I am only able to get 78% accuraacy can anyone help to improve it

#1

my code is

def dist(x1, x2):

n = len(x1)

s = 0.0

for i in range(n):

    d = (x1[i] - x2[i])**2

    s+=d

return np.sqrt(s)

def knn(x, y, x_t, k = 11):

vals = []

m = x.shape[0]

for i in range(m):

    d = dist(x[i], x_t)

    vals.append((d, y[i]))

vals.sort()


vals = vals[:k]


vals = np.array(vals)

new_vals = np.unique(vals[:, 1], return_counts = True)

index= new_vals[1].argmax()

pred = new_vals[0][index]

return pred

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np

x = pd.read_csv(‘Diabetes_XTrain.csv’).values

y = pd.read_csv(‘Diabetes_YTrain.csv’).values.reshape((-1,))

m = np.mean(x, axis = 0)

standard= np.std(x, axis = 0)

x = (x-m)/standard
x_t = pd.read_csv(‘Diabetes_Xtest.csv’).values

x_t = (x_t - m)/standard

y_t = []

n = x_t.shape[0]

for i in range(n):

t = knn(x, y, x_t[i])

y_t.append(int(t))

y_t = np.array(y_t)

df = pd.DataFrame(data = y_t, columns = [“Outcome”])

df.to_csv(‘OUTPUT.csv’, index = False)

#2

Hey @yashb4569, congrats you achieved maximum accuracy that can be achieved by knn.

Those who achieved more accuracy, have used different technique. No need to ponder on that, proceed further with the course.

Hope this resolved your doubt.
Plz mark the doubt as resolved in my doubs section :blush: