The Wide View

A hard Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
hard
Seniority
L4

Problem

A reporting export arrives in long format: one record per cell, each carrying a row label, a col label, and a numeric val. Reshape it into a nested lookup keyed by row then column, so result[row][col] returns that cell's value. Rows are sparse, holding only the columns that actually appear for them, and when the same row and column pair shows up more than once the later record wins.

Long format is easy. Wide format is useful.

Example 1

Input:

records = [{"col":"math","row":"Alice","val":90},{"col":"sci","row":"Alice","val":85},{"col":"math","row":"Bob","val":78}]

Output:

{"Bob":{"math":78},"Alice":{"sci":85,"math":90}}

Example 2

Input:

records = [{"col":"c1","row":"r1","val":1}]

Output:

{"r1":{"c1":1}}

Worked solution and explanation

What this really is

This is a sparse two-level dict build wearing a spreadsheet costume. The skill being probed: can you grow a nested dict whose inner dict may not exist yet, without clobbering rows that share an outer key? Everyone writes the loop. The tell is whether you create result[row] exactly once and keep adding columns to it as more records for that same row arrive. Get that wrong and Alice's sci entry wipes out her math entry.

Break down the requirements

Step 1: The row label is the outer key

Each record's row field is the outer key. Different records can share the same row, so you cannot treat each record as a fresh row.

Step 2: Columns map to values inside each row

Within a row, set result[row][col] = val. Because rows are sparse, you only ever write the columns that show up; there is no dense matrix to fill.

Step 3: Initialize the inner dict once

Create the inner dict the first time a row is seen, and never again. This is the one line that separates a correct answer from one that silently drops columns.

The solution

Two-level dict construction from flat records
def pivot_table(records):
    result = {}
    for record in records:
        row = record['row']
        col = record['col']
        val = record['val']
        if row not in result:
            result[row] = {}
        result[row][col] = val
    return result

One pass, inner dict created only on first sight of a row.

Common follow-up questions