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ggplot2 Recreation Challenge: 12 Plots to Rebuild from Scratch

Twelve finished ggplot2 charts, described in words. Your job is to rebuild each one from scratch as a plot object named p. Because a picture cannot be auto-graded, every problem ends with a short fingerprint probe that inspects your plot and prints a signature: match that printed line and your recreation is correct. Solutions stay hidden until you reveal them.

By Selva Prabhakaran  ·  Published July 22, 2026  ·  Last updated July 22, 2026

Work in order. The first four are single-geom warm-ups (scatter, histogram, bar, line); the middle four add colour, facets, and a position adjustment; the last four layer on statistics, flipped coordinates, an annotation, and a custom theme. Read each spec carefully, because the dataset, the geoms, the aesthetic mappings, and any labels are all part of the target.

Every fingerprint is one line you run after building p. It reports things like the geom class, the number of layers, how many rows the plot's internal data has, or a label you set. None of them draw the chart, so the only thing printed to the console is the signature you are matching. Match it exactly, spaces and all.

RRun this once before any exercise
library(ggplot2)

  

  





Section 1. Single-geom warm-ups (4 plots)






Exercise 1.1: Rebuild a weight-versus-mileage scatterplot



Task: Recreate a simple scatterplot from the mtcars dataset: put car weight wt on the x axis and fuel economy mpg on the y axis, drawn with a single point layer, then relabel the axes to "Weight (1000 lbs)" and "Miles per gallon". Save the plot to p, then run cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), p$labels$x).





Expected result:


#> GeomPoint 32 Weight (1000 lbs)





Difficulty: Beginner






  

    
RYour turn

    
p <- # your code here

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), p$labels$x)

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(mtcars, aes(x = wt, y = mpg)) +
  geom_point() +
  labs(x = "Weight (1000 lbs)", y = "Miles per gallon")

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), p$labels$x)
#> GeomPoint 32 Weight (1000 lbs)

    

      
      
    

    

  

  





Explanation: geom_point() draws one mark per row, so the plot's built data has 32 rows, matching the 32 cars in mtcars. The fingerprint reads the class of the first layer's geom, that row count, and the x-axis label you set with labs(). A frequent beginner mistake is quoting the column names inside aes(); variables in aes() are unquoted because they refer to columns, not strings.












Exercise 1.2: Recreate a 20-bin histogram of eruption times



Task: Using the built-in faithful dataset, build a histogram of the eruptions column with exactly 20 bins and give it the title "Old Faithful eruptions". The bars should show how eruption durations cluster into two groups. Save the plot to p, then run cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), p$labels$title).





Expected result:


#> GeomBar 20 Old Faithful eruptions





Difficulty: Beginner






  

    
RYour turn

    
p <- # your code here

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), p$labels$title)

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(faithful, aes(x = eruptions)) +
  geom_histogram(bins = 20) +
  labs(title = "Old Faithful eruptions")

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), p$labels$title)
#> GeomBar 20 Old Faithful eruptions

    

      
      
    

    

  

  





Explanation: A histogram is drawn with rectangles, so under the hood geom_histogram() uses the bar geom, which is why the fingerprint reports GeomBar rather than a dedicated histogram class. Setting bins = 20 produces exactly 20 rows in the built data, one per bin. Leaving bins unset falls back to 30 and prints a "pick better value" message, so always choose a bin count deliberately.












Exercise 1.3: Recreate a count bar chart of diamond cuts



Task: From the diamonds dataset, draw a bar chart that counts how many diamonds fall into each cut category, and fill every bar with the constant colour "steelblue" rather than mapping colour to a variable. Save the plot to p, then run cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), ggplot_build(p)$data[[1]]$fill[1]).





