Binning continuous variables in Data Analysis Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of pandas cut with custom bins

What is the output of the following code snippet?

Data Analysis Python

import pandas as pd
import numpy as np

values = np.array([1, 5, 10, 15, 20])
bins = [0, 5, 10, 15]
categories = pd.cut(values, bins)
print(categories)

A

[(0, 5], (0, 5], (5, 10], (10, 15], NaN]
Categories (3, interval[int64, right]): [(0, 5] < (5, 10] < (10, 15]]

B

[(0, 5], NaN, (5, 10], (10, 15], NaN]
Categories (3, interval[int64, right]): [(0, 5] < (5, 10] < (10, 15]]

C

[NaN, (0, 5], (5, 10], (10, 15], NaN]
Categories (3, interval[int64, right]): [(0, 5] < (5, 10] < (10, 15]]

D

[NaN, (0, 5], (5, 10], (10, 15], (15, 20]]
Categories (4, interval[int64, right]): [(0, 5] < (5, 10] < (10, 15] < (15, 20]]

💡 Hint

Remember that pd.cut assigns values to bins based on intervals defined by the bins list. Values equal to the left edge are excluded, right edge included by default.

The bins are (0,5], (5,10], and (10,15]. The value 1 falls into (0,5], 5 also falls into (0,5] because the right edge is included, 10 falls into (5,10], 15 falls into (10,15], and 20 is outside the bins so it is NaN.

❓ data_output

intermediate

Number of bins created by qcut

Given the following code, how many unique bins will be created?

Data Analysis Python

import pandas as pd
import numpy as np

values = np.array([1, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10])
categories = pd.qcut(values, 4)
unique_bins = categories.unique()
print(len(unique_bins))

A 3

B 6

C 5

D 4

💡 Hint

qcut tries to create equal-sized bins but duplicates in data can reduce the number of bins.

Because the value 2 appears twice, qcut merges bins to avoid empty bins, resulting in 3 unique bins instead of 4.

❓ visualization

advanced

Visualizing binning effect on data distribution

Which option shows the correct histogram with bins created by pd.cut for the data below?

Data Analysis Python

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np

np.random.seed(0)
data = np.random.normal(loc=50, scale=10, size=1000)
bins = [20, 40, 60, 80]
categories = pd.cut(data, bins)
plt.hist(data, bins=bins, edgecolor='black')
plt.title('Histogram with bins [20, 40, 60, 80]')
plt.show()

A Histogram with 4 bars showing counts of data in intervals [20,40), [40,60), [60,80), [80,100)

B Histogram with 3 bars showing counts of data in intervals (20,40], (40,60], (60,80]

C Histogram with 3 bars showing counts of data in intervals [20,40), [40,60), [60,80)

D Histogram with 4 bars showing counts of data in intervals (20,40], (40,60], (60,80], (80,100]

💡 Hint

pd.cut by default includes the right edge of intervals and excludes the left edge.

The bins define intervals (20,40], (40,60], and (60,80]. The histogram shows 3 bars corresponding to these intervals.

🧠 Conceptual

advanced

Effect of right parameter in pd.cut

What is the effect of setting right=False in pd.cut when binning data?

A Bins include the right edge and exclude the left edge of intervals.

B Bins include both edges of intervals.

C Bins include the left edge and exclude the right edge of intervals.

D Bins exclude both edges of intervals.

💡 Hint

By default, pd.cut includes the right edge. Changing right changes which edge is included.

Setting right=False means intervals are left-inclusive and right-exclusive, so the left edge is included in the bin.

🔧 Debug

expert

Identify the error in binning code

What error will the following code raise?

Data Analysis Python

import pandas as pd
values = [1, 2, 3, 4, 5]
bins = [0, 2, 4]
categories = pd.cut(values, bins, labels=['Low', 'Medium', 'High'])

A No error, code runs successfully

B TypeError: 'list' object is not callable

C IndexError: list index out of range

D ValueError: Bin labels must be one fewer than the number of bin edges

💡 Hint

Check the length of labels compared to bins.

The labels list has 3 elements but bins define only 2 intervals, so labels length must be 2, causing a ValueError.