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[qu2cu] Use NamedTuple for solution

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This commit is contained in:
Behdad Esfahbod 2023-02-20 08:17:24 -07:00
parent f1086ddb65
commit b3be1883c8
1 changed files with 14 additions and 6 deletions
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20
Lib/fontTools/qu2cu/qu2cu.py
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@ -23,6 +23,7 @@ except ImportError:
from fontTools.misc import cython
from fontTools.misc.bezierTools import splitCubicAtTC
from typing import NamedTuple
__all__ = ["quadratic_to_curves", "quadratics_to_curves"]
@ -210,6 +211,13 @@ def quadratic_to_curves(q, tolerance=0.5, all_cubic=False):
return curves
class Solution(NamedTuple):
num_points: int
error: float
start_index: int
is_cubic: bool
def spline_to_curves(q, costs, tolerance=0.5, all_cubic=False):
assert len(q) >= 3, "quadratic spline requires at least 3 points"
@ -220,18 +228,18 @@ def spline_to_curves(q, costs, tolerance=0.5, all_cubic=False):
# Dynamic-Programming to find the solution with fewest number of
# cubic curves, and within those the one with smallest error.
sols = [(0, 0, 0, False)] # (best_num_points, best_error, start_index, cubic)
sols = [Solution(0, 0, 0, False)]
for i in range(1, len(elevated_quadratics) + 1):
best_sol = (len(q) + 2, 0, 1, False)
best_sol = Solution(len(q) + 2, 0, 1, False)
for j in range(0, i):
j_sol_count, j_sol_error, _, _ = sols[j]
j_sol_count, j_sol_error = sols[j].num_points, sols[j].error
if not all_cubic:
# Solution with quadratics between j:i
i_sol_count = j_sol_count + costs[2 * i] - costs[2 * j]
i_sol_error = j_sol_error
i_sol = (i_sol_count, i_sol_error, i - j, False)
i_sol = Solution(i_sol_count, i_sol_error, i - j, False)
if i_sol < best_sol:
best_sol = i_sol
@ -273,7 +281,7 @@ def spline_to_curves(q, costs, tolerance=0.5, all_cubic=False):
# Save best solution
i_sol_count = j_sol_count + 3
i_sol_error = max(j_sol_error, error)
i_sol = (i_sol_count, i_sol_error, i - j, True)
i_sol = Solution(i_sol_count, i_sol_error, i - j, True)
if i_sol < best_sol:
best_sol = i_sol
@ -288,7 +296,7 @@ def spline_to_curves(q, costs, tolerance=0.5, all_cubic=False):
cubic = []
i = len(sols) - 1
while i:
_, _, count, is_cubic = sols[i]
count, is_cubic = sols[i].start_index, sols[i].is_cubic
splits.append(i)
cubic.append(is_cubic)
i -= count