Expected result:


#> GeomBar 5 steelblue





Difficulty: Beginner






  

    
RYour turn

    
p <- # your code here

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), ggplot_build(p)$data[[1]]$fill[1])

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(diamonds, aes(x = cut)) +
  geom_bar(fill = "steelblue")

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), ggplot_build(p)$data[[1]]$fill[1])
#> GeomBar 5 steelblue

    

      
      
    

    

  

  





Explanation: geom_bar() applies a counting statistic, so a single categorical x with five cut levels yields five bars and five rows in the built data. The fill is placed outside aes() because it is a fixed constant, not data to be mapped; putting fill = "steelblue" inside aes() would treat the literal word as a one-level variable and draw a redundant legend. Use geom_col() instead when you already have the heights.












Exercise 1.4: Recreate the unemployment time-series line



Task: Using the economics dataset, plot a line of total unemploy (thousands of people) over date so the reader sees the long-run trend and recessions as peaks. Label the y axis "Unemployed (thousands)". Save the plot to p, then run cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), p$labels$y).





Expected result:


#> GeomLine 574 Unemployed (thousands)





Difficulty: Beginner






  

    
RYour turn

    
p <- # your code here

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), p$labels$y)

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(economics, aes(x = date, y = unemploy)) +
  geom_line() +
  labs(y = "Unemployed (thousands)")

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]), p$labels$y)
#> GeomLine 574 Unemployed (thousands)

    

      
      
    

    

  

  





Explanation: geom_line() connects the observations in ascending x order, which is exactly what you want for a monthly time series; economics holds 574 months, so the built data keeps all 574 vertices. The fingerprint confirms the geom, the number of points, and the y label you set. Reach for geom_line() over geom_path() here, because geom_path() connects rows in table order rather than sorting by the x value.









Section 2. Colour, facets, and position (4 plots)






Exercise 2.1: Recreate a scatter coloured by vehicle class



Task: Using the mpg dataset, plot highway mileage hwy against engine displacement displ as points, and map the point colour to the class variable so each vehicle type gets its own colour and a legend appears. Save the plot to p, then run cat(class(p$layers[[1]]$geom)[1], length(unique(ggplot_build(p)$data[[1]]$group)), nrow(ggplot_build(p)$data[[1]])).





Expected result:


#> GeomPoint 7 234





Difficulty: Intermediate






  

    
RYour turn

    
p <- # your code here

cat(class(p$layers[[1]]$geom)[1], length(unique(ggplot_build(p)$data[[1]]$group)), nrow(ggplot_build(p)$data[[1]]))

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(mpg, aes(x = displ, y = hwy, colour = class)) +
  geom_point()

cat(class(p$layers[[1]]$geom)[1], length(unique(ggplot_build(p)$data[[1]]$group)), nrow(ggplot_build(p)$data[[1]]))
#> GeomPoint 7 234

    

      
      
    

    

  

  





Explanation: Mapping colour = class inside aes() tells ggplot the colour depends on the data, so it creates one group per class (seven here) and adds a legend automatically. All 234 rows of mpg are still plotted, so the built data keeps 234 rows while the internal group column takes seven distinct values. The classic mistake is writing colour = class outside aes(), which paints every point one fixed colour and produces no legend.












Exercise 2.2: Recreate a small-multiples grid of scatterplots



Task: Take the same mpg scatter of hwy against displ with plain points, but split it into a grid of small panels, one panel per vehicle class, so each type is shown separately with shared axes. Save the plot to p, then run cat(class(p$facet)[1], nrow(ggplot_build(p)$layout$layout)).





Expected result:


#> FacetWrap 7





Difficulty: Intermediate






  

    
RYour turn

    
p <- # your code here

cat(class(p$facet)[1], nrow(ggplot_build(p)$layout$layout))

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(mpg, aes(x = displ, y = hwy)) +
  geom_point() +
  facet_wrap(~ class)

cat(class(p$facet)[1], nrow(ggplot_build(p)$layout$layout))
#> FacetWrap 7

    

      
      
    

    

  

  





Explanation: facet_wrap(~ class) builds one panel for each of the seven classes and wraps them into a grid, sharing the x and y scales so panels are directly comparable. The plot's internal layout table holds one row per panel, which is why the count is seven. Use facet_wrap() for a single faceting variable and facet_grid() when you want a full rows-by-columns matrix from two variables.












