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# numpy.linalg.lstsq

`numpy.linalg.``lstsq`(a, b, rcond=-1)

  Return the least-squares solution to a linear matrix equation.

  Solves the equation a x = b by computing a vector x that minimizes the Euclidean 2-norm || b - a x ||^2. The equation may be under-, well-, or over- determined (i.e., the number of linearly independent rows of a can be less than, equal to, or greater than its number of linearly independent columns). If a is square and of full rank, then x (but for round-off error) is the “exact” solution of the equation.

  Parameters : | a : (M, N) array_like “Coefficient” matrix. b : {(M,), (M, K)} array_like Ordinate or “dependent variable” values. If b is two-dimensional, the least-squares solution is calculated for each of the K columns of b. rcond : float, optional Cut-off ratio for small singular values of a. Singular values are set to zero if they are smaller than rcond times the largest singular value of a.
  Returns : | x : {(M,), (M, K)} ndarray Least-squares solution. The shape of x depends on the shape of b. residuals : {(), (1,), (K,)} ndarray Sums of residuals; squared Euclidean 2-norm for each column in `b - a*x`. If the rank of a is < N or > M, this is an empty array. If b is 1-dimensional, this is a (1,) shape array. Otherwise the shape is (K,). rank : int Rank of matrix a. s : (min(M, N),) ndarray Singular values of a.
  Raises : | LinAlgError : If computation does not converge.

  Notes

  If b is a matrix, then all array results are returned as matrices.

  Examples

  Fit a line, `y = mx + c`, through some noisy data-points:

      >>> x = np.array([0, 1, 2, 3])
      >>> y = np.array([-1, 0.2, 0.9, 2.1])

  By examining the coefficients, we see that the line should have a gradient of roughly 1 and cut the y-axis at, more or less, -1.

  We can rewrite the line equation as `y = Ap`, where `A = [[x 1]]` and `p = [[m], [c]]`. Now use `lstsq` to solve for p:

      >>> A = np.vstack([x, np.ones(len(x))]).T
      >>> A
      array([[ 0.,  1.],
             [ 1.,  1.],
             [ 2.,  1.],
             [ 3.,  1.]])

      >>> m, c = np.linalg.lstsq(A, y)[0]
      >>> print m, c
      1.0 -0.95

  Plot the data along with the fitted line:

      >>> import matplotlib.pyplot as plt
      >>> plt.plot(x, y, 'o', label='Original data', markersize=10)
      >>> plt.plot(x, m*x + c, 'r', label='Fitted line')
      >>> plt.legend()
      >>> plt.show()

  (Source code, png, pdf)

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