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Pivot a long metrics table to wide · Python scenario · D8LooP

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Pivot a long metrics table to wide

Metrics are stored long — one row per day, per metric, per value — because that shape survives new metrics being added. The dashboard needs them wide: one row per day with a column per metric. Not every metric is reported every day, and the collector occasionally sends the same day and metric twice. Write pivot_rows(rows, index_field, column_field, value_field).

Function to write

pivot_rows(rows: list[dict], index_field: str, column_field: str, value_field: str) -> dict

A dict with columns — the sorted column names — and rows, one rectangular wide row per index value.

How to approach it

Collect the two axes and the cells in one pass, then build the grid from them.

Sample cases

+ 2 held back until you submit

a metrics table

A repeated day and metric pair, a metric only one day has, a null value and a null index.

Input

Argument 1

day	metric	value
2026-03-01	orders	10
2026-03-01	revenue	250.5
2026-03-02	orders	4
2026-03-02	orders	3
2026-03-03	refunds	1
2026-03-01	refunds	None
None	orders	99

Argument 2

'day'

Argument 3

'metric'

Argument 4

'value'

Returns

{
  'columns': [
    'orders',
    'refunds',
    'revenue'
  ],
  'rows': [
    {
      'day': '2026-03-01',
      'orders': 10,
      'refunds': None,
      'revenue': 250.5
    },
    {
      'day': '2026-03-02',
      'orders': 7,
      'refunds': None,
      'revenue': None
    },
    {
      'day': '2026-03-03',
      'orders': None,
      'refunds': 1,
      'revenue': None
    }
  ]
}

nothing to pivot

No rows means no columns and no wide rows, with both keys still present.

Input

Argument 1

[] (empty list)

Argument 2

'day'

Argument 3

'metric'

Argument 4

'value'

Returns

{
  'columns': [],
  'rows': []
}

one metric only

A single column still produces a wide table rather than the input unchanged.

Input

Argument 1

day	metric	value
2026-03-01	orders	1
2026-03-02	orders	2

Argument 2

'day'

Argument 3

'metric'

Argument 4

'value'

Returns

{
  'columns': [
    'orders'
  ],
  'rows': [
    {
      'day': '2026-03-01',
      'orders': 1
    },
    {
      'day': '2026-03-02',
      'orders': 2
    }
  ]
}

Constraints

1One output row per distinct index value, ordered ascending as text.
2The columns are every distinct value of the column field, as text, sorted ascending. Return them alongside the rows.
3Every output row carries every column, so the rows are rectangular.
4A cell with no measurement is None, not zero. Nothing measured and a measured zero are different facts.
5When the same index and column pair appears more than once, sum the values.
6A value that is not a number, or is a boolean, contributes nothing — but the index value still gets a row and the column still exists.
7A row with a null index or a null column cannot be placed, and is skipped.
8Each output row starts with the index field under its own name.

Worked example

2026-03-02 reports orders twice, as 4 and 3. They are summed to 7 — the collector sent two partial counts, and keeping only one of them loses whichever it did not keep. A pivot that assigns rather than accumulates silently keeps the last one written, which depends on file order. refunds is reported on 2026-03-03 and nowhere else, and on 2026-03-01 it arrives with a null value. Both days need a refunds cell — one empty, one holding 1 — and every other day needs one too. Filling those with zero would say refunds were measured and found to be none, which is a claim the data does not make and which makes an average over the column wrong.

What this tests

Building a rectangular result from sparse input, and the discipline of keeping absent distinct from zero. The two-pass structure — discover the axes, then fill the grid — is what makes the rectangle possible at all.

PYTHONpivot_rows(rows: list[dict], index_field: str, column_field: str, value_field: str) -> dict

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