burgle

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The goal of burgle is to “steal” only the necessary parts of model objects for applications in simuation.

Installation

install.packages("burgle")

Or you can install the development version of burgle like so:

devtools::install_github("ClevelandClinicQHS/burgle")

Pros of burgle

  1. The reduction in size can save on memory and storage space. A great advantage since R works with physical memory
  2. The removal of data from the model objects can allow them to be shared freely, if there are data sharing concerns or requirements.
  3. A streamlined method to simulate response for parameter uncertainty and probabilistic sampling

Linear Model Example

set.seed(287453)
library(burgle)
fit <- lm(Sepal.Length ~., data = iris)
bfit <- burgle(fit)
pryr::object_size(fit)
#> 39.43 kB
pryr::object_size(bfit)
#> 2.88 kB

as.numeric(pryr::object_size(bfit)/pryr::object_size(fit))*100
#> [1] 7.303713

Our burgle_lm is roughly 7.3% the size of the original lm object, the iris dataset has 150 observations and 5 columns.

Another example is the using the nycflights13::flights dataset.

fit2 <- lm(arr_delay ~ as.factor(month) + dep_delay + origin + distance + hour, data = nycflights13::flights)
b_fit2 <- burgle(fit2)

as.numeric(pryr::object_size(b_fit2)/pryr::object_size(fit2))*100
#> [1] 0.008213793

Our burgle_lm is roughly 0.01% the size of the original lm object. This dataset has 336776 observations and our model has used 5 of the 19 columns as predictors.

Simulation Massive Example

Here one can see a simulated dataset of 10 million data points with 3 random covariates.

N <- 1e7
df <- data.frame(y = rnorm(N), x1 = runif(N), x2 = runif(N, -1, 1), x3 = runif(N, -2, 2))
mfit <- lm(y~., data = df)
b_mfit <- burgle(mfit)

m0 <- pryr::object_size(mfit)
print(m0, units = "Gb")
#> 1.60 GB
pryr::object_size(b_mfit)
#> 2.15 kB

The lm is 1.6 Gb while the burgle_lm object is 2152 bytes. A reduction of size by 10^6!

Predictions

The new predict methods for our burgle objects allow for one to easily predict new values and multiple simluated responses of newdata. The structure is as follows:

If one wants to predict using the original model simply set original = TRUE.

Depending on the model object there are different types of predictions. By default it will return the linear predictor (lp). If one wants to see the response (type = "response"), which makes more sense when using glmobjects and survival models. The se = FALSE is an argument on whether to include the standard error of the model when simluating responses. TRUE means to use the model standard error when sampling. We recommend setting it to TRUE when doing more than one simulation or when setting type = "response".

predict(bfit, newdata = head(iris), original = TRUE, draws = 1, se = FALSE, type = "lp")
#>       [,1]
#> 1 5.004788
#> 2 4.756844
#> 3 4.773097
#> 4 4.889357
#> 5 5.054377
#> 6 5.388886
predict(bfit, newdata = head(iris), original = FALSE, draws = 5, type = "lp")
#>          [,1]     [,2]     [,3]     [,4]     [,5]
#> [1,] 5.056657 5.007215 4.961257 4.932543 5.050141
#> [2,] 4.790295 4.746012 4.713736 4.754813 4.785725
#> [3,] 4.822688 4.775977 4.736837 4.739272 4.814687
#> [4,] 4.917719 4.872770 4.839147 4.876992 4.915412
#> [5,] 5.109929 5.059456 5.010762 4.968089 5.103024
#> [6,] 5.454726 5.424994 5.301047 5.286944 5.422597
## These two should be similar
predict(bfit, newdata = head(iris), original = FALSE, draws = 5, sims = 5, se = TRUE, type = "response")
#> [[1]]
#>          [,1]     [,2]     [,3]     [,4]     [,5]
#> [1,] 5.188335 5.064758 4.528114 5.460978 4.938227
#> [2,] 4.301446 4.554181 4.743173 4.829344 4.793417
#> [3,] 4.441790 4.637052 4.831230 4.536184 4.383245
#> [4,] 4.739983 4.891668 4.808410 4.820148 5.297950
#> [5,] 4.470888 5.130100 4.996598 5.327621 5.217366
#> [6,] 5.012921 5.122429 5.258801 4.947419 5.906721
#> 
#> [[2]]
#>          [,1]     [,2]     [,3]     [,4]     [,5]
#> [1,] 4.617066 5.588269 4.633249 4.743949 5.383410
#> [2,] 4.956657 4.558627 4.992207 4.682602 4.068407
#> [3,] 4.600649 4.620536 4.568879 4.919372 4.894229
#> [4,] 4.550183 4.777452 5.177517 4.658162 5.435255
#> [5,] 5.061492 4.952346 4.724817 4.587407 4.911647
#> [6,] 5.589748 5.249421 5.258343 5.781827 4.594522
#> 
#> [[3]]
#>          [,1]     [,2]     [,3]     [,4]     [,5]
#> [1,] 4.654951 5.018870 5.111338 4.881092 5.267062
#> [2,] 4.979906 5.620321 4.146499 5.021050 4.759204
#> [3,] 4.666988 4.154987 5.114534 4.697866 4.957075
#> [4,] 4.611058 4.369505 4.767237 4.834120 4.395112
#> [5,] 4.653835 5.552697 5.733886 4.925832 4.933249
#> [6,] 5.918499 5.859100 5.146537 5.276244 5.499586
#> 
#> [[4]]
#>          [,1]     [,2]     [,3]     [,4]     [,5]
#> [1,] 4.957008 5.086826 5.090159 4.612207 5.024133
#> [2,] 4.917264 4.299553 4.726446 4.003881 4.799311
#> [3,] 5.164780 4.434195 5.413705 4.613030 4.858791
#> [4,] 4.609427 5.069844 4.472140 4.443499 4.939135
#> [5,] 4.441486 4.840654 5.358090 4.544883 4.997390
#> [6,] 5.557312 5.554476 5.046674 5.480583 5.295779
#> 
#> [[5]]
#>          [,1]     [,2]     [,3]     [,4]     [,5]
#> [1,] 5.306194 4.787267 4.607056 5.086712 4.689726
#> [2,] 4.593383 4.595724 4.741960 4.961092 4.503969
#> [3,] 4.868304 5.384595 4.949947 5.243014 4.305317
#> [4,] 4.795677 4.308122 5.015002 4.419806 5.059970
#> [5,] 5.347532 5.804453 5.136792 4.602399 4.960713
#> [6,] 6.070301 5.485667 5.432137 4.758103 5.662043

