I have the percentage data of six error types in four different experiments. How can I visualise the distribution of each error type across four conditions? Thank you!
this is how my data look like
I want to have six histograms showing the distribution of each error type in the four conditions on the same panel using facet_grid function and compare the differences and changes in the distribution. I tried the following but did not work.
p <- ggplot(data = error_percentage_sep, aes(x = factor(error_type))) +
geom_histogram(binwidth = 100)
p + facet_wrap(~color)
p
error_percentage_sep %>% #this is the dataset I have
mutate(error_percentage_sep = factor(error_type)) %>%
group_by(error_type) %>%
summarise(mean_percent = rount(mean(percentage), 2))
ggplot(error_percentage_sep, aes(x=error_type, y=mean_percent)) +
geom_bar(stat = "identity")
Related
I am trying to visualise a friedman's test followed by pairwise comparisons using a boxplot with p-values.
Here is an example of how it should look like:
[example graph downloaded from the internet][1]
However, since there are way too many significant comparisons in my case, my graph currently looks like this:
[my graph][2]
[1]: https://i.stack.imgur.com/DO6Vz.png
[2]: https://i.stack.imgur.com/94OXK.png
Here is the code I used to generate the graph with p-value
pwc_IFX_plot <- pwc_IFX %>% add_xy_position(x = "Variant")
ggboxplot(IFX_variant, x = "Variant", y = "Concentration", add = "point") +
stat_pvalue_manual(pwc_IFX_plot, hide.ns = TRUE)+
labs(
subtitle = get_test_label(res.fried_IFX, detailed = TRUE),
caption = get_pwc_label(pwc_IFX)
)+scale_y_log10(breaks = trans_breaks("log10", function(x) 10^x),
labels = trans_format("log10", math_format(10^.x)))
I hope to only show the comparisons of each group to my control group, rather than all the intergroup comparisons.
Thank you for your time.
Any suggestions would be highly appreciated!
I am working with a subset of the 'Ames Housing' dataset and have originally 17 features. Using the 'recipes' package, I have engineered the original set of features and created dummy variables for nominal predictors with the following code. That has resulted in 35 features in the 'baked_train' dataset below.
blueprint <- recipe(Sale_Price ~ ., data = _train) %>%
step_nzv(Street, Utilities, Pool_Area, Screen_Porch, Misc_Val) %>%
step_impute_knn(Gr_Liv_Area) %>%
step_integer(Overall_Qual) %>%
step_normalize(all_numeric_predictors()) %>%
step_other(Neighborhood, threshold = 0.01, other = "other") %>%
step_dummy(all_nominal_predictors(), one_hot = FALSE)
prepare <- prep(blueprint, data = ames_train)
baked_train <- bake(prepare, new_data = ames_train)
baked_test <- bake(prepare, new_data = ames_test)
Now, I am trying to train random forests with the 'ranger' package using the following code.
cv_specs <- trainControl(method = "repeatedcv", number = 5, repeats = 5)
param_grid_rf <- expand.grid(mtry = seq(1, 35, 1),
splitrule = "variance",
min.node.size = 2)
rf_cv <- train(blueprint,
data = ames_train,
method = "ranger",
trControl = cv_specs,
tuneGrid = param_grid_rf,
metric = "RMSE")
I have set the grid of 'mtry' values based on the number of features in the 'baked_train' data. It is my understanding that 'caret' will apply the blueprint within each resample of 'ames_train' creating a baked version at each CV step.
The text Hands-On Machine Learning with R by Boehmke & Greenwell says on section 3.8.3,
Consequently, the goal is to develop our blueprint, then within each resample iteration we want to apply prep() and bake() to our resample training and validation data. Luckily, the caret package simplifies this process. We only need to specify the blueprint and caret will automatically prepare and bake within each resample.
However, when I run the code above I get an error,
mtry can not be larger than number of variables in data. Ranger will EXIT now.
