Description Usage Arguments Details Value Author(s) References See Also Examples

This function computes the solution path for the trend filtering
problem of an arbitrary polynomial order. When the order is set to
zero, trend filtering is equivalent to the 1d fused lasso, see
`fusedlasso1d`

.

1 2 3 | ```
trendfilter(y, pos, X, ord = 1, approx = FALSE, maxsteps = 2000,
minlam = 0, rtol = 1e-07, btol = 1e-07, eps = 1e-04,
verbose = FALSE)
``` |

`y` |
a numeric response vector. |

`pos` |
an optional numeric vector specifying the positions of the
observations, and missing |

`X` |
an optional matrix of predictor variables, with observations along
the rows, and variables along the columns. If the passed |

`ord` |
an integer specifying the desired order of the piecewise polyomial produced by the solution of the trend filtering problem. Must be non-negative, and the default to 1 (linear trend filtering). |

`approx` |
a logical variable indicating if the approximate solution path
should be used (with no dual coordinates leaving the boundary).
Default is |

`maxsteps` |
an integer specifying the maximum number of steps for the algorithm to take before termination. Default is 2000. |

`minlam` |
a numeric variable indicating the value of lambda at which the path should terminate. Default is 0. |

`rtol` |
a numeric variable giving the tolerance for determining the rank of a matrix: if a diagonal value in the R factor of a QR decomposition is less than R, in absolute value, then it is considered zero. Hence making rtol larger means being less stringent with determination of matrix rank. In general, do not change this unless you know what you are getting into! Default is 1e-7. |

`btol` |
a numeric variable giving the tolerance for accepting "late" hitting and leaving times: future hitting times and leaving times should always be less than the current knot in the path, but sometimes for numerical reasons they are larger; any computed hitting or leaving time larger than the current knot + btol is thrown away. Hence making btol larger means being less stringent withthe determination of hitting and leaving times. Again, in general, do not change this unless you know what you are getting into! Default is 1e-7. |

`eps` |
a numeric variable indicating the multiplier for the ridge penalty,
in the case that |

`verbose` |
a logical variable indicating if progress should be reported after each knot in the path. |

When the predictor matrix is the identity, trend filtering fits a piecewise polynomial to linearly ordered observations. The result is similar to that of a polynomial regression spline or a smoothing spline, except the knots in the piecewise polynomial (changes in the (k+1)st derivative, if the polynomial order is k) are chosen adaptively based on the observations. This is in contrast to regression splines, where the knots are prespecified, and smoothing splines, which place a knot at every data point.

With a nonidentity predictor matrix, the trend filtering problem enforces piecewise polynomial smoothness along successive components of the coefficient vector. This can be used to fit a kind of varying coefficient model.

We note that, in the signal approximator (identity predictor matrix)
case, fitting trend filtering estimate with arbitrary positions `pos`

is theoretically no harder than doing so on an evenly spaced grid. However
in practice, with differing gaps between points, the algorithm can
become numerically unstable even for large (or moderately large) problems.
This is especially true as the polynomial order increases. Hence, use the
positions argument `pos`

with caution.

Returns an object of class "trendfilter", a subclass of "genlasso". This is a list with at least following components:

`lambda` |
values of lambda at which the solution path changes slope, i.e., kinks or knots. |

`beta` |
a matrix of primal coefficients, each column corresponding to a knot in the solution path. |

`fit` |
a matrix of fitted values, each column corresponding to a knot in the solution path. |

`u` |
a matrix of dual coefficients, each column corresponding to a knot in the solution path. |

`hit` |
a vector of logical values indicating if a new variable in the dual
solution hit the box contraint boundary. A value of |

`df` |
a vector giving an unbiased estimate of the degrees of freedom of the fit at each knot in the solution path. |

`y` |
the observed response vector. Useful for plotting and other methods. |

`completepath` |
a logical variable indicating whether the complete path was
computed (terminating the path early with the |

`bls` |
the least squares solution, i.e., the solution at lambda = 0. This
can be |

`ord` |
the order of the piecewise polyomial that has been fit. |

`call` |
the matched call. |

Taylor B. Arnold and Ryan J. Tibshirani

Tibshirani, R. J. and Taylor, J. (2011), "The solution path of the generalized lasso", Annals of Statistics 39 (3) 1335–1371.

Tibshirani, R. J. (2014), "Adaptive piecewise polynomial estimation via trend filtering", Annals of Statistics 42 (1): 285–323.

Arnold, T. B. and Tibshirani, R. J. (2014), "Efficient implementations of the generalized lasso dual path algorithm", arXiv: 1405.3222.

Kim, S.-J., Koh, K., Boyd, S. and Gorinevsky, D. (2009), "l1 trend filtering", SIAM Review 51 (2), 339–360.

`fusedlasso1d`

, `genlasso`

,
`cv.trendfilter`

, `plot.trendfilter`

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 | ```
# Constant trend filtering (the 1d fused lasso)
set.seed(0)
n = 100
beta0 = rep(sample(1:10,5),each=n/5)
y = beta0 + rnorm(n,sd=0.8)
a = fusedlasso1d(y)
plot(a)
# Linear trend filtering
set.seed(0)
n = 100
beta0 = numeric(n)
beta0[1:20] = (0:19)*4/19+2
beta0[20:45] = (25:0)*3/25+3
beta0[45:80] = (0:35)*9/35+3
beta0[80:100] = (20:0)*4/20+8
y = beta0 + rnorm(n)
a = trendfilter(y,ord=1)
plot(a,df=c(2,3,4,10))
# Cubic trend filtering
set.seed(0)
n = 100
beta0 = numeric(100)
beta0[1:40] = (1:40-20)^3
beta0[40:50] = -60*(40:50-50)^2 + 60*100+20^3
beta0[50:70] = -20*(50:70-50)^2 + 60*100+20^3
beta0[70:100] = -1/6*(70:100-110)^3 + -1/6*40^3 + 6000
beta0 = -beta0
beta0 = (beta0-min(beta0))*10/diff(range(beta0))
y = beta0 + rnorm(n)
a = trendfilter(y,ord=3)
plot(a,nlam=5)
``` |

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