从这个问题
Finding the best trade-off point on a curve
我正在使用python解决方案,如果我自己运行它,它会很好地工作。
但我有一本关于这些曲线的字典:
苏格兰文{}
{'00': [4.998436383771836,
0.0160165895672664,
0.004512422186993107,
0.0013171501024742112,
0.000788783358847752,
0.0005498425886621068],
'67':[0.13598930783101504,
0.04717783658889547,
0.027547125931827038,
0.021440839841617088,
0.016775671441391703,
0.013185864754748117,
0.010318462898609907],
等等…
如果我把它运行得很好:
curve = sse['67']
nPoints = len(curve)
allCoord = np.vstack((range(nPoints), curve)).T
np.array([range(nPoints), curve])
firstPoint = allCoord[0]
lineVec = allCoord[-1] - allCoord[0]
lineVecNorm = lineVec / np.sqrt(np.sum(lineVec**2))
vecFromFirst = allCoord - firstPoint
scalarProduct = np.sum(vecFromFirst * np.matlib.repmat(lineVecNorm, nPoints, 1), axis=1)
vecFromFirstParallel = np.outer(scalarProduct, lineVecNorm)
vecToLine = vecFromFirst - vecFromFirstParallel
distToLine = np.sqrt(np.sum(vecToLine ** 2, axis=1))
idxOfBestPoint = np.argmax(distToLine)
print(idxOfBestPoint)
但当我试图使它成为一个循环函数时,我会得到一些奇怪的值,这些值在单独运行时不匹配:
for k in sse:
curve = sse[str(k)]
nPoints = len(curve)
allCoord = np.vstack((range(nPoints), curve)).T
np.array([range(nPoints), curve])
firstPoint = allCoord[0]
lineVeceVeceVec = allCoord[-1] - allCoord[0]
lineVecNorm = lineVec / np.sqrt(np.sum(lineVec**2))
vecFromFirst = allCoord - firstPoint
scalarProduct = np.sum(vecFromFirst * np.matlib.repmat(lineVecNorm, nPoints, 1), axis=1)
vecFromFirstParallel = np.outer(scalarProduct, lineVecNorm)
vecToLine = vecFromFirst - vecFromFirstParallel
distToLine = np.sqrt(np.sum(vecToLine ** 2, axis=1))
idxOfBestPoint = np.argmax(distToLine)
print(idxOfBestPoint, k)
它会吐出这样的东西:
7 57
98 11
6 45
4 50
98 91
98 00
1 62
98 79
7 48
98 12
98 38
98 23
5 37
98 56
98 67
5 25
7 46
98 22
98 49
2 47
98 41
98 78
98 35
98 68
98 14
98 24
1 0
98 42
我不知道是否有某个变量没有重置,或者添加一个简单的循环会导致它失败?
要清楚的是,它确实在运行,但是Calc将'98'作为循环的肘部,但是当单独运行时,它将是'67'曲线列表的7,而不是98。