如何让pandas dataframe与numpy和scipy用户定义函数一起工作?

如有任何帮助,我将不胜感激,因为我在使用熊猫与以下功能时遇到错误。

下面是我尝试使用的示例设置:

example_data = {'age': [37,37,27,22,32,22,42,32,37,22], 'target': [0,0,2,0,0,0,0,0,2,0]}
example_df = pd.DataFrame(data=example_data)
example_df

我调用了rdc函数,如下所示:

ldc(x=example_data['age'],y=example_data['target'])

但是,我遇到了一个问题:

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-248-c4a80b9f6e55> in <module>()
----> 1 ldc(x=example_data['age'],y=example_data['target'])

<ipython-input-231-78a770305f24> in ldc(x, y, f, k, s, n)
     29         return np.median(values)
     30 
---> 31     if len(x.shape) == 1: x = x.values.reshape((-1, 1))
     32     if len(y.shape) == 1: y = y.values.reshape((-1, 1))
     33 

AttributeError: 'list' object has no attribute 'shape'

下面是我正在使用的函数本身:

"""
Implements the Randomized Dependence Coefficient
David Lopez-Paz, Philipp Hennig, Bernhard Schoelkopf
http://papers.nips.cc/paper/5138-the-randomized-dependence-coefficient.pdf
"""
import numpy as np
from scipy.stats import rankdata

def rdc(x, y, f=np.sin, k=20, s=1/6., n=1):
    """
    Computes the Randomized Dependence Coefficient
    x,y: numpy arrays 1-D or 2-D
         If 1-D, size (samples,)
         If 2-D, size (samples, variables)
    f:   function to use for random projection
    k:   number of random projections to use
    s:   scale parameter
    n:   number of times to compute the RDC and
         return the median (for stability)
    According to the paper, the coefficient should be relatively insensitive to
    the settings of the f, k, and s parameters.
    """
    if n > 1:
        values = []
        for i in range(n):
            try:
                values.append(rdc(x, y, f, k, s, 1))
            except np.linalg.linalg.LinAlgError: pass
        return np.median(values)

    if len(x.shape) == 1: x = x.values.reshape((-1, 1))
    if len(y.shape) == 1: y = y.values.reshape((-1, 1))

    # Copula Transformation
    cx = np.column_stack([rankdata(xc, method='ordinal') for xc in x.T])/float(x.size)
    cy = np.column_stack([rankdata(yc, method='ordinal') for yc in y.T])/float(y.size)

    # Add a vector of ones so that w.x + b is just a dot product
    O = np.ones(cx.shape[0])
    X = np.column_stack([cx, O])
    Y = np.column_stack([cy, O])

    # Random linear projections
    Rx = (s/X.shape[1])*np.random.randn(X.shape[1], k)
    Ry = (s/Y.shape[1])*np.random.randn(Y.shape[1], k)
    X = np.dot(X, Rx)
    Y = np.dot(Y, Ry)

    # Apply non-linear function to random projections
    fX = f(X)
    fY = f(Y)

    # Compute full covariance matrix
    C = np.cov(np.hstack([fX, fY]).T)

    # Due to numerical issues, if k is too large,
    # then rank(fX) < k or rank(fY) < k, so we need
    # to find the largest k such that the eigenvalues
    # (canonical correlations) are real-valued
    k0 = k
    lb = 1
    ub = k
    while True:

        # Compute canonical correlations
        Cxx = C[:k, :k]
        Cyy = C[k0:k0+k, k0:k0+k]
        Cxy = C[:k, k0:k0+k]
        Cyx = C[k0:k0+k, :k]

        eigs = np.linalg.eigvals(np.dot(np.dot(np.linalg.inv(Cxx), Cxy),
                                        np.dot(np.linalg.inv(Cyy), Cyx)))

        # Binary search if k is too large
        if not (np.all(np.isreal(eigs)) and
                0 <= np.min(eigs) and
                np.max(eigs) <= 1):
            ub -= 1
            k = (ub + lb) / 2
            continue
        if lb == ub: break
        lb = k
        if ub == lb + 1:
            k = ub
        else:
            k = (ub + lb) / 2

    return np.sqrt(np.max(eigs))