Python function to merge two sorted linear graphs
-1
$begingroup$
I have code for creating a function to calculator the addition of two sorted (lists) that each element represent x, y values.
A:[(3,3), (5,5), (6,3), (7,5)]
B:[(0,0), (2,2), (4,3), (5,3), (10,2)]
explanation
when x = 0 at B line, y=0, at A line y = 0 => (0, 0)
when x = 2 at B line, y=2, at A line y = 0 => (0, 0)
when x = 4 at B line, y=3, at A line y = (5-3 / 5-3 * 4) => (4, 7)
when x = 5 at B line, y=3, at A line y = 5 => (5, 8)
when x = 6 at B line, y= (2 - 3)/(10-5) * 6, at A line y = 3 => (6, 0)
when x = 7 at B line, y= (2 - 3)/(10-5) * 7, at A line y = 5 => (7, 1.5)
when x = 10 at B line, y=3, at A line y = (5-3 / 5-3 * 4) => (4, 7)
=> [(0, 0), (2, 2), (3, 5.5), (4, 7.0), (5, 8), (6, 5.8), (7, 7.6), (10, 2)]
Is there a better way for me to not using blueforce way to calculate x, y values?
def graph_addition(A, B):
if not A or not B: return A or B
res =
i = j = 0
while i < len(A) and j < len(B):
if A[i][0] < B[0][0]:
x, y = A[i]
i += 1
elif B[j][0] < A[0][0]:
x, y = B[j]
j += 1
elif A[i][0] < B[j][0]:
x = A[i][0]
y = (B[j][1] - B[j - 1][1]) / (B[j][0] - B[j - 1][0]) * (x - B[j - 1][0]) + B[j - 1][1] + A[i][1]
i += 1
elif A[i][0] > B[j][0]:
x = B[j][0]
y = (A[i][1] - A[i - 1][1]) / (A[i][0] - A[i - 1][0]) * (x - A[i - 1][0]) + A[i - 1][1] + B[j][1]
j += 1
else:
x = A[i][0]
y = A[i][1] + B[j][1]
i += 1
j += 1
res.append((x, y))
if A[i:]: res += A[i:]
if B[j:]: res += B[j:]
return res
python python-3.x
edited 2 hours ago
Jamal♦
30.4k11121227
asked 7 hours ago
A.LeeA.Lee
492
$endgroup$
add a comment |
-1
$begingroup$
I have code for creating a function to calculator the addition of two sorted (lists) that each element represent x, y values.
A:[(3,3), (5,5), (6,3), (7,5)]
B:[(0,0), (2,2), (4,3), (5,3), (10,2)]
explanation
when x = 0 at B line, y=0, at A line y = 0 => (0, 0)
when x = 2 at B line, y=2, at A line y = 0 => (0, 0)
when x = 4 at B line, y=3, at A line y = (5-3 / 5-3 * 4) => (4, 7)
when x = 5 at B line, y=3, at A line y = 5 => (5, 8)
when x = 6 at B line, y= (2 - 3)/(10-5) * 6, at A line y = 3 => (6, 0)
when x = 7 at B line, y= (2 - 3)/(10-5) * 7, at A line y = 5 => (7, 1.5)
when x = 10 at B line, y=3, at A line y = (5-3 / 5-3 * 4) => (4, 7)
=> [(0, 0), (2, 2), (3, 5.5), (4, 7.0), (5, 8), (6, 5.8), (7, 7.6), (10, 2)]
Is there a better way for me to not using blueforce way to calculate x, y values?
def graph_addition(A, B):
if not A or not B: return A or B
res =
i = j = 0
while i < len(A) and j < len(B):
if A[i][0] < B[0][0]:
x, y = A[i]
i += 1
elif B[j][0] < A[0][0]:
x, y = B[j]
j += 1
elif A[i][0] < B[j][0]:
x = A[i][0]
y = (B[j][1] - B[j - 1][1]) / (B[j][0] - B[j - 1][0]) * (x - B[j - 1][0]) + B[j - 1][1] + A[i][1]
i += 1
elif A[i][0] > B[j][0]:
x = B[j][0]
y = (A[i][1] - A[i - 1][1]) / (A[i][0] - A[i - 1][0]) * (x - A[i - 1][0]) + A[i - 1][1] + B[j][1]
j += 1
else:
x = A[i][0]
y = A[i][1] + B[j][1]
i += 1
j += 1
res.append((x, y))
if A[i:]: res += A[i:]
if B[j:]: res += B[j:]
return res
python python-3.x
edited 2 hours ago
Jamal♦
30.4k11121227
asked 7 hours ago
A.LeeA.Lee
492
$endgroup$
add a comment |
-1
-1
-1
$begingroup$
I have code for creating a function to calculator the addition of two sorted (lists) that each element represent x, y values.
