Heritability estimation

Heritability

limix.her.estimate(y, lik, K, M=None, verbose=True)[source]

Estimate the so-called narrow-sense heritability.

It supports Normal, Bernoulli, Probit, Binomial, and Poisson phenotypes. Let \(N\) be the sample size and \(S\) the number of covariates.

Parameters:
  • y (array_like) – Either a tuple of two arrays of N individuals each (Binomial phenotypes) or an array of N individuals (Normal, Poisson, or Bernoulli phenotypes). If a continuous phenotype is provided (i.e., a Normal one), make sure they have been normalised in such a way that its values are not extremely large; it might cause numerical errors otherwise. For example, by using limix.qc.mean_standardize() or limix.qc.quantile_gaussianize().
  • lik ("normal", "bernoulli", "probit", binomial", "poisson") – Sample likelihood describing the residual distribution.
  • K (array_like) – \(N\)-by-\(N\) covariance matrix. It might be, for example, the estimated kinship relationship between the individuals. The provided matrix will be normalised via the function limix.qc.normalise_covariance().
  • M (array_like, optional) – \(N\) individuals by \(S\) covariates. It will create a \(N\)-by-\(1\) matrix M of ones representing the offset covariate if None is passed. If an array is passed, it will used as is. Defaults to None.
  • verbose (bool, optional) – True to display progress and summary; False otherwise.
Returns:

Estimated heritability.
Return type:

float

Examples

>>> from numpy import dot, exp, sqrt
>>> from numpy.random import RandomState
>>> from limix.her import estimate
>>>
>>> random = RandomState(0)
>>>
>>> G = random.randn(150, 200) / sqrt(200)
>>> K = dot(G, G.T)
>>> z = dot(G, random.randn(200)) + random.randn(150)
>>> y = random.poisson(exp(z))
>>>
>>> print('%.3f' % estimate(y, 'poisson', K, verbose=False))  
0.183

Notes

It will raise a ValueError exception if non-finite values are passed. Please, refer to the limix.qc.mean_impute() function for missing value imputation.