I get the same error when I specify 'tuneLength = 20' instead of the 'tuneGrid'. Although the code works fine when the grid of 'mtry' values is specified to be from 1 to 17 (the number of features in the original training data 'ames_train').
When I specify a grid of 'mtry' values from 1 to 17, info about the final model after CV is shown below. Notice that it mentions Number of independent variables: 35 which corresponds to the 'baked_train' data, although specifying a grid from 1 to 35 throws an error.
Type: Regression
Number of trees: 500
Sample size: 618
Number of independent variables: 35
Mtry: 15
Target node size: 2
Variable importance mode: impurity
Splitrule: variance
OOB prediction error (MSE): 995351989
R squared (OOB): 0.8412147
What am I missing here? Specifically, why do I have to specify the number of features in 'ames_train' instead of 'baked_train' when essentially 'caret' is supposed to create a baked version before fitting and evaluating the model for each resample?
Thanks.
I am unable to reproduce the only example I can find of using h2o with iml (https://www.r-bloggers.com/2018/08/iml-and-h2o-machine-learning-model-interpretability-and-feature-explanation/) as detailed here (Error when extracting variable importance with FeatureImp$new and H2O). Can anyone point to a workaround or other examples of using iml with h2o?
Reproducible example:
library(rsample) # data splitting
library(ggplot2) # allows extension of visualizations
library(dplyr) # basic data transformation
library(h2o) # machine learning modeling
library(iml) # ML interprtation
library(modeldata) #attrition data
# initialize h2o session
h2o.no_progress()
h2o.init()
# classification data
data("attrition", package = "modeldata")
df <- rsample::attrition %>%
mutate_if(is.ordered, factor, ordered = FALSE) %>%
mutate(Attrition = recode(Attrition, "Yes" = "1", "No" = "0") %>% factor(levels = c("1", "0")))
# convert to h2o object
df.h2o <- as.h2o(df)
# create train, validation, and test splits
set.seed(123)
splits <- h2o.splitFrame(df.h2o, ratios = c(.7, .15), destination_frames =
c("train","valid","test"))
names(splits) <- c("train","valid","test")
# variable names for resonse & features
y <- "Attrition"
x <- setdiff(names(df), y)
# elastic net model
glm <- h2o.glm(
x = x,
y = y,
training_frame = splits$train,
validation_frame = splits$valid,
family = "binomial",
seed = 123
)
# 1. create a data frame with just the features
features <- as.data.frame(splits$valid) %>% select(-Attrition)
# 2. Create a vector with the actual responses
response <- as.numeric(as.vector(splits$valid$Attrition))
# 3. Create custom predict function that returns the predicted values as a
# vector (probability of purchasing in our example)
pred <- function(model, newdata) {
results <- as.data.frame(h2o.predict(model, as.h2o(newdata)))
return(results[[3L]])
}
# create predictor object to pass to explainer functions
predictor.glm <- Predictor$new(
model = glm,
data = features,
y = response,
predict.fun = pred,
class = "classification"
)
imp.glm <- FeatureImp$new(predictor.glm, loss = "mse")
Error obtained:
Error in `[.data.frame`(prediction, , self$class, drop = FALSE): undefined columns
selected
traceback()
1. FeatureImp$new(predictor.glm, loss = "mse")
2. .subset2(public_bind_env, "initialize")(...)
3. private$run.prediction(private$sampler$X)
4. self$predictor$predict(data.frame(dataDesign))
5. prediction[, self$class, drop = FALSE]
6. `[.data.frame`(prediction, , self$class, drop = FALSE)
7. stop("undefined columns selected")
In the iml package documentation, it says that the class argument is "The class column to be returned.". When you set class = "classification", it's looking for a column called "classification" which is not found. At least on GitHub, it looks like the iml package has gone through a fair amount of development since that blog post, so I imagine some functionality may not be backwards compatible anymore.