Exercise 2.3: Recreate boxplots of mileage by drive type



Task: From the mpg dataset, draw side-by-side boxplots of highway mileage hwy for each drive train drv (four-wheel, front, and rear), and fill each box by mapping fill to drv as well. Save the plot to p, then run cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]])).





Expected result:


#> GeomBoxplot 3





Difficulty: Intermediate






  

    
RYour turn

    
p <- # your code here

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]))

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(mpg, aes(x = drv, y = hwy, fill = drv)) +
  geom_boxplot()

cat(class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]))
#> GeomBoxplot 3

    

      
      
    

    

  

  





Explanation: There are three drive types, so ggplot computes one box per group and the built data holds three rows, each carrying the median, hinges, and whisker limits from the boxplot statistic. Mapping fill to the same variable as x simply colours each box, which aids scanning even though it duplicates the x information. If you only need one box per category, mapping fill is cosmetic; drop it when a legend would be redundant with the axis.












Exercise 2.4: Recreate a 100 percent stacked bar of clarity within cut



Task: Using the diamonds dataset, build a bar chart with cut on the x axis and clarity mapped to fill, but stack the bars to a constant height of one so each bar shows the proportion of clarity grades within that cut rather than raw counts. Save the plot to p, then run cat(class(p$layers[[1]]$geom)[1], class(p$layers[[1]]$position)[1], nrow(ggplot_build(p)$data[[1]])).





Expected result:


#> GeomBar PositionFill 40





Difficulty: Intermediate






  

    
RYour turn

    
p <- # your code here

cat(class(p$layers[[1]]$geom)[1], class(p$layers[[1]]$position)[1], nrow(ggplot_build(p)$data[[1]]))

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(diamonds, aes(x = cut, fill = clarity)) +
  geom_bar(position = "fill")

cat(class(p$layers[[1]]$geom)[1], class(p$layers[[1]]$position)[1], nrow(ggplot_build(p)$data[[1]]))
#> GeomBar PositionFill 40

    

      
      
    

    

  

  





Explanation: position = "fill" stacks the clarity segments and then rescales every bar to height one, converting counts into within-cut proportions so cuts of different sizes become comparable. With five cut levels and eight clarity grades the built data has forty segments. Swap the position to "stack" for raw counts or "dodge" to place the clarity bars side by side; the geom and mappings stay the same, only the arrangement changes.









Section 3. Stats, coordinates, and polish (4 plots)






Exercise 3.1: Recreate a scatter with a linear trend line



Task: Using the mpg dataset, plot highway mileage hwy against displacement displ as points, then add a second layer that fits and draws a straight linear trend line through the cloud with method lm and formula y ~ x. Save the plot to p, then run cat(length(p$layers), class(p$layers[[2]]$geom)[1], class(p$layers[[2]]$stat)[1]).





Expected result:


#> 2 GeomSmooth StatSmooth





Difficulty: Advanced






  

    
RYour turn

    
p <- # your code here

cat(length(p$layers), class(p$layers[[2]]$geom)[1], class(p$layers[[2]]$stat)[1])

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(mpg, aes(x = displ, y = hwy)) +
  geom_point() +
  geom_smooth(method = "lm", formula = y ~ x)

cat(length(p$layers), class(p$layers[[2]]$geom)[1], class(p$layers[[2]]$stat)[1])
#> 2 GeomSmooth StatSmooth

    

      
      
    

    

  

  





Explanation: The plot has two layers: the points and the smoother. geom_smooth() pairs the smooth geom with the smoothing statistic, which is why the fingerprint reports GeomSmooth and StatSmooth on the second layer. Passing method = "lm" fits a linear model and formula = y ~ x states the model explicitly, avoiding the automatic message ggplot prints when it guesses. Layer order matters: add the line after the points so it sits on top.












