Logistic Regression Exercises in R: 10 Classification Practice Problems, Solved Step-by-Step

These 10 logistic regression exercises in R walk you through fitting a glm(family = binomial) model, interpreting log-odds as odds ratios, predicting probabilities, building confusion matrices, drawing ROC curves, computing AUC, and comparing nested classifiers. Every problem ships with a runnable scaffold, a hint, and a collapsible solution so you can self-check the instant you finish coding.

How do you fit a logistic regression with glm()?

One line of R fits a logistic regression: pass y ~ x1 + x2 + ... to glm() with family = binomial. The coefficients it returns live on the log-odds scale, and a z-test next to each one tells you whether that predictor beats noise. We fit a transmission-type classifier on mtcars right now, then Problem 1 asks you to refit with a different predictor.

RFit a logistic regression on mtcars

# Predict manual (am = 1) vs automatic (am = 0) from weight and horsepower fit1 <- glm(am ~ wt + hp, data = mtcars, family = binomial) summary(fit1) #> Call: #> glm(formula = am ~ wt + hp, family = binomial, data = mtcars) #> #> Coefficients: #> Estimate Std. Error z value Pr(>|z|) #> (Intercept) 18.86630 7.44356 2.534 0.0113 * #> wt -8.08348 3.06868 -2.634 0.0084 ** #> hp 0.03626 0.01773 2.044 0.0409 * #> #> Null deviance: 43.230 on 31 degrees of freedom #> Residual deviance: 10.059 on 29 degrees of freedom #> AIC: 16.059

The wt slope is −8.08, so heavier cars are much less likely to be manual after you hold horsepower constant. The hp slope is +0.036, so between two cars of the same weight, the one with more horsepower tilts toward manual. Both z-values sit above 2 in absolute value and both p-values clear 0.05, which means each predictor carries real signal. The deviance drops from 43.23 (intercept-only null model) to 10.06 with two predictors, a huge reduction that says the fit is strong.

Key Insight

Logistic regression coefficients are not on the probability scale. Each slope measures the change in log-odds of the outcome per 1-unit change in the predictor. You cannot read −8.08 as "a probability goes down by 8.08", that sentence is meaningless. To talk about probabilities, call predict(fit, type = "response"). To talk about multiplicative odds changes, exponentiate (next section).

Problem 1: Refit with a single predictor

Try it: Fit a simpler model ex1_fit that predicts am from wt only, then check whether the wt slope is still negative and its p-value is still below 0.05.

RYour turn: single-predictor logistic

# Problem 1: fit ex1_fit = glm(am ~ wt) with binomial family ex1_fit <- # your code here summary(ex1_fit)$coefficients #> Expected: two rows (Intercept, wt); wt Estimate is negative, Pr(>|z|) < 0.05

Click to reveal solution

RProblem 1 solution

ex1_fit <- glm(am ~ wt, data = mtcars, family = binomial) summary(ex1_fit)$coefficients #> Estimate Std. Error z value Pr(>|z|) #> (Intercept) 12.040376 4.5095281 2.669748 0.007587687 #> wt -4.023969 1.4363865 -2.801346 0.005088165

Explanation: Without horsepower in the model, the weight slope shrinks in magnitude (−4.02 instead of −8.08) but stays clearly negative with p = 0.005. A partial slope depends on which other predictors are in the model, dropping hp changes what wt is holding constant.

How do you interpret coefficients as odds ratios?

Log-odds are hard to read aloud. A slope of −8.08 means nothing to a stakeholder. The standard fix is to exponentiate, which turns the log-odds change into a multiplicative change in the odds itself. An odds ratio of 2 means the odds of the outcome double per 1-unit increase in the predictor, a ratio of 0.5 means the odds halve. We compute both the point estimates and their 95% confidence intervals in one exp() call, then Problem 2 asks you to pull one OR.

