Question

我有数千个“观测”。每个观测值都由位置（x，y）和传感器读数（z）组成，请参见下面的示例。

我想将双线性表面拟合到x，y和z数据。我目前正在使用amroamroamro/gist的代码段进行此操作：

def bi2Dlinter(xdata, ydata, zdata, gridrez):
    X,Y = np.meshgrid(
             np.linspace(min(x), max(x), endpoint=True, num=gridrez),
             np.linspace(min(y), max(y), endpoint=True, num=gridrez))  
    A = np.c_[xdata, ydata, np.ones(len(zdata))]
    C,_,_,_ = scipy.linalg.lstsq(A, zdata)
    Z = C[0]*X + C[1]*Y + C[2]
    return Z

我当前的方法是循环浏览DataFrame的行。（这对于1000个观测值非常有用，但不适用于较大的数据集。）

ZZ = []
for index, row in df2.iterrows():
    x=row['x1'], row['x2'], row['x3'], row['x4'], row['x5']
    y=row['y1'], row['y2'], row['y3'], row['y4'], row['y5']
    z=row['z1'], row['z2'], row['z3'], row['z4'], row['z5']
    ZZ.append(np.median(bi2Dlinter(x,y,z,gridrez)))
df2['ZZ']=ZZ

如果没有更有效的方法来做到这一点，我会感到惊讶。有没有办法向量化线性插值？

我放置了代码here，该代码也生成虚拟条目。谢谢

Answer 1

通常不建议像这样在DataFrames上进行循环。相反，您应该选择尽可能地对代码进行向量化。

首先，我们为您的输入创建一个数组

x_vals = df2[['x1','x2','x3','x4','x5']].values
y_vals = df2[['y1','y2','y3','y4','y5']].values
z_vals = df2[['z1','z2','z3','z4','z5']].values

接下来，我们需要创建一个用于处理矢量输入的bi2Dlinter函数，这涉及更改linspace / meshgrid以用于数组，以及更改minimum_squares函数。通常scipy.linalg函数可在数组上工作，但据我所知.lstsq方法不起作用。相反，我们可以使用.SVD在数组上复制相同的功能。

def create_ranges(start, stop, N, endpoint=True):
    if endpoint==1:
        divisor = N-1
    else:
        divisor = N
    steps = (1.0/divisor) * (stop - start)
    return steps[:,None]*np.arange(N) + start[:,None]

def linspace_nd(x,y,gridrez):
    a1 = create_ranges(x.min(axis=1), x.max(axis=1), N=gridrez, endpoint=True)
    a2 = create_ranges(y.min(axis=1), y.max(axis=1), N=gridrez, endpoint=True)
    out_shp = a1.shape + (a2.shape[1],)
    Xout = np.broadcast_to(a1[:,None,:], out_shp)
    Yout = np.broadcast_to(a2[:,:,None], out_shp)
    return Xout, Yout

def stacked_lstsq(L, b, rcond=1e-10):
    """
    Solve L x = b, via SVD least squares cutting of small singular values
    L is an array of shape (..., M, N) and b of shape (..., M).
    Returns x of shape (..., N)
    """
    u, s, v = np.linalg.svd(L, full_matrices=False)
    s_max = s.max(axis=-1, keepdims=True)
    s_min = rcond*s_max
    inv_s = np.zeros_like(s)
    inv_s[s >= s_min] = 1/s[s>=s_min]
    x = np.einsum('...ji,...j->...i', v,
                  inv_s * np.einsum('...ji,...j->...i', u, b.conj()))
    return np.conj(x, x)

def vectorized_bi2Dlinter(x_vals, y_vals, z_vals, gridrez):

    X,Y = linspace_nd(x_vals, y_vals, gridrez)
    A = np.stack((x_vals,y_vals,np.ones_like(z_vals)), axis=2)
    C = stacked_lstsq(A, z_vals)
    n_bcast = C.shape[0]
    return C.T[0].reshape((n_bcast,1,1))*X + C.T[1].reshape((n_bcast,1,1))*Y + C.T[2].reshape((n_bcast,1,1))

在对n = 10000行的数据进行测试时，矢量化函数明显更快。

%%timeit
ZZ = []
for index, row in df2.iterrows():
    x=row['x1'], row['x2'], row['x3'], row['x4'], row['x5']
    y=row['y1'], row['y2'], row['y3'], row['y4'], row['y5']
    z=row['z1'], row['z2'], row['z3'], row['z4'], row['z5']
    ZZ.append((bi2Dlinter(x,y,z,gridrez)))
df2['ZZ']=ZZ

Out: 5.52 s ± 17.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
res = vectorized_bi2Dlinter(x_vals,y_vals,z_vals,gridrez)

Out: 74.6 ms ± 159 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

您应该仔细注意此向量化函数中发生的事情，并熟悉numpy中的广播。我不能相信前三个功能，相反，我将从堆栈溢出中链接它们的答案，以便您了解。

Vectorized NumPy linspace for multiple start and stop values

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如何在Pandas DataFrame中逐行有效地插值数据？

1 个答案: