ROC curve with Leave-One-Out Cross validation in s

I want to plot a ROC curve of a classifier using leave-one-out cross validation.

It seems that a similar question has been asked here but without any answer.

In another question here is was stated:

In order to obtain a meaningful ROC AUC with LeaveOneOut, you need to calculate probability estimates for each fold (each consisting of just one observation), then calculate the ROC AUC on the set of all these probability estimates.

Additionally, in the official scikit-learn website there is a similar example but using KFold cross validation (https://scikit-learn.org/stable/auto_examples/model_selection/plot_roc_crossval.html#sphx-glr-auto-examples-model-selection-plot-roc-crossval-py).

So for the leave-one-out cross validation case, I am thinking of gathering all the probability prediction on the test sets (one sample at the time) and after having the predicted probabilities for all my folds, to compute and plot the ROC curve.

Does this seems okay? I do not see any other way to achieve my goal.

Here is my code:

from sklearn.svm import SVC
import numpy as np, matplotlib.pyplot as plt,  pandas as pd
from sklearn.model_selection import cross_val_score,cross_val_predict,  KFold,  LeaveOneOut, StratifiedKFold
from sklearn.metrics import roc_curve, auc
from sklearn import datasets

# Import some data to play with
iris = datasets.load_iris()
X_svc = iris.data
y = iris.target
X_svc, y = X_svc[y != 2], y[y != 2]

clf = SVC(kernel='linear', class_weight='balanced', probability=True, random_state=0)
kf = LeaveOneOut()

all_y = []
all_probs=[]
for train, test in kf.split(X_svc, y):
    all_y.append(y[test])
    all_probs.append(clf.fit(X_svc[train], y[train]).predict_proba(X_svc[test])[:,1])
all_y = np.array(all_y)
all_probs = np.array(all_probs)

fpr, tpr, thresholds = roc_curve(all_y,all_probs)
roc_auc = auc(fpr, tpr)
plt.figure(1, figsize=(12,6))
plt.plot(fpr, tpr, lw=2, alpha=0.5, label='LOOCV ROC (AUC = %0.2f)' % (roc_auc))
plt.plot([0, 1], [0, 1], linestyle='--', lw=2, color='k', label='Chance level', alpha=.8)
plt.xlim([-0.05, 1.05])
plt.ylim([-0.05, 1.05])
plt.xlabel('False Positive Rate')
plt.ylabel('True Positive Rate')
plt.title('Receiver operating characteristic example')
plt.legend(loc="lower right")
plt.grid()
plt.show()

I believe the code is correct and the splitting too. I've added a few lines for validation purposes of both the implementation and the results:

from sklearn.model_selection import cross_val_score,cross_val_predict,  KFold,  LeaveOneOut, StratifiedKFold
from sklearn.metrics import roc_curve, auc
from sklearn import datasets

# Import some data to play with
iris = datasets.load_iris()
X_svc = iris.data
y = iris.target
X_svc, y = X_svc[y != 2], y[y != 2]

clf = SVC(kernel='linear', class_weight='balanced', probability=True, random_state=0)
kf = LeaveOneOut()
if kf.get_n_splits(X_svc) == len(X_svc):
    print("They are the same length, splitting correct")
else:
    print("Something is wrong")
all_y = []
all_probs=[]
for train, test in kf.split(X_svc, y):
    all_y.append(y[test])
    all_probs.append(clf.fit(X_svc[train], y[train]).predict_proba(X_svc[test])[:,1])
all_y = np.array(all_y)
all_probs = np.array(all_probs)
#print(all_y) #For validation 
#print(all_probs) #For validation

fpr, tpr, thresholds = roc_curve(all_y,all_probs)
print(fpr, tpr, thresholds) #For validation
roc_auc = auc(fpr, tpr)
plt.figure(1, figsize=(12,6))
plt.plot(fpr, tpr, lw=2, alpha=0.5, label='LOOCV ROC (AUC = %0.2f)' % (roc_auc))
plt.plot([0, 1], [0, 1], linestyle='--', lw=2, color='k', label='Chance level', alpha=.8)
plt.xlim([-0.05, 1.05])
plt.ylim([-0.05, 1.05])
plt.xlabel('False Positive Rate')
plt.ylabel('True Positive Rate')
plt.title('Receiver operating characteristic example')
plt.legend(loc="lower right")
plt.grid()
plt.show()

The If line is meant to only make sure that the splitting is made n times, where n is the number of observations for the given dataset. This is because as the documentation states, LeaveOneOut works the same as Kfold(n_splits=n) and LeaveOneOut(p=1). Also when printing the predicted proba values they were good, making sense of the curve. Congratz on your 1.00AUC!