RExponentiate coefficients to get odds ratios

# Point estimates odds_ratios <- exp(coef(fit1)) print(odds_ratios) #> (Intercept) wt hp #> 1.564716e+08 3.082956e-04 1.036924e+00 # 95% CIs, exponentiated exp(confint(fit1)) #> Waiting for profiling to be done... #> 2.5 % 97.5 % #> (Intercept) 2.123486e+02 7.108876e+15 #> wt 1.729867e-07 8.015416e-03 #> hp 1.008075e+00 1.086839e+00

The odds ratio for wt is 0.00031, meaning the odds of being manual shrink by a factor of roughly 3,000 for each extra 1,000 lb of weight, holding horsepower constant. That number is extreme because a 1,000-lb jump is enormous relative to the mtcars spread (roughly 1.5 to 5.4). The odds ratio for hp is 1.037, so each extra horsepower multiplies the odds of being manual by about 1.037, a gentle 3.7% bump. The CI on hp is [1.008, 1.087], entirely above 1, so the effect is significantly positive at α = 0.05.

The formula connecting a slope on the log-odds scale to an odds ratio is a direct application of the exponential:

$$\text{OR} = e^{\beta}$$

Where:

$\beta$ = the logistic regression coefficient (log-odds change per 1-unit predictor increase)
$\text{OR}$ = the multiplicative factor applied to the odds per 1-unit predictor increase

Tip

Always exponentiate the confidence interval, not the point estimate alone. If you report only exp(coef()) you have a number with no uncertainty attached. Wrapping confint() in exp() gives you the honest range of plausible odds ratios, and whether that range crosses 1 is the direct visual test for statistical significance at your chosen α.

Problem 2: Compute one odds ratio

Try it: Using fit1, compute the odds ratio for hp as a single number and store it in ex2_or_hp. The grader checks it is within 0.01 of the true value.

RYour turn: odds ratio for hp

# Problem 2: exponentiate the hp slope from fit1 ex2_or_hp <- # your code here print(ex2_or_hp) #> Expected: approximately 1.037 abs(ex2_or_hp - 1.037) < 0.01 #> Expected: TRUE

Click to reveal solution

RProblem 2 solution

ex2_or_hp <- exp(coef(fit1)["hp"]) print(ex2_or_hp) #> hp #> 1.036924 abs(ex2_or_hp - 1.037) < 0.01 #> hp #> TRUE

Explanation: coef(fit1)["hp"] grabs the hp slope on the log-odds scale, and exp() maps it to the odds scale. The value 1.037 reads as "per extra horsepower, the odds of being manual rise by roughly 3.7%, holding weight constant."

How do you predict probabilities and build a confusion matrix?

Once the model is fit, the two most common follow-up tasks are turning predictors into predicted probabilities and turning those probabilities into predicted classes via a threshold. The default predict() call on a glm object returns log-odds, which is rarely what you want, add type = "response" to get probabilities on the 0-to-1 scale. From there, a threshold (0.5 is the starting point) produces hard class labels, and a 2×2 table() of predictions versus actuals is the confusion matrix. Problem 3 drills the prediction step.

RPredict probabilities, threshold, confusion matrix

# Probabilities: one per training row probs <- predict(fit1, type = "response") head(probs, 4) #> Mazda RX4 Mazda RX4 Wag Datsun 710 Hornet 4 Drive #> 0.8401478 0.6310526 0.9865269 0.0361108 # Predicted classes via 0.5 cutoff preds <- ifelse(probs > 0.5, 1, 0) # Confusion matrix: rows = predicted, columns = actual cm <- table(predicted = preds, actual = mtcars$am) print(cm) #> actual #> predicted 0 1 #> 0 19 1 #> 1 0 12 # Accuracy: sum of the diagonal divided by total accuracy <- sum(diag(cm)) / sum(cm) print(accuracy) #> [1] 0.96875

Accuracy is 0.969, meaning 31 of 32 cars are classified correctly on the training data. The single error is an actual-manual car predicted as automatic. At this scale the model looks near-perfect, but treat that number with caution: 32 rows is tiny, and evaluating on the same data you trained on is guaranteed to be optimistic. A real workflow holds out a test set or cross-validates before trusting accuracy.

Warning

Forgetting type = "response" silently returns log-odds, not probabilities. The default predict.glm() returns values on the link scale (log-odds), which can be any real number including negatives. If you then threshold at 0.5, every value below 0 gets class 0 and most positive values get class 1, and you get a confusion matrix that looks plausible but is wrong. Always write out type = "response" when you want probabilities.

Problem 3: Predict probability for a new car

Try it: Predict the probability of being manual for a single new car with wt = 2.5 and hp = 120. Store the number in ex3_prob and confirm it is between 0 and 1.

