Reference · Statistics

stats.covariance

stats.covariance(list, list) -> float

What it does

The sample covariance of two lists of numbers of the same length: how much they vary together, as Python's statistics.covariance. Positive when the pairs rise together, negative when one falls as the other rises, near zero when they are unrelated. The sum of the products of the distances from each mean is divided by one less than the count. Its scale is the data's, so for a scale-free measure use stats.correlation.

Examples

let x = [1, 2, 3, 4, 5, 6, 7, 8, 9]let y = [1, 2, 3, 1, 2, 3, 1, 2, 3]let z = [9, 8, 7, 6, 5, 4, 3, 2, 1]print(stats.covariance(x, y), stats.covariance(x, z))

0.75 -7.5

Edge cases

• The order of the two lists does not matter.
• A list with itself gives its sample variance.
• A constant list gives 0.0.
• Two points are enough.
• Ints and floats can mix; the answer is always a float.

print(stats.covariance([1, 2, 3], [3, 2, 1]), stats.covariance([3, 2, 1], [1, 2, 3]))print(stats.covariance([1, 2, 3], [1, 2, 3]) == stats.variance([1, 2, 3]))print(stats.covariance([1, 2], [3, 3]), stats.covariance([1, 2], [2, 4.0]))

-1.0 -1.0true0.0 1.0

Errors

Lists of different lengths, fewer than two points, or an item that is not a number are each a runtime error, not an error value.

print(stats.covariance([1, 2, 3], [1, 2]))

line 1: `stats.covariance` needs lists of the same length (3 and 2)

print(stats.covariance([1], [2]))

line 1: `stats.covariance` requires at least two data points

Chaining

In a pipeline the first list comes first and the second is the argument:

let ads = [10, 20, 30, 40]let sales = [110, 190, 310, 390]print(ads | stats.covariance(sales))print((ads | stats.covariance(sales)) / stats.variance(ads))

1600.09.600000000000001

See also

stats.correlation, stats.variance, stats.linear_regression

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