Due Tuesday, Nov 6 at 18:00 (although you should be able to finish in lecture)
suppressPackageStartupMessages(library(tidyverse))
In this course, we’ve looked at four layers of data vis:
ggplot2 and other niche R packages; also matplotlib, pandas, and seaborn in Python.In this “half lab” assignment, you create a graph in a team of 2-3 that touches on all of these components, by choosing one of five analysis scenarios. Your submission will include the following parts:
The following section outlines the evaluation of this assignment. The scenarios follow.
rubric={mechanics:5}
To get the marks for tidy submission:
rubric={writing:5}
To get these marks, you must use proper English, spelling, and grammar in your submission.
rubric={vis:10}
To get these marks, your graph must be publication quality.
rubric={reasoning:40}
To get these marks, briefly describe the decisions you made in designing this graph to be effective. Don’t write a lot. Only focus on the big picture things here. Point form is fine.
rubric={accuracy:15, quality:15}
To get these marks, your code must work (accuracy rubric) and must be high quality (quality rubric: readable and reasonably efficient).
To get these marks, you should accompany your graph with a figure caption that:
Every day, you run a model that predicts tomorrow’s river discharge of the Bow River at Banff. On some days, the model produces an invalid forecast (this makes sense in the context of probabilistic forecasts, something you’ll learn in DSCI 562). You suspect that invalid forecasts (called error in the data) are more likely to happen when the actual discharge that materializes the next day (called outcome in the data) is larger. Your task is to produce a visual to explore this, while still giving the viewer a sense of the overall chance of an invalid forecast.
(sA <- read_csv("data/sA-error_prop.csv"))
## Parsed with column specification:
## cols(
## outcome = col_double(),
## error = col_logical()
## )
## # A tibble: 546 x 2
## outcome error
## <dbl> <lgl>
## 1 9.96 FALSE
## 2 9.98 FALSE
## 3 10.3 FALSE
## 4 10.3 FALSE
## 5 10.3 FALSE
## 6 10.4 FALSE
## 7 10.5 FALSE
## 8 10.5 FALSE
## 9 10.6 FALSE
## 10 10.6 FALSE
## # ... with 536 more rows
Suppose you have snowmelt (in millimeters) and river discharge (in m^3/s) data that are recorded daily, although some days have missing data. You’d like to show your audience when you have data on record, for both variables.
The data are given in two forms for your data wrangling convenience:
sBa, each row corresponds to an available observation of data type Type on date date.sBb, each row corresponds to an (inclusive) range of dates where data are available.(sBa <- read_csv("data/sB-time_intervals-pointwise.csv"))
## Parsed with column specification:
## cols(
## date = col_date(format = ""),
## Type = col_character()
## )
## # A tibble: 23,070 x 2
## date Type
## <date> <chr>
## 1 1980-01-01 Discharge
## 2 1980-01-02 Discharge
## 3 1980-01-03 Discharge
## 4 1980-01-04 Discharge
## 5 1980-01-05 Discharge
## 6 1980-01-06 Discharge
## 7 1980-01-07 Discharge
## 8 1980-01-08 Discharge
## 9 1980-01-09 Discharge
## 10 1980-01-10 Discharge
## # ... with 23,060 more rows
(sBb <- read_csv("data/sB-time_intervals-range.csv"))
## Parsed with column specification:
## cols(
## beginning = col_date(format = ""),
## ending = col_date(format = ""),
## Type = col_character()
## )
## # A tibble: 89 x 3
## beginning ending Type
## <date> <date> <chr>
## 1 1980-01-01 2014-12-31 Discharge
## 2 1984-01-05 1984-02-15 Snowmelt
## 3 1984-02-28 1984-03-19 Snowmelt
## 4 1984-03-22 1984-06-15 Snowmelt
## 5 1984-06-17 1984-09-30 Snowmelt
## 6 1984-10-03 1984-10-10 Snowmelt
## 7 1984-10-13 1984-11-02 Snowmelt
## 8 1984-11-06 1984-11-21 Snowmelt
## 9 1984-11-24 1984-11-29 Snowmelt
## 10 1985-01-03 1985-01-31 Snowmelt
## # ... with 79 more rows
Suppose you’re writing a report to justify a prediction model you’ve built, which predicts river discharge (in m^3/s) using snowmelt (in millimeters). Part of your investigation involves trying out different (increasing) weight functions on snowmelt (you’ll learn more about weight functions later in the program, but their meaning is not important here). You want to convince your readers that your choice of weight functions do a fairly good job spanning the choices of sensible weight functions.