Generalized Linear Model

The framework should work with all glm family options, just the binomial example is demonstrated below.

b_glm <- burgle(glm(I(Species == "versicolor") ~ ., family = "binomial", data = iris))

predict(b_glm, head(iris), original = FALSE, se = TRUE, draws = 5, type = "lp")
#>            [,1]       [,2]       [,3]      [,4]      [,5]
#> [1,]  1.1931513  0.8544506 -0.9125312 -4.449063 -5.371412
#> [2,]  2.2506299  1.8325844 -1.2188106 -2.859110 -5.147537
#> [3,] -1.2826411  1.7282266 -2.2889517 -2.672046 -3.255157
#> [4,] -0.5582609  0.2016295 -3.3602143 -2.508221 -1.622880
#> [5,] -2.6288233 -4.6355719 -2.9090909 -2.429126 -2.970703
#> [6,]  1.1098697 -5.8229683 -2.4733745 -3.157545 -6.473457
predict(b_glm, head(iris), original = FALSE, se = TRUE, draws = 5, type = "response")
#>      [,1] [,2] [,3] [,4] [,5]
#> [1,]    0    1    0    0    1
#> [2,]    0    1    0    0    1
#> [3,]    1    0    0    0    1
#> [4,]    0    1    0    0    0
#> [5,]    1    0    0    0    0
#> [6,]    0    0    0    0    0

Cox Proporiontal Hazards Model

library(survival)
lung <- survival::lung
lung$status <- lung$status - 1

cox <- coxph(Surv(time, status) ~ age + sex + ph.ecog + ph.karno + pat.karno, data = lung)
cox_sm <- coxph(Surv(time, status) ~ age + sex + ph.ecog + ph.karno + pat.karno, data = lung, x = FALSE, y = FALSE)

b_cox <- burgle(cox)
pryr::object_size(cox)
#> 36.02 kB
pryr::object_size(cox_sm)
#> 31.21 kB
pryr::object_size(b_cox)
#> 5.59 kB

as.numeric(pryr::object_size(b_cox)/pryr::object_size(cox))*100
#> [1] 15.52643
as.numeric(pryr::object_size(b_cox)/pryr::object_size(cox_sm))*100
#> [1] 17.91848

Our burgle_coxph model is 17.92% the size of the original Cox proportional hazards model even after setting x=FALSE and y= FALSE. The lung dataset has 228 observations.

One way to further reduce the size of the burgle_coxph is to reduce the number of unique time points in the data, since the it contains the baseline hazard of the model. The lung dataset as 186 unique values. If we were to round these to the nearest 14 days that would reduce the number of timepoints to 58.

lung$time2 <- plyr::round_any(lung$time, 14)
cox2 <- coxph(Surv(time2, status) ~ age + sex + ph.ecog + ph.karno + pat.karno, data = lung, x = TRUE, y = FALSE)
b_cox2 <- burgle(cox2)

pryr::object_size(cox2)
#> 42.50 kB
pryr::object_size(b_cox2)
#> 3.87 kB
as.numeric(pryr::object_size(b_cox2)/pryr::object_size(cox2))*100
#> [1] 9.109731

This reduce the size to 9.11% of the original coxph object.

