pbiecek / breakdown Goto Github PK

For your xgBoost example found here: https://pbiecek.github.io/breakDown/articles/break_xgboost.html
an error is raised when executing:
explain_xgb <- broken(HR_xgb_model, new_observation = nobs, data = model_matrix_train)

The error is:
Error in new_observation[rep(1, nrow(data)), ] : incorrect number of dimensions

Any idea how I might overcome this? I love your package but I need it to work for xgBoost. Thanks!

Somehow now the values given by the broken function are always rounded vs exactv (tested lm model):

f.e.
contribution
(Intercept) 2300
tenure = 3 -1900
MonthlyCharges = 74.4 350
SeniorCitizen = 0 14
final_prognosis 720
baseline: 0

while: predict(model, data_in_test[analysed_user,]) gives 721.206

Hi Biecek,

breakDown R Pkg is great
and easy to use.

I followed your Wine quality example lm() model
in:
https://pbiecek.github.io/breakDown/

But my question is
on how to interpret / understand the results,
shown in plot(br) of the WINE example.

Q1) If the var "residual.sugar" is = 1.2 (highest positive value of all vars),
does it mean, it is the most significant var
contributing?.

Q2) Why do some vars have positive values,
(ie: var "residual.sugar" is = 1.2)
and other vars in the plot show negative values?.

Q3) What is the meaning of the plot bar colors?.
Some vars in the plot(br) are yellow, some are green.

and finally,
Q4) What is the meaning of "final_prognosis" = 5.6? (gray color bar).

Just trying to learn and understand what the results mean in practical terms.
Thanks for your help Biecek!

RAY
San Francisco
using latest Rstudio, R and Ubuntu Linux

Pls info syntax for random forest

For models with a larger number of variables plot become unreadable, allowing user to pick top variables (by absolute value of contribution) would be a quick fix

I am having difficulty with interpretation in the situation where the outcome is continuous and the predictors are Boolean.

If I were to think of effects in terms of a linear model I would turn coefficients on or off depending on whether the predictors were a 1 or a 0. Breakdown does not seem to do this.

In the example below I have chosen a point that has 0s assigned to the predictors. Again, in a linear model prediction this would simply result in setting the coefficients of these predictors to 0. With breakdown I am seeing a negative effect for versicolor and setosa.

1. How am I supposed to interpret the output in this situation?

2. Is there a way to show that since these values are 0 for this observation that they are not contributing to the final prediction?

library(reprex)
library(tidyverse)
library(breakDown)

# data prep
iris_dummy <- iris %>% 
  mutate(setosa = ifelse(Species == "setosa",1,0),
         versicolor = ifelse(Species == "versicolor",1,0)) %>% 
  select(-Species)

# fit model
fit <- lm(Sepal.Length ~Sepal.Width + Petal.Length + setosa + versicolor, data = iris_dummy)

set.seed(42)

# pick an observation
no <- iris_dummy[sample(nrow(iris_dummy), 1), ]

# use broken
br <- broken(fit, no)

# the example `no` is not setosa or versicolor yet has breakdown effects
no
#>     Sepal.Length Sepal.Width Petal.Length Petal.Width setosa versicolor
#> 138          6.4         3.1          5.5         1.8      0          0

plot(br)

Created on 2018-10-19 by the reprex package (v0.2.0).

I will fix #18 , let me know if anything else should be fixed

as is presented in the https://arxiv.org/abs/1804.01955

Do we need to add/correct anything before v 1.3 will go to CRAN?

Hi, recently I tried to run the breakdown plots on random forest model (ranger from caret package) trained on sparse data (TFiDF matrix). The model is not doing really good job, but still...

