Numpy.linalg.lstsq() method is “used to calculate the optimal solution for a given set of data points, making it a valuable tool for data analysis and modeling.”
Syntax
Numpy.linalg.lstsq(a, b, rcond=’warn’)Parameters
- a: It depicts a coefficient matrix.
- b: It depicts Ordinate or “dependent variable” values. If the parameter is a two-dimensional matrix, the least square is calculated for each K column of that specific matrix.
- Rcond: It is of float datatype. It is a cut-off ratio for smaller singular values of a. In rank determination, singular values are treated as zero if smaller than rcond times a’s largest singular value.
Return Value
- X: It depicts the least-squares solution. If the input was a two-dimensional matrix, the solution is always in the K columns of x.
- Residuals: It depicts the Sum of residuals or a squared Euclidean 2-norm for each column in b-a*x. If a rank is <N or M<=Nshowspicts an empty array. If b is 1-dimensional, it will be a (1,) shape array. Otherwise, the shape becomes (k,)
- Rank: It returns in Int datatype and depicts the rank of matrix A.
- S: It depicts the singular values of a.
Note
If b is a matrix, the results are in matrix forms.
Example 1: How to Use numpy.linalg.lstsq() Method
import numpy as np
import matplotlib.pyplot as plt
# x co-ordinates
x = np.arange(0, 9)
A = np.array([x, np.ones(9)])
# linearly generated sequence
y = [19, 20, 20.5, 21.5, 22, 23, 23, 25.5, 24]
# obtaining the parameters of regression line
w = np.linalg.lstsq(A.T, y, rcond=None)[0]
print(w)Output
[0.71666667 19.18888889]Example 2: Plotting the graph
import numpy as np
import matplotlib.pyplot as plt
# x co-ordinates
x = np.arange(0, 9)
A = np.array([x, np.ones(9)])
# linearly generated sequence
y = [19, 20, 20.5, 21.5, 22, 23, 23, 25.5, 24]
# obtaining the parameters of regression line
w = np.linalg.lstsq(A.T, y, rcond=None)[0]
print(w)
line = w[0]*x + w[1] # regression line
plt.plot(x, line, 'r-')
plt.plot(x, y, 'o')
plt.show()Output
Example 3
import numpy as np
# Define the coefficient matrix 'a'
# For a system of equations like:
# 3*x0 + 2*x1 = 9
# x0 + 2*x1 = 8
# The coefficient matrix would be:
a = np.array([[3, 2], [1, 2]])
# Define the ordinate values 'b'
b = np.array([9, 8])
# Use np.linalg.lstsq() to solve for 'x'
x, residuals, rank, s = np.linalg.lstsq(a, b, rcond=None)
print("Solution:", x)Output
Solution: [0.5 3.75]That’s it.
Ankit Lathiya is a Master of Computer Application by education and Android and Laravel Developer by profession and one of the authors of this blog. He is also expert in JavaScript and Python development.