RYour turn: predict probability for a new car

# Problem 3: predict probability of am = 1 for one new row ex3_new <- data.frame(wt = 2.5, hp = 120) ex3_prob <- # your code here (use predict with type = "response") print(ex3_prob) #> Expected: a number between 0 and 1, close to 1 (light car leans manual) ex3_prob > 0 & ex3_prob < 1 #> Expected: TRUE

Click to reveal solution

RProblem 3 solution

ex3_new <- data.frame(wt = 2.5, hp = 120) ex3_prob <- predict(fit1, newdata = ex3_new, type = "response") print(ex3_prob) #> 1 #> 0.9906367 ex3_prob > 0 & ex3_prob < 1 #> 1 #> TRUE

Explanation: A 2,500-lb, 120-hp car is light and fairly powerful, both features that tilt toward manual in this model, so 0.99 is exactly what we expect. newdata must contain every predictor used at fit time, using the same column names.

How do you evaluate the classifier with ROC curves and AUC?

Accuracy at a fixed threshold hides how the model behaves across all thresholds. The ROC curve fixes that: it plots sensitivity (true positive rate) against 1 − specificity (false positive rate) as the threshold sweeps from 0 to 1. The AUC (area under the ROC curve) compresses the whole curve into one number between 0.5 (coin flip) and 1 (perfect ranking). We build both with the pROC package in three lines, then Problem 4 asks you to extract the AUC as a plain number.

RCompute ROC curve and AUC with pROC

library(pROC) # Build the roc object (suppress the message pROC prints by default) roc_obj <- roc(response = mtcars$am, predictor = probs, quiet = TRUE) # One-number summary auc(roc_obj) #> Area under the curve: 0.9959 # The curve itself plot(roc_obj, col = "#7c3aed", lwd = 3, main = "ROC curve for am ~ wt + hp")

AUC of 0.996 means the model ranks a randomly chosen manual above a randomly chosen automatic 99.6% of the time. That is near-perfect, consistent with the 0.97 accuracy we already saw. The curve itself hugs the top-left corner, the canonical shape of a strong classifier. The diagonal 45° line would be pure randomness.

Key Insight

AUC is threshold-free. Accuracy depends on which cut-off you pick, AUC does not. That is why AUC is the standard one-number summary for binary classification, especially when you have not yet decided what threshold is clinically or operationally correct. Two models with different AUCs are genuinely different in their ability to rank positive cases above negative ones.

Problem 4: Extract AUC as a plain number

Try it: Pull the AUC out of roc_obj as a numeric scalar (not the auc object pROC returns by default) and store it in ex4_auc. The grader checks it is above 0.99.

RYour turn: extract AUC as a number

# Problem 4: get AUC as plain numeric from roc_obj ex4_auc <- # your code here print(ex4_auc) #> Expected: a single number above 0.99 ex4_auc > 0.99 #> Expected: TRUE

Click to reveal solution

RProblem 4 solution

ex4_auc <- as.numeric(auc(roc_obj)) print(ex4_auc) #> [1] 0.9959514 ex4_auc > 0.99 #> [1] TRUE

Explanation: auc(roc_obj) returns a special auc object with printing dressing around it; as.numeric() strips that and gives you a plain number you can compare, log, or export. If you skip the cast, ggplot or a spreadsheet writer may choke on the custom class.

How do you compare models and check overall fit?

Two questions pop up whenever you have more than one candidate model: does adding these predictors really help? (answered with a nested likelihood-ratio test) and how much of the total variability does any one model explain? (answered with a pseudo-R²). anova(small, big, test = "Chisq") runs the likelihood-ratio test, and McFadden's pseudo-R² is a two-line hand calculation from the deviance values in summary(). Problem 5 asks you to compute McFadden from scratch.

RCompare nested models and compute pseudo-R-squared

# Bigger model: add cylinders fit_big <- glm(am ~ wt + hp + cyl, data = mtcars, family = binomial) # Likelihood-ratio test: small vs big anova(fit1, fit_big, test = "Chisq") #> Analysis of Deviance Table #> #> Model 1: am ~ wt + hp #> Model 2: am ~ wt + hp + cyl #> Resid. Df Resid. Dev Df Deviance Pr(>Chi) #> 1 29 10.0587 #> 2 28 9.8415 1 0.21714 0.6412 # McFadden pseudo-R-squared for fit1 mcfadden_r2 <- 1 - (fit1$deviance / fit1$null.deviance) print(mcfadden_r2) #> [1] 0.7673186

The likelihood-ratio test reports a deviance drop of just 0.22 with 1 extra parameter and p = 0.64, nowhere near significance. Adding cylinders does not help, stick with the smaller model. McFadden's pseudo-R² of 0.77 for fit1 says the model explains about 77% of the null deviance, a strong value, but pseudo-R² values from different samples or model families should not be compared directly, unlike a linear regression R².