When using DALEX package and after creating explainer without any problems I got for this call:

variable_attribution(rf_explainer,
                               new_observation = x_df_test[ind_to_check, ],
                               type = "break_down")

the following error:

Error in `[.data.frame`(out, , obsLevels, drop = FALSE) : 
  undefined columns selected

Then, I switched to this breakDown package. First of all, after calling it like this:

broken(rf_mod, x_df_test[ind_to_check, ])

It tells me that:

Error in "data.frame" %in% class(data) : 
  argument "data" is missing, with no default

Thus, I changed my call to:

broken(rf_mod, x_df_test[ind_to_check, ], data = x_df_test)

and this time:

Error in yhats[[which.max(yhats_diff)]] : 
  attempt to select less than one element in get1index

The whole code is here: https://github.com/CaRdiffR/tidy_thursdays/blob/master/april_30_2020/predict_gross_clf.R

Strangely, it worked well on exactly same pipeline but with a regression problem.

I use R 4.0.0 and latest version of the DALTEX, breakDown packages.

Might be related to #29 .

Hi,

I was wondering if breakDown can be used to examine the tree votes for non-probability ranger predictions?

I'm getting weird break downs in that the contributions of the predicted value are entirely based on the intercept -- omitting the actual factors which I assumed are used in the training. I'm asking because the examples only deal with probabilities.

                                                          contribution
(Intercept)                                                      6.154
- run = -1.68749498470884                                        0.000
- kernel_time = 393.481201                                       0.000
- total_time = 445.312100278895                                  0.000
- device = xeon_es-2697v2                                        0.000
- invocation = 0                                                 0.000
- branch_entropy_average_linear = 0.0625153                      0.000
- branch_entropy_yokota = -0.164396323864394                     0.000
- ninety_percent_branch_instructions = -0.457287868935426        0.000
- total_unique_branch_instructions = -0.279132672222695          0.000
- local_memory_address_entropy_10 = 4.85307                      0.000
- local_memory_address_entropy_9 = 5.78479                       0.000
- local_memory_address_entropy_8 = 0.286085532384769             0.000
- local_memory_address_entropy_7 = 0.355682140016823             0.000
- local_memory_address_entropy_6 = 0.404053324038373             0.000
- local_memory_address_entropy_5 = 0.372824371077543             0.000
- local_memory_address_entropy_4 = 0.391075370713963             0.000
- local_memory_address_entropy_3 = 0.408194624910907             0.000
- local_memory_address_entropy_2 = 0.430309441064799             0.000
- local_memory_address_entropy_1 = 0.425171485537762             0.000
- global_memory_address_entropy = 0.442245377596217              0.000
- ninety_percent_memory_footprint = -0.189804453106315           0.000
- total_memory_footprint = -0.246916191110329                    0.000
- stddev_simd_width = 0                                          0.000
- mean_simd_width = 1                                            0.000
- max_simd_width = -0.284600989261938                            0.000
- median_instructions_to_barrier = -0.109768646102735            0.000
- max_instructions_to_barrier = 435                              0.000
- min_instructions_to_barrier = 435                              0.000
- total_barriers_hit = 0                                         0.000
- operand_sum = 111360                                           0.000
- workitems = 256                                                0.000
- total_instruction_count = 111360                               0.000
- instructions_per_operand = 1                                   0.000
- barriers_per_instruction = 8.97989e-06                         0.000
- granularity = -0.483677232475196                               0.000
- opcode = -0.714987553238371                                    0.000
- kernel = invert_mapping                                        0.000
- size = tiny                                                    0.000
- application = kmeans                                           0.000
final_prognosis                                                  6.154
baseline:  0

This was generated with the following:

library(breakDown)
library(ranger)
library(ggplot2)

load("./train_dat.Rdf")
load("./test_dat.Rdf")

#build the model
rgd.aiwc <- ranger(log(kernel_time)~.,
                   data = train_dat,
                   num.trees = 505,
                   mtry = 30,
                   min.node.size = 9,
                   importance = "impurity",
                   splitrule = 'variance',
                   respect.unordered.factors = 'order')

