linest-matrix.RdGenerate matrix specifying linear estimate.
LE_matrix(object, effect = NULL, at = NULL)
# S3 method for default
LE_matrix(object, effect = NULL, at = NULL)
aggregate_linest_list(linest_list)
get_linest_list(object, effect = NULL, at = NULL)Model object
A vector of variables. For each configuration of these the estimate will be calculated.
Either NULL, a list or a dataframe. 1) If a list, then the list must consist of covariates (including levels of some factors) to be used in the calculations. 2) If a dataframe, the dataframe is split rowwise and the function is invoked on each row.
Linear estimate list (as generated by get_linest_list).
Check this
## Two way anova:
data(warpbreaks)
## An additive model
m0 <- lm(breaks ~ wool + tension, data=warpbreaks)
## Estimate mean for each wool type, for tension="M":
K <- LE_matrix(m0, at=list(wool=c("A", "B"), tension="M"))
K
#> (Intercept) woolB tensionM tensionH
#> [1,] 1 0 1 0
#> [2,] 1 1 1 0
## Vanilla computation:
K %*% coef(m0)
#> [,1]
#> [1,] 29.27778
#> [2,] 23.50000
## Alternative; also providing standard errors etc:
linest(m0, K)
#> estimate std.error statistic df p.value
#> [1,] 29.2778 3.1618 9.2599 50.0000 0
#> [2,] 23.5000 3.1618 7.4325 50.0000 0
esticon(m0, K)
#> estimate std.error statistic p.value beta0 df
#> [1,] 2.9278e+01 3.1618e+00 9.2599e+00 2.0017e-12 0.0000e+00 50
#> [2,] 2.3500e+01 3.1618e+00 7.4325e+00 1.2661e-09 0.0000e+00 50
## Estimate mean for each wool type when averaging over tension;
# two ways of doing this
K <- LE_matrix(m0, at=list(wool=c("A", "B")))
K
#> (Intercept) woolB tensionM tensionH
#> [1,] 1 0 0.3333333 0.3333333
#> [2,] 1 1 0.3333333 0.3333333
K <- LE_matrix(m0, effect="wool")
K
#> (Intercept) woolB tensionM tensionH
#> [1,] 1 0 0.3333333 0.3333333
#> [2,] 1 1 0.3333333 0.3333333
linest(m0, K)
#> estimate std.error statistic df p.value
#> [1,] 31.0370 2.2357 13.8824 50.0000 0
#> [2,] 25.2593 2.2357 11.2981 50.0000 0
## The linear estimate is sometimes called to "least squares mean"
# (LSmeans) or popupulation means.
# Same as
LSmeans(m0, effect="wool")
#> estimate std.error statistic df p.value
#> [1,] 31.0370 2.2357 13.8824 50.0000 0
#> [2,] 25.2593 2.2357 11.2981 50.0000 0
## Without mentioning 'effect' or 'at' an average across all
#predictors are calculated:
K <- LE_matrix(m0)
K
#> (Intercept) woolB tensionM tensionH
#> [1,] 1 0.5 0.3333333 0.3333333
linest(m0, K)
#> estimate std.error statistic df p.value
#> [1,] 28.1481 1.5809 17.8052 50.0000 0
## Because the design is balanced (9 observations per combination
#of wool and tension) this is the same as computing the average. If
#the design is not balanced, the two quantities are in general not
#the same.
mean(warpbreaks$breaks)
#> [1] 28.14815
## Same as
LSmeans(m0)
#> estimate std.error statistic df p.value
#> [1,] 28.1481 1.5809 17.8052 50.0000 0
## An interaction model
m1 <- lm(breaks ~ wool * tension, data=warpbreaks)
K <- LE_matrix(m1, at=list(wool=c("A", "B"), tension="M"))
K
#> (Intercept) woolB tensionM tensionH woolB:tensionM woolB:tensionH
#> [1,] 1 0 1 0 0 0
#> [2,] 1 1 1 0 1 0
linest(m1, K)
#> estimate std.error statistic df p.value
#> [1,] 24.0000 3.6468 6.5812 48.0000 0
#> [2,] 28.7778 3.6468 7.8913 48.0000 0
K <- LE_matrix(m1, at=list(wool=c("A", "B")))
K
#> (Intercept) woolB tensionM tensionH woolB:tensionM woolB:tensionH
#> [1,] 1 0 0.3333333 0.3333333 0.0000000 0.0000000
#> [2,] 1 1 0.3333333 0.3333333 0.3333333 0.3333333
linest(m1, K)
#> estimate std.error statistic df p.value
#> [1,] 31.0370 2.1055 14.7412 48.0000 0
#> [2,] 25.2593 2.1055 11.9970 48.0000 0
K <- LE_matrix(m1, effect="wool")
K
#> (Intercept) woolB tensionM tensionH woolB:tensionM woolB:tensionH
#> [1,] 1 0 0.3333333 0.3333333 0.0000000 0.0000000
#> [2,] 1 1 0.3333333 0.3333333 0.3333333 0.3333333
linest(m1, K)
#> estimate std.error statistic df p.value
#> [1,] 31.0370 2.1055 14.7412 48.0000 0
#> [2,] 25.2593 2.1055 11.9970 48.0000 0
LSmeans(m1, effect="wool")
#> estimate std.error statistic df p.value
#> [1,] 31.0370 2.1055 14.7412 48.0000 0
#> [2,] 25.2593 2.1055 11.9970 48.0000 0
K <- LE_matrix(m1)
K
#> (Intercept) woolB tensionM tensionH woolB:tensionM woolB:tensionH
#> [1,] 1 0.5 0.3333333 0.3333333 0.1666667 0.1666667
linest(m1, K)
#> estimate std.error statistic df p.value
#> [1,] 28.1481 1.4888 18.9068 48.0000 0
LSmeans(m1)
#> estimate std.error statistic df p.value
#> [1,] 28.1481 1.4888 18.9068 48.0000 0