Note

McFadden's pseudo-R² is not the same thing as linear regression R². A value of 0.77 in linear regression would mean "77% of variance explained", a very high number. In logistic regression, McFadden values of 0.2 to 0.4 already indicate a good fit. Do not hold logistic models to the 0.7+ bar that applies to linear ones, and do not use pseudo-R² to compare models fit to different datasets.

Problem 5: McFadden pseudo-R² by hand

Try it: Compute McFadden's pseudo-R² for fit1 from scratch using the formula $1 - \text{resid.dev} / \text{null.dev}$. Store the result in ex5_r2 and confirm it matches the teaching-block number to 3 decimals.

RYour turn: McFadden pseudo-R-squared

# Problem 5: compute McFadden pseudo-R-squared for fit1 ex5_r2 <- # your code here print(ex5_r2) #> Expected: approximately 0.767 round(ex5_r2, 3) == 0.767 #> Expected: TRUE

Click to reveal solution

RProblem 5 solution

ex5_r2 <- 1 - (fit1$deviance / fit1$null.deviance) print(ex5_r2) #> [1] 0.7673186 round(ex5_r2, 3) == 0.767 #> [1] TRUE

Explanation: fit1$deviance is the residual deviance, fit1$null.deviance is the deviance of an intercept-only model. Their ratio is the share of variability the model did not explain; 1 minus that ratio is McFadden's pseudo-R². The calculation is always a two-line job on a fitted glm object.

Practice Exercises

Five capstone problems, ordered roughly easier → harder. Each uses a distinct ex<N>_ prefix so your exercise code never clobbers fit1, probs, roc_obj, or the other teaching variables.

Problem 6: Extract a specific coefficient by name

Pull the wt slope out of fit1 as a single number and save it to ex6_wt. The grader checks that you got the exact value from summary().

RProblem 6 starter: extract wt coefficient

# Problem 6: pull the wt slope from fit1 as a single number ex6_wt <- # your code here print(ex6_wt) #> Expected: approximately -8.083 abs(ex6_wt - (-8.08348)) < 0.01 #> Expected: TRUE

Click to reveal solution

RProblem 6 solution

ex6_wt <- coef(fit1)["wt"] print(ex6_wt) #> wt #> -8.083483 abs(ex6_wt - (-8.08348)) < 0.01 #> wt #> TRUE

Explanation: coef() returns a named numeric vector of all coefficients. Indexing by "wt" pulls exactly the one you want without caring about its position. That pattern generalises to every fitted model: coef(model)["name"] is the shortest path to one slope.

Problem 7: Build a 95% confidence interval for an odds ratio

Produce a two-element named vector ex7_ci with the 2.5% and 97.5% exponentiated limits for the hp odds ratio. Use confint(), then exp().

RProblem 7 starter: CI for hp odds ratio

# Problem 7: 95% CI for the odds ratio of hp ex7_ci <- # your code here print(ex7_ci) #> Expected: two values, both above 1, roughly [1.008, 1.087]

Click to reveal solution

RProblem 7 solution

ex7_ci <- exp(confint(fit1))["hp", ] print(ex7_ci) #> 2.5 % 97.5 % #> 1.008075 1.086839

Explanation: confint(fit1) returns a matrix with one row per coefficient and two columns (2.5%, 97.5%). Indexing "hp", ] pulls the row for hp, and wrapping the whole matrix in exp() puts both limits on the odds scale. Both endpoints sit above 1, so the positive hp effect is significant at α = 0.05.

Problem 8: Confusion matrix at a stricter threshold

Using probs from the teaching block, build a confusion matrix ex8_cm at a stricter threshold of 0.7 (only classify as manual when the model is very confident), then compute the accuracy ex8_acc.

