Spearman correlation in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the function to calculate Spearman correlation.

SciPy

from scipy.stats import [1]

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

Akendalltau

Bspearmanr

Cpearsonr

Dttest_ind

Attempts:

3 left

💡 Hint

Common Mistakes

Importing 'pearsonr' instead of 'spearmanr'.

Using unrelated functions like 'ttest_ind'.

The spearmanr function from scipy.stats calculates Spearman correlation.

2fill in blank

medium

Complete the code to calculate Spearman correlation between two lists x and y.

SciPy

corr, p_value = spearmanr([1], y)

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

Ax

Brange(10)

C[1, 2, 3]

Dy

Attempts:

3 left

💡 Hint

Common Mistakes

Passing 'y' twice.

Passing a range instead of the variable.

You need to pass the first data list x to the spearmanr function.

3fill in blank

hard

Fix the error in the code to correctly calculate Spearman correlation for lists a and b.

SciPy

result = spearmanr(a, [1])

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Ab

Bc

Ca

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Passing 'a' twice.

Passing an undefined variable like 'c'.

The second argument must be the second data list b to compute correlation with a.

4fill in blank

hard

Fill both blanks to create a dictionary of Spearman correlations for each pair of columns in data.

SciPy

correlations = {col1: {col2: spearmanr(data[col1], data[1])[0] for col2 in data if col1 [2] col2} for col1 in data}

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A[col2]

B(col2)

C!= col2

D== col2

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets for column access.

Using '==' instead of '!=' in the condition.

Use square brackets to access the column col2 in the DataFrame. The condition excludes pairs where columns are the same using !=.

5fill in blank

hard

Fill all three blanks to filter pairs with correlation above 0.5 and create a result dictionary.

SciPy

result = [1]: corr for col1, corr in correlations.items() for col2, corr in corr.items() if corr [2] 0.5 and col1 [3] 'target'}

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

Acol1

B>

C!=

Dcol2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'col1' as the key instead of 'col2'.

Using '<' instead of '>' for filtering.

Using '==' instead of '!=' to exclude 'target'.

The dictionary key should be col2, correlation must be greater than 0.5, and exclude the 'target' column using '!='.