Problems

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Binning

Feature EngineeringData Processing

Easy

Binning (also called discretization) converts continuous numeric features into discrete categories by dividing the value range into equal-width intervals. Each value is assigned to its corresponding bin. This technique can reduce noise, handle outliers, and make features compatible with algorithms that work better with categorical data.

Given a list of numeric values and a number of bins, assign each value to a bin index using equal-width binning.

Algorithm

Compute the bin width from the data range:

w = \frac{\max(x) - \min(x)}{\text{num\_bins}}

Assign each value to a bin:

\text{bin}(x_i) = \min\left(\left\lfloor \frac{x_i - \min(x)}{w} \right\rfloor, \; \text{num\_bins} - 1\right)

The maximum value is clamped to the last bin. If all values are equal, all are assigned to bin 0.

Return one integer from zero through num_bins minus one for each value.

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Examples

Input: values = [0, 25, 50, 75, 100], num_bins = 4

Output: [0, 1, 2, 3, 3]

Explanation: The width is 25, and the maximum is clamped into the final bin.

Input: values = [1, 2, 3, 4, 5, 6], num_bins = 2

Output: [0, 0, 0, 1, 1, 1]

Hint 1

Compute bin width from the observed minimum and maximum.

Hint 2

Convert each offset into an integer bin and clamp the maximum to the final index.

Requirements

Divide the range [min, max] into num_bins equal-width intervals
Assign each value a 0-indexed bin number
Clamp the maximum value to the last bin (num_bins - 1)
If all values are identical, assign all to bin 0
Return a list of integers

Constraints

values has at least 1 element
num_bins >= 1
Return a list of integers in [0, num_bins - 1]
Time limit: 300 ms

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Binning

Feature EngineeringData Processing

Easy

Given a list of numeric values and a number of bins, assign each value to a bin index using equal-width binning.

Algorithm

Compute the bin width from the data range:

w = \frac{\max(x) - \min(x)}{\text{num\_bins}}

Assign each value to a bin:

\text{bin}(x_i) = \min\left(\left\lfloor \frac{x_i - \min(x)}{w} \right\rfloor, \; \text{num\_bins} - 1\right)

The maximum value is clamped to the last bin. If all values are equal, all are assigned to bin 0.

Return one integer from zero through num_bins minus one for each value.

Loading visualization...

Examples

Input: values = [0, 25, 50, 75, 100], num_bins = 4

Output: [0, 1, 2, 3, 3]

Explanation: The width is 25, and the maximum is clamped into the final bin.

Input: values = [1, 2, 3, 4, 5, 6], num_bins = 2

Output: [0, 0, 0, 1, 1, 1]

Hint 1

Compute bin width from the observed minimum and maximum.

Hint 2

Convert each offset into an integer bin and clamp the maximum to the final index.

Requirements

Divide the range [min, max] into num_bins equal-width intervals
Assign each value a 0-indexed bin number
Clamp the maximum value to the last bin (num_bins - 1)
If all values are identical, assign all to bin 0
Return a list of integers

Constraints

values has at least 1 element
num_bins >= 1
Return a list of integers in [0, num_bins - 1]
Time limit: 300 ms

Try Similar Problems

Rank Transform Zscore Standardization Polynomial Features Interaction Features Log Transform

Case 1

Case 2

values =

Accepts: array

num_bins =

Accepts: number

Given a list of numeric values and a number of bins, assign each value to a bin index using equal-width binning.

Algorithm

Compute the bin width from the data range:

w = \frac{\max(x) - \min(x)}{\text{num\_bins}}

Assign each value to a bin:

\text{bin}(x_i) = \min\left(\left\lfloor \frac{x_i - \min(x)}{w} \right\rfloor, \; \text{num\_bins} - 1\right)

The maximum value is clamped to the last bin. If all values are equal, all are assigned to bin 0.

Return one integer from zero through num_bins minus one for each value.