def testChooseSimple(self):
# select a single value from a list of values.
options = [
[0, 1, 2],
[3, 4, 5],
[6, 7, 8]
]
for i in range(len(options)):
self.assertEqual(choose(i, *options), options[i])
def hsb_to_rgb(h, s, v, *rest):
i = floor(6 * h)
f = 6 * h - i
i = mod(i, 6)
p = v * (1 - s)
q = v * (1 - f * s)
t = v * (1 - (1 - f) * s)
r, g, b = choose(i, (v, t, p), (q, v, p), (p, v, t), (p, q, v), (t, p, v),
(v, p, q))
return (r, g, b) + rest
def hsb_to_rgb(h, s, v, *rest):
i = floor(6 * h)
f = 6 * h - i
i = mod(i, 6)
p = v * (1 - s)
q = v * (1 - f * s)
t = v * (1 - (1 - f) * s)
r, g, b = choose(i,
(v, t, p),
(q, v, p),
(p, v, t),
(p, q, v),
(t, p, v),
(v, p, q))
return (r, g, b) + rest
def testChooseComplex(self):
def m(i):
return [[10 * i + 0, 10 * i + 1],
[10 * i + 2, 10 * i + 3]]
selector = [[0, 1],
[1, 2]]
a = choose(selector,
(m(1), m(2), m(3)),
(m(4), m(5), m(6)),
(m(7), m(8), m(9)))
# choose() operates on column after column of the options matrix, i.e.
# in the example above, the selector is first applied to (m(1), m(4), m(7)),
# then to (m(2), m(5), m(8)), and so on. let's calculate the result for the
# first column:
# selector (integers indicate from which option to take a value):
# [0 1]
# [1 2]
# option #0 / column 1 (i.e. m(1)):
# [10 11]
# [12 13]
# option #1 / column 1 (i.e. m(4)):
# [40 41]
# [42 43]
# option #2 / column 1 (i.e. m(7)):
# [70 71]
# [72 73]
# choose() now picks the right value from each options depending on selector:
# [10 41]
# [42 73]
self.assertEqual(len(a), 3)
self.assertEqualArrays(a[0], [[10, 41], [42, 73]])
self.assertEqualArrays(a[1], [[20, 51], [52, 83]])
self.assertEqualArrays(a[2], [[30, 61], [62, 93]])