R's equivalence of numpy.linalg.lstsq - r

Question

I have multiple linear regressions of the form vc = x1 * va + x2 * vb.
(Now, a too minimal example follows - it has the same values, which leads to warnings in R. Below a second data set illustrating my issue)
In Python, I programmed
#!/usr/bin/env python3
import numpy as np
va = np.array([1, 2, 3, 4, 5])
vb = np.array([1, 2, 3, 4, 5])
vc = np.array([1, 2, 3, 4, 5])
A = np.vstack([va, vb]).T
print(A)
result = np.linalg.lstsq(A, vc)
print(result)
Output:
(array([ 0.5, 0.5]), array([], dtype=float64), 1, array([ 1.04880885e+01, 3.14018492e-16]))
I thought, following code would be identical:
#!/usr/bin/Rscript
va <- c(1, 2, 3, 4, 5)
vb <- c(1, 2, 3, 4, 5)
vc <- c(1, 2, 3, 4, 5)
reg <- lm(vc ~ va + vb)
reg
summary(reg)
However, I get following output (excerpt):
Coefficients:
A1 A2
1 NA
esidual standard error: 7.022e-16 on 4 degrees of freedom
In summary.lm(reg) : essentially perfect fit: summary may be unreliable
Even if I adjust the numbers somehow, R still keeps complaining.
I assume, I am doing something basic wrong, but I can't figure out. I also tried to construct a matrix A (containg vb and vc as colums) and then use reg <- lm(vc ~ 0 + A). There, I get 3 degrees of freedom, but with the same Coefficients.
2nd data set
va = np.array([1, 2, 3, 4, 5])
vb = np.array([2, 2, 2, 2, 2])
vc = np.array([3.1, 3.2, 3.3, 3.4, 3.5])
va <- c(1, 2, 3, 4, 5)
vb <- c(2, 2, 2, 2, 2)
vc <- c(3.1, 3.2, 3.3, 3.4, 3.5)
If I add 0 + (which results in lm(vc ~ 0 + va + vb)), I geed 3 degrees of freedom and the same result. Looks good.
The 0 + removes the "implied intercept term" (whatever this this). Source

Answer 1

The problem is that you have a singular fit, and multiple combinations of coefficients will represent it equally well. IMHO, both numpy and R should really throw an error in this case by default. You can get R to give you an error by adding singular.ok = FALSE to your arguments. Additionally, altough your intercept in this case is zero, your regression equation indicates that you're not looking to fit one. To fit a linear model without an intercept in R, use a formula in the form:
lm(vc ~ va + vb - 1)
So, to (properly) return an error in this singular fit, you would call:
reg <- lm(vc ~ va + vb - 1, singular.ok = FALSE)