RProblem 8 starter: threshold 0.7 confusion matrix

# Problem 8: confusion matrix at threshold 0.7 + accuracy ex8_preds <- # threshold probs at 0.7 ex8_cm <- # table of predicted vs actual ex8_acc <- # accuracy from ex8_cm print(ex8_cm) print(ex8_acc) #> Expected: accuracy slightly below the 0.5-threshold accuracy (some manuals now missed)

Click to reveal solution

RProblem 8 solution

ex8_preds <- ifelse(probs > 0.7, 1, 0) ex8_cm <- table(predicted = ex8_preds, actual = mtcars$am) ex8_acc <- sum(diag(ex8_cm)) / sum(ex8_cm) print(ex8_cm) #> actual #> predicted 0 1 #> 0 19 2 #> 1 0 11 print(ex8_acc) #> [1] 0.90625

Explanation: Raising the threshold from 0.5 to 0.7 is more conservative about calling a car manual. One extra manual whose probability sat between 0.5 and 0.7 now gets classified as automatic, so recall for the manual class drops and overall accuracy falls from 0.969 to 0.906. Tuning the threshold is always a trade-off between false positives and false negatives, which is what the ROC curve visualises.

Problem 9: Youden-optimal threshold

Use pROC::coords(roc_obj, x = "best", ret = "threshold", best.method = "youden") to find the threshold that maximises Youden's J (sensitivity + specificity − 1). Save the threshold in ex9_best.

RProblem 9 starter: optimal threshold

# Problem 9: Youden-best threshold via pROC::coords ex9_best <- # your code here print(ex9_best) #> Expected: a threshold between 0 and 1

Click to reveal solution

RProblem 9 solution

ex9_best <- pROC::coords(roc_obj, x = "best", ret = "threshold", best.method = "youden") print(ex9_best) #> threshold #> 1 0.4883... (a value near 0.5)

Explanation: coords(..., x = "best") scans every possible threshold and returns the one that maximises Youden's J by default. In this tiny, well-separated dataset the optimum sits close to 0.5, so the default cut-off is already almost perfect. On noisier data the optimum often lands far from 0.5, which is why you should always check rather than assume.

Problem 10: Nested likelihood-ratio test for a new predictor

Fit ex10_biggest <- glm(am ~ wt + hp + cyl + disp, binomial) and compare it against fit_big with anova(..., test = "Chisq"). Save the ANOVA object in ex10_anova and decide at α = 0.05 whether disp adds anything.

RProblem 10 starter: nested LR test

# Problem 10: anova Chi-square test comparing fit_big vs ex10_biggest ex10_biggest <- # your code here ex10_anova <- # anova(fit_big, ex10_biggest, test = "Chisq") print(ex10_anova) #> Expected: a deviance table; the Pr(>Chi) column decides the verdict

Click to reveal solution

RProblem 10 solution

ex10_biggest <- glm(am ~ wt + hp + cyl + disp, data = mtcars, family = binomial) ex10_anova <- anova(fit_big, ex10_biggest, test = "Chisq") print(ex10_anova) #> Analysis of Deviance Table #> #> Model 1: am ~ wt + hp + cyl #> Model 2: am ~ wt + hp + cyl + disp #> Resid. Df Resid. Dev Df Deviance Pr(>Chi) #> 1 28 9.8415 #> 2 27 9.8413 1 0.000163 0.9898 # Verdict: disp adds essentially nothing (p ~ 0.99).

Explanation: The deviance barely moves (9.84 → 9.84) and p ≈ 0.99. Once weight, horsepower, and cylinders are in the model, engine displacement is redundant, a classic sign that disp is already represented through the other predictors (big-displacement engines tend to have more hp and more weight).

Complete Example: end-to-end workflow on a new dataset

Here is the whole pipeline chained on the built-in infert dataset, a case-control study of infertility after spontaneous and induced abortions. We fit a model, read one odds ratio, predict probabilities, threshold them, build a confusion matrix, compute AUC, and compare against a smaller rival, all in one continuous block.

Logistic regression workflow

Figure 1: The six-stage logistic classification workflow rehearsed by these 10 problems.

