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# 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.

12/21/2020 5:39:57 AM

Bat-Ochir Artur

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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)

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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.]

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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)

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Have you completed Arash? How is the test case 3? Do you mind to share it?

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Thanks Arash! :)

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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)

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Bat-Ochir Artur Please share the complete question