In other words, the weight functions should increase from 0 to 1 at different rates and at different positions, and this increase should happen in the range of sensible snowmelt values (snowmelt data are recorded in sC-weight_justif.csv).
You’ve chosen nine weight functions. They’re stored in the list wfun below. (If you’re interested, they are logistic functions with all combinations of three choices of the location parameter x0, and three choices of the rate parameter r. This information does not help you jusify your prediction model in this imaginary report.)
(sC <- read_csv("data/sC-weight_justif.csv"))
## Parsed with column specification:
## cols(
## snowmelt = col_double()
## )
## # A tibble: 658 x 1
## snowmelt
## <dbl>
## 1 3.01
## 2 11.3
## 3 13.4
## 4 11.4
## 5 1.72
## 6 5.36
## 7 4.21
## 8 0.223
## 9 0.719
## 10 4.86
## # ... with 648 more rows
## Define weight functions
logistic <- function(x0, r) function(x) 1 / (1 + exp(-r*(x-x0)))
wfun <- crossing(
x0 = c(0.0, 7.5, 15.0),
r = c(0.05, 0.20, 0.70)
) %>%
mutate(f = map2(x0, r, logistic)) %>%
`[[`("f")
You have daily snowmelt data (in millimeters) at a particular location near Canmore, Alberta. You notice that the distribution of snowmelt on a given day changes across the year, so you’ve estimated the distribution for every day of the year.
Someone’s given you many years worth of new data. Make a graph to show whether the new data “agrees” with your distribution estimates – that is, whether or not the data could have plausibly been drawn from the distributions you estimated. It would also be useful to show that your distribution estimates match up with the original data, too.
About the data:
sD-dist_melt.csv has the snowmelt data, new and original.sD-dist_melt.Rdata (too large to gift: find in students repo, or click here) loads a list with variable name dist_melt containing the distributions. Density estimates are not available, but the cdf and quantile functions are, and are stored as list elements 1 and 2, respectively.
x: vectorized. The variable of the distribution.d: day of the year; integer acceptable from 1 to 366.p: vectorized. Probabilities to find quantiles of.d: day of the year; integer acceptable from 1 to 366.(sF <- read_csv("data/sD-dist_melt.csv"))
## Parsed with column specification:
## cols(
## data = col_character(),
## date = col_date(format = ""),
## snowmelt = col_double()
## )
## # A tibble: 3,185 x 3
## data date snowmelt
## <chr> <date> <dbl>
## 1 original 1980-04-19 NA
## 2 original 1982-04-20 NA
## 3 original 1983-04-20 NA
## 4 original 1984-04-19 -2.33
## 5 original 1986-04-20 -0.984
## 6 original 1987-04-20 -3.45
## 7 original 1988-04-19 -0.772
## 8 original 1990-04-20 12.3
## 9 original 1991-04-20 2.53
## 10 original 1993-04-20 -2.15
## # ... with 3,175 more rows
load("data/sD-dist_melt.Rdata")
str(dist_melt)
## List of 2
## $ cdf:function (x, d)
## ..- attr(*, "srcref")= 'srcref' int [1:8] 27 16 43 9 16 9 27 43
## .. ..- attr(*, "srcfile")=Classes 'srcfilecopy', 'srcfile' <environment: 0x7ffce8b852b8>
## $ qf :function (p, d)
## ..- attr(*, "srcref")= 'srcref' int [1:8] 52 11 60 5 11 5 52 60
## .. ..- attr(*, "srcfile")=Classes 'srcfilecopy', 'srcfile' <environment: 0x7ffce8b852b8>