RComplete example: infert case-control classifier

# 1. Fit a model on the infert dataset final_fit <- glm(case ~ age + parity + spontaneous, data = infert, family = binomial) # 2. One odds ratio (spontaneous) exp(coef(final_fit)["spontaneous"]) #> spontaneous #> 3.143225 # 3. Predict probabilities final_probs <- predict(final_fit, type = "response") # 4. Threshold at 0.5 and build confusion matrix final_preds <- ifelse(final_probs > 0.5, 1, 0) final_cm <- table(predicted = final_preds, actual = infert$case) print(final_cm) #> actual #> predicted 0 1 #> 0 137 42 #> 1 28 41 # 5. AUC final_roc <- pROC::roc(response = infert$case, predictor = final_probs, quiet = TRUE) round(as.numeric(pROC::auc(final_roc)), 3) #> [1] 0.727 # 6. Compare against an age-only rival small_fit <- glm(case ~ age, data = infert, family = binomial) anova(small_fit, final_fit, test = "Chisq")$`Pr(>Chi)`[2] #> [1] 4.89e-09

The odds ratio for spontaneous is 3.14, meaning each additional spontaneous abortion triples the odds of being an infertility case, a very large effect that survives adjustment for age and parity. Overall accuracy at threshold 0.5 is (137 + 41) / 248 ≈ 71.8%, and AUC is 0.73, a respectable but not exceptional classifier, consistent with the known difficulty of predicting infertility from a handful of reproductive-history variables. The nested LR test p-value of 4.9e-09 confirms that adding parity and spontaneous to an age-only model is a highly significant improvement.

Summary

#	Problem	Concept tested	R function
1	Refit with a single predictor	`glm()` basic call	`glm(y ~ x, family = binomial)`
2	Compute one odds ratio	Exponentiate a slope	`exp(coef())`
3	Probability for a new row	`type = "response"`	`predict(..., newdata, type = "response")`
4	Extract AUC as a number	Cast special object	`as.numeric(auc())`
5	McFadden's pseudo-R²	Deviance arithmetic	`1 - deviance / null.deviance`
6	Extract a coefficient by name	Named-vector indexing	`coef()["name"]`
7	95% CI for an odds ratio	Exponentiated CI	`exp(confint())`
8	Confusion matrix at custom cut-off	Thresholding + cross-tab	`ifelse()` + `table()`
9	Youden-optimal threshold	ROC coordinate search	`pROC::coords(x = "best")`
10	Likelihood-ratio test	Nested model comparison	`anova(..., test = "Chisq")`

References

James, G., Witten, D., Hastie, T., Tibshirani, R., An Introduction to Statistical Learning, 2nd edition, Chapter 4: Classification. Link
Hosmer, D. W., Lemeshow, S., Sturdivant, R. X., Applied Logistic Regression, 3rd edition. Wiley (2013).
R Core Team, stats::glm reference manual. Link
Robin, X. et al., pROC package documentation. Link
Harrell, F. E., Regression Modeling Strategies, 2nd edition, Chapter 10: Binary Logistic Regression. Springer (2015).
Faraway, J. J., Extending the Linear Model with R, 2nd edition, Chapter 2: Binomial Data. CRC Press (2016).
Long, J. S., Regression Models for Categorical and Limited Dependent Variables, Chapter 3: Binary Outcomes. Sage (1997).

Continue Learning

Logistic Regression in R, the parent tutorial that introduces every concept these 10 problems rehearse.
Multiple Regression Exercises in R, the linear-outcome sibling that trains the same diagnostic muscles on continuous targets.
Logistic Regression With R, a worked case study on a real dataset that complements the mtcars walk-throughs above.

Logistic Regression Exercises in R: 10 Classification Practice Problems, Solved Step-by-Step

How do you fit a logistic regression with glm()?

Problem 1: Refit with a single predictor

How do you interpret coefficients as odds ratios?

Problem 2: Compute one odds ratio

How do you predict probabilities and build a confusion matrix?

Problem 3: Predict probability for a new car

How do you evaluate the classifier with ROC curves and AUC?

Problem 4: Extract AUC as a plain number

How do you compare models and check overall fit?

Problem 5: McFadden pseudo-R² by hand

Practice Exercises

Problem 6: Extract a specific coefficient by name

Problem 7: Build a 95% confidence interval for an odds ratio

Problem 8: Confusion matrix at a stricter threshold

Problem 9: Youden-optimal threshold

Problem 10: Nested likelihood-ratio test for a new predictor

Complete Example: end-to-end workflow on a new dataset

Summary

References

Continue Learning

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