#make predictions over all devices we've used for training
predict.function <- function(model, new_observation) predict(model,data=new_observation,type='response')$predictions[1]

#remove the measured time from the data to break down the random forest model
test_data <- subset(test_dat, select = -kernel_time)

explain <- broken(rgd.aiwc,
                  test_dat[1,], #selected element to understand which factors drive its prediction
                  data = test_dat,
                  predict.function = predict.function,
                  direction = "down"
                  )

print(explain)

p <- plot(explain) + ggtitle("Breakdown of AIWC contributions")

pdf("contributions.pdf")
print(p)
dev.off()

Your suggestions would be greatly appreciated!

via number of trees per class
https://www.jstatsoft.org/article/view/v077i01

any new features before the package will be submitted?

Hi, I would like to know if they can be calculated the confidence intervals of the broken.default function?
Thanks

Hi
I am following the example for Random Forests : https://pbiecek.github.io/breakDown/articles/break_randomForest.html
I am still not sure how to translate the output of breakDown.
in the example about random forests we get

what can be said about the final prediction? is it that for this employee there is 88% probability that she left and that is because of each feature contribution to the prediction.

thanks

In broken() function documentation there isn't any information about the direction of the exploration strategy (that information is only in the broken.default()).
Also in DALEX prediction_breakdown() function we don't have direction argument.

Hi

There is a misspelling of the word cumulative -- it appears as "cummulative".

That leads to a ggplot with a typo on one of the axis.

Is it possible to fix it in the next version?

Hi Biecek,

The package is great; but I got a non-sense issue. I have a xgboost binary classification model (objective = "binary:logistic", metric = "auc) with roughly 1000 trees and 200 variables. When I am trying to breakdowm the model for a new observation using "broken", it takes 1.5 hour per each observation!! Is it normal or something is wrong?

Should I specify the prediction function? DMatrix or model.matrix objects could be source of the problem?
In this implementation, can I use DMatrix object without using "model.matrix"? If so how?

Thanks,

Amir

right now the linear predictors are plotted and this does not look good

Hi !

Can I have help to fix the error below?

Error in yhats[[which.min(yhats_diff)]] :
attempt to select less than one element in get1index

I only code this:

contribution <- broken(model = model, new_observation = df[1,features], data = train[,features])

and my model is a randomForest.

It seems that is trying to access to the zero position, and it doesn't exist.

Thanks, Margarida

It's unclear where the wineQuality on https://pbiecek.github.io/breakDown/ originates from. It should either ship with the package or the reference should be clarified.

Hi, I don't understand how the broken function calculates the coefficients? (or something is off?)

In the lm function this is my test result:

summary(model)

Call:
lm(formula = TotalCharges ~ ., data = data_in_test)

Residuals:
Min 1Q Median 3Q Max
-1943.33 -453.71 -94.64 490.26 1887.26

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2162.4583 21.9717 -98.420 < 2e-16 ***
MonthlyCharges 36.1234 0.3080 117.301 < 2e-16 ***
tenure 65.3606 0.3683 177.476 < 2e-16 ***
SeniorCitizen -86.7050 24.3449 -3.562 0.000371 ***

Test user:
-2162.4583 + (data_in_test[analysed_user,]$MonthlyCharges * 36.1234) +
data_in_test[analysed_user,]$tenure65.3606 +
data_in_test[analysed_user,]$SeniorCitizen(-86.7050)

[1] 721.2045

While you get: (u can see that the intercept is different)

lm_br
contribution
(Intercept) 2283.300
tenure = 3 -1923.025
MonthlyCharges = 74.4 346.850
SeniorCitizen = 0 14.081
final_prognosis 721.206
baseline: 0

• strangely the final prognosis is now the same for both lm and broken but broken does not have the same coefficients as the summary(model) when doing calculations

Obviously one would expect that contributions of a waterfall plot would be simply Y=intercept + beta*value ... etc. from the summary output?