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Data preprocessing and handling noisy data
Short answer
Preprocessing tasks: cleaning, integration, transformation, reduction, discretization. Noisy data handled by binning (sort, partition, smooth by bin means/medians/boundaries), regression, clustering, or human inspection. Example: equal-frequency binning of prices smooths values within each bin.
What kind of data preprocessing do we need before applying a data mining algorithm to any dataset? Explain the techniques (including the binning method) to handle noisy data.
Data preprocessing
Real-world data is dirty (incomplete, noisy, inconsistent), and poor data gives poor mining results. Preprocessing makes the data fit for mining. The major tasks are:
- Data cleaning: fill missing values, smooth noisy data, identify outliers, resolve inconsistencies.
- Data integration: merge data from several sources, remove redundancy.
- Data transformation: normalisation, aggregation, generalisation, attribute construction.
- Data reduction: reduce volume but keep the analytical result: dimensionality reduction, numerosity reduction, compression.
- Data discretisation: replace numeric values by intervals or concept labels.
Noisy data and the techniques to handle it
Noise is random error or variance in a measured variable (e.g. age = 250).
- Binning: sort the data and divide it into bins (equal-frequency or equal-width); then smooth each value by its bin neighbours (local smoothing).
- Regression: fit the data to a function (linear/multiple regression); values are smoothed to the fitted line.
- Clustering: group similar values; values outside the clusters are outliers and are removed.
- Combined computer and human inspection: the computer flags suspicious values, a human checks them.
Binning example
Price data (sorted): 4, 8, 15, 21, 21, 24, 25, 28, 34. Use equal-frequency bins of depth 3.
| Bin | Values | Smoothing by means | By median | Smoothing by boundaries |
|---|---|---|---|---|
| 1 | 4, 8, 15 | 9, 9, 9 | 8, 8, 8 | 4, 4, 15 |
| 2 | 21, 21, 24 | 22, 22, 22 | 21, 21, 21 | 21, 21, 24 |
| 3 | 25, 28, 34 | 29, 29, 29 | 28, 28, 28 | 25, 25, 34 |
Means: (4+8+15)/3 = 9; (21+21+24)/3 = 22; (25+28+34)/3 = 29. In smoothing by boundaries, each value moves to the closer of the bin's minimum and maximum (8 is 4 from 4 and 7 from 15, so 8 → 4).
Larger bin width gives stronger smoothing.
✓ Answer checked by Claude Opus 5.5 (high effort)
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