Survival predictions

Predictions are a slightly different structure for survival or longitudinal models if type = "response" or "risk", since a time point is required. If type = "response" the returned results is a simulated 1 or 0 if the event has been experienced or not at a given time point(s). The structure is as follows:

predict(b_cox, newdata = head(lung), original = TRUE, draws = 1, type = "lp")
#>           [,1]
#> [1,] 1.2620217
#> [2,] 0.7293038
#> [3,] 0.5927067
#> [4,] 1.4729644
#> [5,] 0.7967660
#> [6,] 0.8301417
predict(b_cox, newdata = head(lung), original = TRUE, draws = 1, type = "risk", times = 500)
#>           [,1]
#> [1,] 0.8051989
#> [2,] 0.6171886
#> [3,] 0.5672583
#> [4,] 0.8673345
#> [5,] 0.6420013
#> [6,] 0.6542671
predict(b_cox, newdata = head(lung), original = TRUE, draws = 1, type = "risk", times = c(500, 1000))
#> Warning in predict.burgle_coxph(b_cox, newdata = head(lung), original = TRUE, :
#> times has a value of 1000 which is larger than the maximum time value of 883
#>           [,1]      [,2]
#> [1,] 0.8051989 0.9851588
#> [2,] 0.6171886 0.9155426
#> [3,] 0.5672583 0.8842068
#> [4,] 0.8673345 0.9944786
#> [5,] 0.6420013 0.9289231
#> [6,] 0.6542671 0.9350234
predict(b_cox, newdata = head(lung), original = TRUE, draws = 1, type = "response", times = c(500, 1000))
#> Warning in predict.burgle_coxph(b_cox, newdata = head(lung), original = TRUE, :
#> times has a value of 1000 which is larger than the maximum time value of 883
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    0    1
#> [3,]    1    1
#> [4,]    1    1
#> [5,]    1    1
#> [6,]    1    1
predict(b_cox, newdata = head(lung), original = FALSE, draws = 5, sims = 2, type = "response", times = c(500, 1000))
#> Warning in predict.burgle_coxph(b_cox, newdata = head(lung), original = FALSE,
#> : times has a value of 1000 which is larger than the maximum time value of 883
#> [[1]]
#> [[1]][[1]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    1    1
#> [4,]    1    1
#> [5,]    0    1
#> [6,]    1    1
#> 
#> [[1]][[2]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    0    1
#> [4,]    1    1
#> [5,]    1    1
#> [6,]    0    1
#> 
#> 
#> [[2]]
#> [[2]][[1]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    0    0
#> [4,]    1    1
#> [5,]    1    1
#> [6,]    1    1
#> 
#> [[2]][[2]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    1    1
#> [4,]    1    1
#> [5,]    1    1
#> [6,]    1    1
#> 
#> 
#> [[3]]
#> [[3]][[1]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    1    1
#> [4,]    1    1
#> [5,]    1    1
#> [6,]    1    1
#> 
#> [[3]][[2]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    1    1
#> [4,]    1    1
#> [5,]    1    1
#> [6,]    1    1
#> 
#> 
#> [[4]]
#> [[4]][[1]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    1    1
#> [4,]    1    1
#> [5,]    1    1
#> [6,]    1    1
#> 
#> [[4]][[2]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    1    1
#> [4,]    1    1
#> [5,]    1    1
#> [6,]    1    1
#> 
#> 
#> [[5]]
#> [[5]][[1]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    1    1
#> [4,]    1    1
#> [5,]    0    1
#> [6,]    1    1
#> 
#> [[5]][[2]]
#>      [,1] [,2]
#> [1,]    1    1
#> [2,]    1    1
#> [3,]    1    1
#> [4,]    1    1
#> [5,]    0    1
#> [6,]    1    1

Larger Simulation Example

## The original model at time 500
predict(b_cox, newdata = head(lung), original = TRUE, draws = 1, type = "risk", times = c(500))
#>           [,1]
#> [1,] 0.8051989
#> [2,] 0.6171886
#> [3,] 0.5672583
#> [4,] 0.8673345
#> [5,] 0.6420013
#> [6,] 0.6542671

## Doing 1000 simulations from the original model and calculating the death rate
a0 <- predict(b_cox, newdata = head(lung), original = TRUE, draws = 1, sims = 1000, type = "response", times = c(500)) |> 
  purrr::list_flatten() |>
  Reduce(f = cbind, x = _) |> 
  apply(1, mean)
a0
#> [1] 0.829 0.610 0.553 0.862 0.661 0.635

## Average survival death rate based on 1000 different models
a1 <- predict(b_cox, newdata = head(lung), original = FALSE, draws = 1000, type = "response", times = c(500)) |> 
  purrr::list_flatten() |> 
  Reduce(f = cbind, x = _) |> 
  apply(1, mean)

a1
#> [1] 0.705 0.579 0.573 0.758 0.603 0.610

## Average survival rate based on 100 simlutions for each of the 1000 models
a2 <- predict(b_cox, newdata = head(lung), original = FALSE, draws = 1000, sims = 100, type = "response", times = c(500))

### Average death per model
a3 <- lapply(a2, function(x) apply(Reduce(cbind, x), 1, mean))

## Median death rate across 1000 models and 100 simulations for each model
Reduce(rbind, a3) |> 
  apply(2, median)
#> [1] 0.810 0.625 0.570 0.870 0.645 0.660

This structure has also been implemented for riskRegression::CSC and flexsurv::flexsurvreg objects and numerous others on the works and the plan is to also incorporate rstan objects, and an overall predict.burgle_default method and which will only a mean and covariance matrix as inputs.