+ 1

Help last project in DS with python

I need help with last project of DS with python, Pandas Pandas Pandas named. I’m getting incorrect on 3,4,5 tests case.

21st Dec 2020, 5:39 AM
BatOchir Artur
BatOchir Artur - avatar
7 Réponses
+ 2
yes, here's how i did it at last import math import numpy as np n = int(input()) ce1=[0,0] ce2=[2,2] cl1=np.empty((0,2), float) cl2=np.empty((0,2), float) for i in range(n): x=[float(j) for j in input().split()] d21=(np.array(ce1)-np.array(x))**2 d1=math.sqrt(d21.sum()) d2=math.sqrt(((np.array(ce2)-np.array(x))**2).sum()) if d1<=d2: cl1=np.append(cl1,np.array([x]),axis=0) else: cl2=np.append(cl2,np.array([x]),axis=0) if len(cl1) != 0: nce1=np.mean(cl1,axis=0).round(2) print(nce1) else: print (None) if len(cl2) != 0: nce2=np.mean(cl2,axis=0).round(2) print(nce2) else: print(None)
2nd Jun 2021, 1:43 PM
Arash Moradian
Arash Moradian - avatar
+ 1
Finding the next centroid Unsupervised learning algorithm clustering involves updating the centroid of each cluster. Here we find the next centroids for given data points and initial centroids. Task Assume that there are two clusters among the given two-dimensional data points and two random points (0, 0), and (2, 2) are the initial cluster centroids. Calculate the euclidean distance between each data point and each of the centroid, assign each data point to its nearest centroid, then calculate the new centroid. If there's a tie, assign the data point to the cluster with centroid (0, 0). If none of the data points were assigned to the given centroid, return None. Input Format First line: an integer to indicate the number of data points (n) Next n lines: two numeric values per each line to represent a data point in two dimensional space. Output Format Two lists for two centroids. Numbers are rounded to the second decimal place. Sample Input 3 1 0 0 .5 4 0 Sample Output [0.5 0.25] [4. 0.]
5th Jan 2021, 8:06 PM
Arash Moradian
Arash Moradian - avatar
0
above is the full question below is my code i only fail case 3 please help import math import numpy as np n = int(input()) ce1=[0,0] ce2=[2,2] cl1=np.empty((0,2), float) cl2=np.empty((0,2), float) for i in range(n): x=[float(j) for j in input().split()] d21=(np.array(ce1)-np.array(x))**2 d1=math.sqrt(d21.sum()) d2=math.sqrt(((np.array(ce2)-np.array(x))**2).sum()) if d1<=d2: cl1=np.append(cl1,np.array([x]),axis=0) else: cl2=np.append(cl2,np.array([x]),axis=0) nce1=np.mean(cl1,axis=0).round(2) nce2=np.mean(cl2,axis=0).round(2) print(nce1) print(nce2)
5th Jan 2021, 8:07 PM
Arash Moradian
Arash Moradian - avatar
0
Have you completed Arash? How is the test case 3? Do you mind to share it?
2nd Jun 2021, 1:32 PM
CHIN ELAINE
0
Thanks Arash! :)
4th Jun 2021, 9:40 AM
CHIN ELAINE
0
I have also write my code like this and run well import math import numpy as np n = int(input()) ce1=[0,0] ce2=[2,2] cl1=np.empty([0,2], float) cl2=np.empty([0,2], float) for i in range(n): x = [float(j) for j in input().split()] d21 = (np.array(ce1)-np.array(x))**2 d1=math.sqrt(d21.sum()) d2 = math.sqrt(((np.array(ce2)-np.array(x))**2).sum()) if d1<=d2: cl1 = np.append(cl1, np.array([x]), axis=0) else: cl2 = np.append(cl2, np.array([x]), axis=0) if cl1.shape[0]!=0: print(np.mean(cl1,axis=0).round(2)) else: print(None) if cl2.shape[0]!=0: print(np.mean(cl2,axis=0).round(2)) else: print(None)
10th Nov 2021, 5:31 AM
Isack Alex Masunzu
Isack Alex Masunzu - avatar
0
Arash, i was there! What I did was add the None option if one of the cluster is empty :D
16th Nov 2022, 2:48 PM
Jeanfreddy Gutiérrez
Jeanfreddy Gutiérrez - avatar