KNN Python implementation - python-3.x

Question

this is what shows when i try running my code:
FutureWarning: Unlike other reduction functions (e.g. skew, kurtosis), the default behavior of mode typically preserves the axis it acts along. In SciPy 1.11.0, this behavior will change: the default value of keepdims will become False, the axis over which the statistic is taken will be eliminated, and the value None will no longer be accepted. Set keepdims to True or False to avoid this warning.
lab = mode(labels)
This is my Python code, and i find some difficulties trying find a suited solution:
# Importing the required modules
import numpy as np
from scipy.stats import mode
# Euclidean Distance
def eucledian(p1, p2):
dist = np.sqrt(np.sum((p1 - p2) ** 2))
return dist
# Function to calculate KNN
def predict(x_train, y, x_input, k):
op_labels = []
# Loop through the Datapoints to be classified
for item in x_input:
# Array to store distances
point_dist = []
# Loop through each training Data
for j in range(len(x_train)):
distances = eucledian(np.array(x_train[j, :]), item)
# Calculating the distance
point_dist.append(distances)
point_dist = np.array(point_dist)
# Sorting the array while preserving the index
# Keeping the first K datapoints
dist = np.argsort(point_dist)[:k]
# Labels of the K datapoints from above
labels = y[dist]
** # Majority voting
lab = mode(labels)
lab = lab.mode[0]
op_labels.append(lab)**
return op_labels
# Importing the required modules
# Importing required modules
from sklearn.metrics import accuracy_score
from sklearn.datasets import load_iris
from numpy.random import randint
# Loading the Data
iris= load_iris()
# Store features matrix in X
X= iris.data
# Store target vector in
y = iris.target
# Creating the training Data
train_idx = xxx = randint(0, 150, 100)
X_train = X[train_idx]
y_train = y[train_idx]
# Creating the testing Data
test_idx = xxx = randint(0, 150, 50) # taking 50 random samples
X_test = X[test_idx]
y_test = y[test_idx]
# Applying our function
y_pred = predict(X_train, y_train, X_test, 7)
# Checking the accuracy
accuracy_score(y_test, y_pred)
I am expecting a prediction/accuracy to be the prompt.

Answer 1

KNN can be done like this.
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
url = "https://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data"
# Assign colum names to the dataset
names = ['sepal-length', 'sepal-width', 'petal-length', 'petal-width', 'Class']
# Read dataset to pandas dataframe
dataset = pd.read_csv(url, names=names)
dataset.head()
X = dataset.iloc[:, :-1].values
y = dataset.iloc[:, 4].values
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.20)
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
scaler.fit(X_train)
X_train = scaler.transform(X_train)
X_test = scaler.transform(X_test)
from sklearn.neighbors import KNeighborsClassifier
classifier = KNeighborsClassifier(n_neighbors=5, metric='minkowski')
classifier.fit(X_train, y_train)
y_pred = classifier.predict(X_test)
from sklearn.metrics import classification_report, confusion_matrix
print(confusion_matrix(y_test, y_pred))
print(classification_report(y_test, y_pred))
# Result:
precision recall f1-score support
Iris-setosa 1.00 1.00 1.00 13
Iris-versicolor 1.00 0.89 0.94 9
Iris-virginica 0.89 1.00 0.94 8
accuracy 0.97 30
macro avg 0.96 0.96 0.96 30
weighted avg 0.97 0.97 0.97 30
error = []
# Calculating error for K values between 1 and 40
for i in range(1, 40):
knn = KNeighborsClassifier(n_neighbors=i)
knn.fit(X_train, y_train)
pred_i = knn.predict(X_test)
error.append(np.mean(pred_i != y_test))
plt.figure(figsize=(12, 6))
plt.plot(range(1, 40), error, color='red', linestyle='dashed', marker='o',
markerfacecolor='blue', markersize=10)
plt.title('Error Rate K Value')
plt.xlabel('K Value')
plt.ylabel('Mean Error')