Seaborn style with Matplotlib - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to apply the seaborn style to Matplotlib plots.

Matplotlib

import matplotlib.pyplot as plt
plt.style.use('[1]')
plt.plot([1, 2, 3], [4, 5, 6])
plt.show()

Drag options to blanks, or click blank then click option'

Aggplot

Bclassic

Cseaborn

Dbmh

Attempts:

3 left

💡 Hint

Common Mistakes

Using a style name that is not installed or recognized.

Forgetting to import matplotlib.pyplot as plt.

The plt.style.use function applies a style to Matplotlib plots. Using "seaborn" applies the seaborn style.

2fill in blank

medium

Complete the code to create a scatter plot with seaborn style applied.

Matplotlib

import matplotlib.pyplot as plt
plt.style.use('seaborn')
x = [1, 2, 3, 4]
y = [10, 20, 25, 30]
plt.[1](x, y)
plt.show()

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Ascatter

Bbar

Chist

Dplot

Attempts:

3 left

💡 Hint

Common Mistakes

Using plt.plot which creates a line plot instead of scatter.

Using plt.bar which creates bar charts.

The scatter function creates a scatter plot, which shows points for x and y values.

3fill in blank

hard

Fix the error in the code to correctly apply seaborn style and plot a histogram.

Matplotlib

import matplotlib.pyplot as plt
plt.style.use('seaborn')
data = [1, 2, 2, 3, 3, 3, 4]
plt.hist(data, bins=[1])
plt.show()

Drag options to blanks, or click blank then click option'

A5

B'5'

C'bins=5'

Dbins

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the number of bins as a string instead of an integer.

Passing the parameter name as a string inside the function call.

The bins parameter expects an integer number of bins, not a string or keyword as a string.

4fill in blank

hard

Fill both blanks to create a line plot with seaborn style and add a grid.

Matplotlib

import matplotlib.pyplot as plt
plt.style.use('[1]')
plt.plot([1, 2, 3], [3, 2, 1])
plt.[2](True)
plt.show()

Drag options to blanks, or click blank then click option'

Aseaborn

Bclassic

Cgrid

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong style name.

Forgetting to enable the grid with plt.grid(True).

Use "seaborn" style for the plot and plt.grid(True) to show the grid lines.

5fill in blank

hard

Fill all three blanks to create a seaborn style bar chart with labels.

Matplotlib

import matplotlib.pyplot as plt
plt.style.use('[1]')
categories = ['A', 'B', 'C']
values = [5, 7, 3]
plt.bar(categories, values)
plt.xlabel('[2]')
plt.ylabel('[3]')
plt.show()

Drag options to blanks, or click blank then click option'

Aseaborn

BCategory

CValue

Dclassic

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong style name.

Leaving axis labels empty or unclear.

Use "seaborn" style, label x-axis as "Category" and y-axis as "Value" for clarity.

Practice

1. What does using plt.style.use('seaborn') do in Matplotlib?

easy

A. It resets all plot settings to Matplotlib defaults.

B. It changes the plot style to look like Seaborn's default theme.

C. It imports the Seaborn library for plotting.

D. It saves the current plot as a Seaborn file.

Solution

1. Step 1: Understand plt.style.use function

   This function sets the style for all plots that follow.

2. Step 2: Recognize 'seaborn' style effect

   Using 'seaborn' applies Seaborn's visual theme to Matplotlib plots.

3. Final Answer:

   It changes the plot style to look like Seaborn's default theme. -> Option B

4. Quick Check:

   Seaborn style = changes plot look [OK]

Hint: Remember: plt.style.use sets the plot's visual theme [OK]

Common Mistakes:

• Thinking it imports Seaborn library
• Confusing style setting with saving files
• Assuming it resets to default Matplotlib style

2. Which of the following is the correct way to apply the Seaborn style in Matplotlib?

easy

A. style.use.plt('seaborn')

B. plt.style('seaborn')

C. plt.use.style('seaborn')

D. plt.style.use('seaborn')

Solution

1. Step 1: Recall the correct syntax for style setting

   The correct method is plt.style.use with the style name as a string.

