3024] 0 1 0 0 0 ...
## $ prop_for_regen : num [1:3024] 25.9 0 13.5 0 0 ...
## - attr(*, "spec")=
## .. cols(
## .. Damage = col_double(),
## .. Year = col_double(),
## .. Month = col_double(),
## .. Targetspp = col_character(),
## .. POINT_X = col_double(),
## .. POINT_Y = col_double(),
## .. Bear_abund = col_double(),
## .. Landcover_code = col_double(),
## .. Altitude = col_double(),
## .. Human_population = col_double(),
## .. Dist_to_forest = col_double(),
## .. Dist_to_town = col_double(),
## .. Livestock_killed = col_double(),
## .. Numberdamageperplot = col_double(),
## .. ShannonDivIndex = col_double(),
## .. prop_arable = col_double(),
## .. prop_orchards = col_double(),
## .. prop_pasture = col_double(),
## .. prop_ag_mosaic = col_double(),
## .. prop_seminatural = col_double(),
## .. prop_deciduous = col_double(),
## .. prop_coniferous = col_double(),
## .. prop_mixedforest = col_double(),
## .. prop_grassland = col_double(),
## .. prop_for_regen = col_double()
## .. )
## - attr(*, "problems")=<externalptr># change targetspp to factor dplyr
bear_data <- bear_data %>%
mutate(targetspp = as.factor(targetspp))
levels(bear_data$targetspp)## [1] "alte" "bovine" "ovine"bear_sheep_data <- bear_data %>%
# return only rows for sheep
filter(targetspp == 'ovine') %>%
# select specified columns
select(damage:month, bear_abund:altitude) # most parsimonious way to do it but you could have also done
# select(damage, year, month, bear_abund, landcover_code, altitude)
# or
# select(1:3, 7:9)
head(bear_sheep_data)## # A tibble: 6 × 6
## damage year month bear_abund landcover_code altitude
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 1 2008 9 40 324 570
## 2 1 2008 9 40 324 570
## 3 1 2008 8 36 321 1410
## 4 1 2008 8 22 321 1068
## 5 1 2008 9 40 112 516
## 6 1 2008 8 56 112 533summarise()
function, calculate the mean, sd and SE (the formula
for SE is sd / sqrt(n)) for altitude. This last part might be tricky
at first but give it a try and remember you can always google things and
check the ‘help’ files; and if that fails ‘phone a friend or
me’bear_sheep_data %>%
summarise(mean_alt = mean(altitude),
sd_alt = sd(altitude),
se_alt = sd_alt/sqrt(length(altitude))) # you could also look up the length for altitude in the console and type the number here directly but this is more flexible and readable## # A tibble: 1 × 3
## mean_alt sd_alt se_alt
## <dbl> <dbl> <dbl>
## 1 699. 183. 12.6# or another possible answer
bear_sheep_data %>%
summarise(mean_alt = mean(altitude),
sd_alt = sd(altitude),
se_alt = sd_alt/sqrt(n()))## # A tibble: 1 × 3
## mean_alt sd_alt se_alt
## <dbl> <dbl> <dbl>
## 1 699. 183. 12.6