def test_neville2d():
funcx = numpy.sin
funcy = numpy.exp
nrow = 10
ncol = 10
tol = 1.0e-4
# TODO: As with test_neville; can this not be simplified with
# vectorized code
x = numpy.zeros((nrow, ))
y = numpy.zeros((ncol, ))
fval = numpy.empty((nrow, ncol))
row_tmp = numpy.pi / nrow
# col_tmp = 1.0 / float(ncol)
for row in range(nrow):
x[row] = (row + 1.0) * row_tmp
for col in range(ncol):
y[col] = (col + 1.0) / float(ncol)
fval[row][col] = funcx(x[row]) * funcy(y[col])
for row in range(ncol):
xx = (-0.1 + (row + 1.0) / float(nrow)) * numpy.pi
for col in range(4):
yy = -0.1 + (col + 1.0) / float(ncol)
answer = funcx(xx) * funcy(yy)
val = utils.neville2d(xx, yy, x, y, fval)
assert utils.Knuth_close(answer, val, tol)
def test_neville():
func = numpy.exp
tol = 1.0e-6
num = 10
# TODO: can we not just use vectorized code here?
x = []
y = []
for ii in range(num):
x.append(ii / float(num))
y.append(func(x[ii]))
xx = numpy.array(x)
yy = numpy.array(y)
for ii in range(num):
tmp = 1.01 * (ii / float(num))
answer = func(tmp)
val = utils.neville(tmp, xx, yy)
assert utils.Knuth_close(answer, val, tol)