2. Step 2: Check each option's syntax

   Only plt.style.use('seaborn') matches the correct syntax: plt.style.use('seaborn').

3. Final Answer:

   plt.style.use('seaborn') -> Option D

4. Quick Check:

   Correct syntax = plt.style.use('seaborn') [OK]

Hint: Use plt.style.use('style_name') to set plot style [OK]

Common Mistakes:

• Using plt.style('seaborn') without .use
• Mixing order of style and use
• Incorrect method names or argument order

3. What will be the output style of the plot after running this code?

import matplotlib.pyplot as plt
plt.style.use('seaborn-darkgrid')
plt.plot([1, 2, 3], [4, 5, 6])
plt.show()

medium

A. A plot with a white background and grid lines.

B. A plot with a white background and no grid lines.

C. A plot with default Matplotlib style and no grid.

D. A plot with bright colors but no grid lines.

Solution

1. Step 1: Understand 'seaborn-darkgrid' style

   This style applies a white background with visible grid lines.

2. Step 2: Analyze the plot appearance

   Since plt.style.use('seaborn-darkgrid') is set, the plot will have a white background and grid lines.

3. Final Answer:

   A plot with a white background and grid lines. -> Option A

4. Quick Check:

   seaborn-darkgrid = white background + grid [OK]

Hint: Remember 'darkgrid' means white background with grids [OK]

Common Mistakes:

• Assuming no grid lines appear
• Confusing darkgrid with dark background styles
• Expecting default Matplotlib style

4. Identify the error in this code snippet that tries to apply Seaborn style:

import matplotlib.pyplot as plt
plt.style.use(seaborn)
plt.plot([1, 2, 3], [3, 2, 1])
plt.show()

medium

A. plt.show() is missing parentheses.

B. plt.style.use cannot be used before plt.plot.

C. Missing quotes around 'seaborn' in plt.style.use.

D. plt.plot requires two lists of equal length.

Solution

1. Step 1: Check the argument passed to plt.style.use

   The style name must be a string, so it needs quotes around 'seaborn'.

2. Step 2: Verify other parts of the code

   plt.plot has correct lists, plt.show() has parentheses, and style can be set before plotting.

3. Final Answer:

   Missing quotes around 'seaborn' in plt.style.use. -> Option C

4. Quick Check:

   Style name must be a string [OK]

Hint: Always put style names in quotes in plt.style.use [OK]

Common Mistakes:

• Forgetting quotes around style name
• Thinking plt.show() needs no parentheses
• Believing style must be set after plotting

5. You want to create a Matplotlib plot with Seaborn's 'whitegrid' style but only for one plot without affecting others. Which code snippet achieves this?

hard

A. with plt.style.context('seaborn-whitegrid'):
plt.plot(x, y)
plt.show()

B. plt.style.use('seaborn-whitegrid')
plt.plot(x, y)

C. plt.style.context('seaborn-whitegrid')
plt.plot(x, y)
plt.show()

D. plt.style.use('seaborn-whitegrid')
plt.plot(x, y)
plt.style.reset()

Solution

1. Step 1: Understand style context usage

   Using plt.style.context applies a style temporarily within the with block.

2. Step 2: Check each option for temporary style application

   with plt.style.context('seaborn-whitegrid'): plt.plot(x, y) plt.show() uses with plt.style.context('seaborn-whitegrid') to apply style only to that plot.

3. Final Answer:

   with plt.style.context('seaborn-whitegrid'): plt.plot(x, y) plt.show() -> Option A

4. Quick Check:

   Use plt.style.context for temporary style [OK]

Hint: Use with plt.style.context for one-time style [OK]

Common Mistakes:

• Using plt.style.use without resetting style
• Calling plt.style.context without with statement
• Assuming plt.style.reset() exists