Ordinary Squares project

Question

I’m not understanding this project? It’s in DS with Python 4th module. It’s saying it should multiply by matrix multiplication but it shows me an error when I submit the code?

n, p = [int(x) for x in input().split()]
X = []
for i in range(n):
X.append([float(x) for x in input().split()])

y = [float(x) for x in input().split()]

import numpy as np
print(np.dot(y, X))

Answer 1

n, p = [int(x) for x in input().split()]
X = []
for i in range(n):
X.append([float(x) for x in input().split()])

y = [float(x) for x in input().split()]

import numpy as np

X=np.array(X).reshape(n,p)
y=np.array(y)
b=np.linalg.pinv(X) @ y.transpose()
print(np.around(b,decimals=2))

Answer 2

I dont see how this project is relevant to the module, it’s well beyond the information covered. Does’t make sense at all. I had a couple of challenging ones in the past but this I would have never solved wihtout peeking in the comments.

Answer 3

Robson Marini Instead of doing the calculations explicitly, there's a command to calculate Penrose Inverse directly in numpy. Just use pinv instead of inv.
beta=np.linalg.pinv(X) @ y.T will give answer directly.
Here's mine:

https://code.sololearn.com/cN5LXGeLIYWg/?ref=app

Answer 4

Zhenis Otarbay Robson Marini
binary disorder:
y_true = [int(x) for x in input().split()]
y_pred = [int(x) for x in input().split()]
from sklearn.metrics import confusion_matrix
import numpy as np
s =confusion_matrix(y_true, y_pred)
a=np.flip(s)
s[0][0], s[-1][-1]=s[-1][-1], s[0][0]
s=s.astype(float)
print(s)

Answer 5

The exercise is horribly explained, refers to an image that is not shown, a formula that is not shown and requires instruments not mentioned in the course.

Answer 6

import numpy as np
n, p = [int(x) for x in input().split()]

X = np.empty([n, p])
for i in range(n):
X[i,] = input().split()

y = [float(x) for x in input().split()]

Xarr = np.array(X)
yarr = np.array([y]).T

# beta = inv(Xarr.T * Xarr) * Xarr.T * yarr
# beta = temp1 * Xarr.T * yarr
# beta = temp2 * yarr
# temp1 = inv(Xarr.T * Xarr)
# temp2 = temp1 * Xarr.T
temp1 = np.linalg.inv(np.matmul(Xarr.T, Xarr))
temp2 = np.matmul(temp1, Xarr.T)
beta = np.matmul(temp2, yarr)
print(beta.flatten().round(2))

Answer 7

Bat-Ochir Artur Robson Marini As I had seen this thread earlier, and just completed Binary disorder, I would like to add something.
In Binary disorder the default labeling starts from 0 in confusion matrix, but as 1 is True and 0 is False in python, we could just use them as labels to get the output directly, and we don't have to rotate or flip the matrix.

import numpy as np
from sklearn.metrics import confusion_matrix as cm
y_true = [int(x) for x in input().split()]
y_pred = [int(x) for x in input().split()]
yt=np.array(y_true)
yp=np.array(y_pred)
print(cm(yp,yt,labels=[True,False]).astype(float))

This solves the problem
https://code.sololearn.com/c6f3Gd2zUmaJ/?ref=app

Answer 8

Bat-Ochir Artur Robson Marini could you find out? I am also stack btw

Answer 9

import numpy as np
n, p = [int(x) for x in input().split()]

X = np.empty([n, p])
for i in range(n):
X[i,] = input().split()

y = [float(x) for x in input().split()]

X_arr = np.array(X)
y_arr = np.array([y]).T

temp_1 = np.linalg.inv(np.matmul(X_arr.T, X_arr))
temp_2 = np.matmul(temp_1, X_arr.T)
Abdullah = np.matmul(temp_2, y_arr)
print(Abdullah.flatten().round(2))

Answer 10

You only need to add this line to the code :

print((np.linalg.pinv(X) @ y).round(2))

###############################

Formula:

XB = y ----> B = X^-1 . y

where:
B: beta
^-1: inverse
. : matrix multiplication

Answer 11

help please

Answer 12

Dude. I am trying to solve this one too. My code passes the first test but it doesn't pass the other tests.

The equation that solves beta is:
beta = (X^T . X)^(-1) . X^T . y

where:
^T = transpose
^(-1) = inverse

https://en.wikipedia.org/wiki/Ordinary_least_squares

However, there is something really weird on the second test. "y" has 3 elements, but X is a 2 by 2 matrix. So "y" should have only 2 elements as well. I don't know if there is any hidden trick here, or if it is a sololearn error. Because "y" should have only 2 elements and not 3.

Answer 13

After my complaint, I think Sololearn solved the "Test 2" input error. X was 2x2 and y was 3x1. A clear matching mistake. Now it is correct. And here is my code that passed all tests. I hope it helps.

Answer 14

Robson Marini thanks bro

Answer 15

Robson Marini could you solve binary disorder?

Answer 16

Didn't get there. But I ll try that one soon.

Answer 17

Bat-Ochir Artur thanks bro, now I understand it, did you understand welcome to the matrix in machine learning? I can't pass the last testing case

Answer 18

Didn't do "welcome to the matrix" yet. Bit I just did "Binary Disorder".

import numpy as np

y_true = [int(x) for x in input().split()]
y_pred = [int(x) for x in input().split()]

TP, FP, FN, TN = 0, 0, 0, 0
for i in range(len(y_true)):
if y_true[i] == 1 and y_pred[i] == 1:
TP += 1.0
elif y_true[i] == 1 and y_pred[i] == 0:
FN += 1.0
elif y_true[i] == 0 and y_pred[i] == 0:
TN += 1.0
elif y_true[i] == 0 and y_pred[i] == 1:
FP += 1.0

cm = np.array([[TP, FP],[FN, TN]])
print(cm)

Answer 19

Robson Marini thanks bro plethora of useful tips

Answer 20

Robson Marini also i am stack in Split to Achieve Gain in ML