A:[(3,3), (5,5), (6,3), (7,5)]
B:[(0,0), (2,2), (4,3), (5,3), (10,2)]
explanation
when x = 0 at B line, y=0, at A line y = 0 => (0, 0)
when x = 2 at B line, y=2, at A line y = 0 => (0, 0)
when x = 4 at B line, y=3, at A line y = (5-3 / 5-3 * 4) => (4, 7)
when x = 5 at B line, y=3, at A line y = 5 => (5, 8)
when x = 6 at B line, y= (2 - 3)/(10-5) * 6, at A line y = 3 => (6, 0)
when x = 7 at B line, y= (2 - 3)/(10-5) * 7, at A line y = 5 => (7, 1.5)
when x = 10 at B line, y=3, at A line y = (5-3 / 5-3 * 4) => (4, 7)
=> [(0, 0), (2, 2), (3, 5.5), (4, 7.0), (5, 8), (6, 5.8), (7, 7.6), (10, 2)]
Is there a better way for me to not using blueforce way to calculate x, y values?
def graph_addition(A, B):
if not A or not B: return A or B
res =
i = j = 0
while i < len(A) and j < len(B):
if A[i][0] < B[0][0]:
x, y = A[i]
i += 1
elif B[j][0] < A[0][0]:
x, y = B[j]
j += 1
elif A[i][0] < B[j][0]:
x = A[i][0]
y = (B[j][1] - B[j - 1][1]) / (B[j][0] - B[j - 1][0]) * (x - B[j - 1][0]) + B[j - 1][1] + A[i][1]
i += 1
elif A[i][0] > B[j][0]:
x = B[j][0]
y = (A[i][1] - A[i - 1][1]) / (A[i][0] - A[i - 1][0]) * (x - A[i - 1][0]) + A[i - 1][1] + B[j][1]
j += 1
else:
x = A[i][0]
y = A[i][1] + B[j][1]
i += 1
j += 1
res.append((x, y))
if A[i:]: res += A[i:]
if B[j:]: res += B[j:]
return res
python python-3.x
edited 2 hours ago
Jamal♦
30.4k11121227
asked 7 hours ago
A.LeeA.Lee
492
$endgroup$
I have code for creating a function to calculator the addition of two sorted (lists) that each element represent x, y values.
A:[(3,3), (5,5), (6,3), (7,5)]
B:[(0,0), (2,2), (4,3), (5,3), (10,2)]
explanation
when x = 0 at B line, y=0, at A line y = 0 => (0, 0)
when x = 2 at B line, y=2, at A line y = 0 => (0, 0)
when x = 4 at B line, y=3, at A line y = (5-3 / 5-3 * 4) => (4, 7)
when x = 5 at B line, y=3, at A line y = 5 => (5, 8)
when x = 6 at B line, y= (2 - 3)/(10-5) * 6, at A line y = 3 => (6, 0)
when x = 7 at B line, y= (2 - 3)/(10-5) * 7, at A line y = 5 => (7, 1.5)
when x = 10 at B line, y=3, at A line y = (5-3 / 5-3 * 4) => (4, 7)
=> [(0, 0), (2, 2), (3, 5.5), (4, 7.0), (5, 8), (6, 5.8), (7, 7.6), (10, 2)]
Is there a better way for me to not using blueforce way to calculate x, y values?
def graph_addition(A, B):
if not A or not B: return A or B
res =
i = j = 0
while i < len(A) and j < len(B):
if A[i][0] < B[0][0]:
x, y = A[i]
i += 1
elif B[j][0] < A[0][0]:
x, y = B[j]
j += 1
elif A[i][0] < B[j][0]:
x = A[i][0]
y = (B[j][1] - B[j - 1][1]) / (B[j][0] - B[j - 1][0]) * (x - B[j - 1][0]) + B[j - 1][1] + A[i][1]
i += 1
elif A[i][0] > B[j][0]:
x = B[j][0]
y = (A[i][1] - A[i - 1][1]) / (A[i][0] - A[i - 1][0]) * (x - A[i - 1][0]) + A[i - 1][1] + B[j][1]
j += 1
else:
x = A[i][0]
y = A[i][1] + B[j][1]
i += 1
j += 1
res.append((x, y))
if A[i:]: res += A[i:]
if B[j:]: res += B[j:]
return res
python python-3.x
python python-3.x
edited 2 hours ago
Jamal♦
30.4k11121227
asked 7 hours ago
A.LeeA.Lee
492
edited 2 hours ago
Jamal♦
30.4k11121227
asked 7 hours ago
A.LeeA.Lee
492
edited 2 hours ago
Jamal♦
30.4k11121227
edited 2 hours ago
Jamal♦
30.4k11121227
edited 2 hours ago
Jamal♦
30.4k11121227
30.4k11121227
asked 7 hours ago
A.LeeA.Lee
492
asked 7 hours ago
A.LeeA.Lee
492
asked 7 hours ago
A.LeeA.Lee
492
492
add a comment |
add a comment |
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