手动定义 p 和 q:
问题: 在步骤2中,如何将代码应用于给定的(再次是样本数据集。真实数据集非常大)数据集df:
p = [[45.1024,7.7498],[45.1027,7.7513],[45.1072,7.7568],[45.1076,7.7563]]
q = [[45.0595,7.6829],[45.0595,7.6829],[45.0564,7.6820],[45.0533,7.6796],[45.0501,7.6775]]
- 步骤 1 (fine)
正常的代码部分
def _c(ca, i, j, p, q):
if ca[i, j] > -1:
return ca[i, j]
elif i == 0 and j == 0:
ca[i, j] = np.linalg.norm(p[i]-q[j])
elif i > 0 and j == 0:
ca[i, j] = max(_c(ca, i-1, 0, p, q), np.linalg.norm(p[i]-q[j]))
elif i == 0 and j > 0:
ca[i, j] = max(_c(ca, 0, j-1, p, q), np.linalg.norm(p[i]-q[j]))
elif i > 0 and j > 0:
ca[i, j] = max(
min(
_c(ca, i-1, j, p, q),
_c(ca, i-1, j-1, p, q),
_c(ca, i, j-1, p, q)
),
np.linalg.norm(p[i]-q[j])
)
else:
ca[i, j] = float('inf')
return ca[i, j]
- 问题出在第二步
def frdist(p, q):
# Remove nan values from p
p = np.array([i for i in p if np.any(np.isfinite(i))], np.float64) # ESSENTIAL PART TO REMOVE NaN
q = np.array([i for i in q if np.any(np.isfinite(i))], np.float64) # ESSENTIAL PART TO REMOVE NaN
len_p = len(p)
len_q = len(q)
if len_p == 0 or len_q == 0:
raise ValueError('Input curves are empty.')
# p and q no longer have to be the same length
if len(p[0]) != len(q[0]):
raise ValueError('Input curves do not have the same dimensions.')
ca = (np.ones((len_p, len_q), dtype=np.float64) * -1)
dist = _c(ca, len_p-1, len_q-1, p, q)
return(dist)
frdist(p, q)
0.09754839824415232
问题: 在步骤2中,如何将代码应用于给定的(再次是样本数据集。真实数据集非常大)数据集df:
1 1.1 2 2.1 3 3.1 4 4.1 5 5.1
0 43.1024 6.7498 NaN NaN NaN NaN NaN NaN NaN NaN
1 46.0595 1.6829 25.0695 3.7463 NaN NaN NaN NaN NaN NaN
2 25.0695 5.5454 44.9727 8.6660 41.9726 2.6666 84.9566 3.8484 44.9566 1.8484
3 35.0281 7.7525 45.0322 3.7465 14.0369 3.7463 NaN NaN NaN NaN
4 35.0292 7.5616 45.0292 4.5616 23.0292 3.5616 45.0292 6.7463 NaN
通过取p的第一行和q的第二行,然后计算距离frdist(p,q)
。然后再次将p作为第一行,但现在q是第三行。然后是1和3。
最终我应该得到一个大小为(rows, rows)且对角线为0的矩阵。因为本身的距离为0:
0 1 2 3 4 5 ... 105
0 0
1 0
2 0
3 0
4 0
5 0
... 0
105 0
frdist
是0.09754839824415232
。但是,当我将您的代码应用于真实数据集时,在那里我的前两行是我在这里说明的p和q的数字时,我的frdist
是0.0879574
。 - Mameditertoos
是一个打字错误。已修复。 - Valentinofrdist
直接和我的解决方案都得到了相同的结果。 - Valentino_c
函数是递归的,它来自于那里。 - Valentino