Error bar plots in Matplotlib - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of error bar plot with asymmetric errors

What will be the output of this code snippet that creates an error bar plot with asymmetric errors?

Matplotlib

import matplotlib.pyplot as plt
import numpy as np

x = np.array([1, 2, 3, 4])
y = np.array([2, 3, 5, 7])
yerr = [0.5, 0.2, 0.3, 0.4]
yerr_upper = [0.6, 0.3, 0.4, 0.5]

plt.errorbar(x, y, yerr=[yerr, yerr_upper], fmt='o')
plt.show()

A A plot with points at (1,2), (2,3), (3,5), (4,7) and error bars only on the upper side with lengths 0.6, 0.3, 0.4, 0.5.

B A plot with points at (1,2), (2,3), (3,5), (4,7) and symmetric error bars of length 0.5, 0.2, 0.3, 0.4 above and below each point.

C A plot with points at (1,2), (2,3), (3,5), (4,7) but no error bars because yerr is given as a list of two arrays.

D A plot with points at (1,2), (2,3), (3,5), (4,7) and asymmetric error bars: lower errors 0.5, 0.2, 0.3, 0.4 and upper errors 0.6, 0.3, 0.4, 0.5.

💡 Hint

Check how yerr is passed as a list of two arrays to specify asymmetric errors.

When yerr is a list of two arrays, the first array specifies the lower errors and the second array specifies the upper errors, creating asymmetric error bars.

❓ data_output

intermediate

Number of error bars plotted

Given this code, how many error bars will be drawn on the plot?

Matplotlib

import matplotlib.pyplot as plt
import numpy as np

x = np.arange(5)
y = np.array([1, 4, 9, 16, 25])
errors = np.array([0.1, 0.2, 0.3, 0.4, 0.5])

plt.errorbar(x, y, yerr=errors, fmt='s')
plt.show()

A 1 error bar, because yerr is a single array.

B 10 error bars, two for each point (upper and lower).

C 5 error bars, one for each point.

D 0 error bars, because fmt='s' disables error bars.

💡 Hint

Each point gets one error bar with symmetric upper and lower parts.

Each point has one error bar with symmetric upper and lower error lengths equal to the yerr value. So 5 points mean 5 error bars.

🔧 Debug

advanced

Identify the error in error bar plot code

What error will this code raise when run?

Matplotlib

import matplotlib.pyplot as plt
import numpy as np

x = np.array([1, 2, 3])
y = np.array([2, 4, 6])
yerr = np.array([0.1, 0.2])

plt.errorbar(x, y, yerr=yerr, fmt='o')
plt.show()

A ValueError: x and yerr must have the same size.

B No error, plot will show with truncated error bars.

C TypeError: yerr must be a scalar or array-like of same length as y.

D IndexError: yerr index out of range.

💡 Hint

Check if yerr length matches y length.

The yerr array length (2) does not match y length (3), so matplotlib raises a ValueError about size mismatch.

❓ visualization

advanced

Effect of capsize parameter in error bar plots

Which option best describes the visual effect of setting capsize=10 in this error bar plot code?

Matplotlib

import matplotlib.pyplot as plt
import numpy as np

x = np.arange(4)
y = np.array([3, 5, 7, 9])
errors = np.array([0.5, 0.4, 0.3, 0.2])

plt.errorbar(x, y, yerr=errors, fmt='o', capsize=10)
plt.show()

A Error bars are 10 times longer than the errors array values.

B Error bars have horizontal caps 10 points wide at the ends of each vertical error bar.

C Error bars have no caps because capsize=10 is ignored.

D Error bars have vertical caps 10 points tall at the ends.

💡 Hint

Capsize controls the width of the horizontal line at the error bar ends.

The capsize parameter sets the length of the horizontal caps at the ends of the vertical error bars in points (pixels).

🚀 Application

expert

Calculate mean and plot with error bars showing standard error

You have three groups of data points: Group A: [5, 7, 8, 6] Group B: [6, 9, 7, 10] Group C: [8, 12, 11, 9] Which code correctly calculates the mean and standard error of the mean (SEM) for each group and plots them with error bars?

A

import numpy as np
import matplotlib.pyplot as plt

data = {'A': [5,7,8,6], 'B': [6,9,7,10], 'C': [8,12,11,9]}
means = [np.mean(data[g]) for g in data]
sems = [np.std(data[g], ddof=1)/np.sqrt(len(data[g])) for g in data]
plt.errorbar(['A','B','C'], means, yerr=sems, fmt='o')
plt.show()

B

import numpy as np
import matplotlib.pyplot as plt

data = {'A': [5,7,8,6], 'B': [6,9,7,10], 'C': [8,12,11,9]}
means = [np.mean(data[g]) for g in data]
sems = [np.std(data[g])/np.sqrt(len(data[g])) for g in data]
plt.errorbar(['A','B','C'], means, yerr=sems, fmt='o')
plt.show()

C

import numpy as np
import matplotlib.pyplot as plt

data = {'A': [5,7,8,6], 'B': [6,9,7,10], 'C': [8,12,11,9]}
means = [np.mean(data[g]) for g in data]
sems = [np.var(data[g])/np.sqrt(len(data[g])) for g in data]
plt.errorbar(['A','B','C'], means, yerr=sems, fmt='o')
plt.show()

D

import numpy as np
import matplotlib.pyplot as plt

data = {'A': [5,7,8,6], 'B': [6,9,7,10], 'C': [8,12,11,9]}
means = [np.mean(data[g]) for g in data]
sems = [np.std(data[g])/len(data[g]) for g in data]
plt.errorbar(['A','B','C'], means, yerr=sems, fmt='o')
plt.show()

💡 Hint

Use sample standard deviation (ddof=1) to calculate SEM correctly.

Standard error of the mean is calculated as sample standard deviation (with ddof=1) divided by square root of sample size. Option A uses this formula correctly.