After reading through the package documentation, I think you might want to try something like:
predictor.glm <- Predictor$new(
model = glm,
data = features,
y = "Attrition",
predict.function = pred,
type = "prob"
)
# check ability to predict first
check <- predictor.glm$predict(features)
print(check)
Even better might be to leverage H2O's extensive functionality around machine learning interpretability.
h2o.varimp(glm) will give the user the variable importance for each feature
h2o.varimp_plot(glm, 10) will render a graphic showing the relative importance of each feature.
h2o.explain(glm, as.h2o(features)) is a wrapper for the explainability interface and will by default provide the confusion matrix (in this case) as well as variable importance, and partial dependency plots for each feature.
For certain algorithms (e.g., tree-based methods), h2o.shap_explain_row_plot() and h2o.shap_summary_plot() will provide the shap contributions.
The h2o-3 docs might be useful here to explore more
I am trying to determine lagged predictors to include in my time series model. So I fitted a TSLM with up to lag 3 of the independent variable
lag_models <- data_train %>% model(
ts_lag_0 = TSLM(Y ~ X)
, ts_lag_1 = TSLM(Y ~ X + lag_X_01)
, ts_lag_2 = TSLM(Y ~ X + lag_X_01 + lag_X_02)
, ts_lag_3 = TSLM(Y ~ X + lag_X_01 + lag_X_02 + lag_X_03)
)
data_train contains cross-validation data.
lag_models %>% glance()
Running the code above, I get AIC, AICc, BIC, etc. by lagged predictor model by .id. I am wondering if it's possible to pull out these metrics by model by only the model without using group_by() and summarize().
Thanks very much.
When using cross validation, you are estimating a model on every fold/slice of the data.
As a result, you will receive set of summary statistics (AIC, AICc, BIC, etc.) for every estimated model. If you were to combine them using group_by() and summarise(), you would be combining summary information from models with different response data - this isn't recommended as information criterion are not comparable when the response data varies.
If you wanted to compare the performance of each of the models using cross-validation, you can use out-of-sample accuracy measures using accuracy(). Examples of using fable for cross-validated accuracy evaluation can be found at https://otexts.com/fpp3/tscv.html
for my thesis I have to calculate the number of workers at risk of substitution by machines. I have calculated the probability of substitution (X) and the number of employee at risk (Y) for each occupation category. I have a dataset like this:
X Y
1 0.1300 0
2 0.1000 0
3 0.0841 1513
4 0.0221 287
5 0.1175 3641
....
700 0.9875 4000
I tried to plot a histogram with this command:
hist(dataset1$X,dataset1$Y,xlim=c(0,1),ylim=c(0,30000),breaks=100,main="Distribution",xlab="Probability",ylab="Number of employee")
But I get this error:
In if (freq) x$counts else x$density
length > 1 and only the first element will be used
Can someone tell me what is the problem and write me the right command?
Thank you!
It is worth pointing out that the message displayed is a Warning message, and should not prevent the results being plotted. However, it does indicate there are some issues with the data.
Without the full dataset, it is not 100% obvious what may be the problem. I believe it is caused by the data not being in the correct format, with two potential issues. Firstly, some values have a value of 0, and these won't be plotted on the histogram. Secondly, the observations appear to be inconsistently spaced.
Histograms are best built from one of two datasets:
A dataframe which has been aggregated grouped into consistently sized bins.
A list of values X which in the data
I prefer the second technique. As originally shown here The expandRows() function in the package splitstackshape can be used to repeat the number of rows in the dataframe by the number of observations:
set.seed(123)
dataset1 <- data.frame(X = runif(900, 0, 1), Y = runif(900, 0, 1000))
library(splitstackshape)
dataset2 <- expandRows(dataset1, "Y")
hist(dataset2$X, xlim=c(0,1))
dataset1$bins <- cut(dataset1$X, breaks = seq(0,1,0.01), labels = FALSE)