Exercise 3.2: Recreate a plot of mean mileage per class



Task: Using the mpg dataset, show the average highway mileage for each vehicle class as a single point per class, computed on the fly rather than pre-aggregated, with the point size set to 3. Save the plot to p, then run cat(class(p$layers[[1]]$geom)[1], class(p$layers[[1]]$stat)[1], nrow(ggplot_build(p)$data[[1]])).





Expected result:


#> GeomPoint StatSummary 7





Difficulty: Intermediate






  

    
RYour turn

    
p <- # your code here

cat(class(p$layers[[1]]$geom)[1], class(p$layers[[1]]$stat)[1], nrow(ggplot_build(p)$data[[1]]))

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(mpg, aes(x = class, y = hwy)) +
  stat_summary(fun = mean, geom = "point", size = 3)

cat(class(p$layers[[1]]$geom)[1], class(p$layers[[1]]$stat)[1], nrow(ggplot_build(p)$data[[1]]))
#> GeomPoint StatSummary 7

    

      
      
    

    

  

  





Explanation: stat_summary() collapses each of the seven classes to a single number, here the mean of hwy, and draws it with the point geom, so the built data has exactly seven rows. The fingerprint sees GeomPoint for the geom and StatSummary for the statistic. Computing the mean inside the plot keeps your pipeline short; swap fun = median for a robust centre, or add fun.min and fun.max to turn the same layer into error bars.












Exercise 3.3: Recreate a horizontal bar chart with flipped coordinates



Task: From the mpg dataset, draw a count bar chart of vehicle class, then flip the coordinate system so the bars run horizontally and the class names read comfortably along the vertical axis instead of overlapping on the bottom. Save the plot to p, then run cat(class(p$coordinates)[1], class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]])).





Expected result:


#> CoordFlip GeomBar 7





Difficulty: Advanced






  

    
RYour turn

    
p <- # your code here

cat(class(p$coordinates)[1], class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]))

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(mpg, aes(x = class)) +
  geom_bar() +
  coord_flip()

cat(class(p$coordinates)[1], class(p$layers[[1]]$geom)[1], nrow(ggplot_build(p)$data[[1]]))
#> CoordFlip GeomBar 7

    

      
      
    

    

  

  





Explanation: coord_flip() swaps the x and y axes at draw time, so the seven counted bars stay GeomBar with seven rows of data while the coordinate object becomes CoordFlip. Flipping is the easy way to make long category labels legible without rotating text. A modern alternative that avoids coord_flip() entirely is to map the categorical variable directly to y, for example aes(y = class), which ggplot now supports natively for bars.












Exercise 3.4: Recreate an annotated, minimally themed scatterplot



Task: Rebuild the mtcars scatter of mpg against wt with points, then add a fixed text annotation reading "Heavier cars, lower mpg" at position x equals 4 and y equals 30, give the plot the title "Fuel economy vs weight", and apply the minimal theme. Save the plot to p, then run cat(length(p$layers), class(p$layers[[2]]$geom)[1], p$labels$title).





Expected result:


#> 2 GeomText Fuel economy vs weight





Difficulty: Advanced






  

    
RYour turn

    
p <- # your code here

cat(length(p$layers), class(p$layers[[2]]$geom)[1], p$labels$title)

    

      
      
    

    

  

  






Click to reveal solution



  

    
RSolution

    
p <- ggplot(mtcars, aes(x = wt, y = mpg)) +
  geom_point() +
  annotate("text", x = 4, y = 30, label = "Heavier cars, lower mpg") +
  labs(title = "Fuel economy vs weight") +
  theme_minimal()

cat(length(p$layers), class(p$layers[[2]]$geom)[1], p$labels$title)
#> 2 GeomText Fuel economy vs weight

    

      
      
    

    

  

  





Explanation: annotate("text", ...) adds a single text layer at fixed data coordinates, so the plot now has two layers and the second one is a GeomText. Because annotations carry their own inline values rather than a mapped column, they never spawn a legend, which is why they beat geom_text() for one-off labels. theme_minimal() strips the grey panel background for a cleaner look, and the fingerprint reads the title straight from the labels you set with labs().









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