Question

假设我有两个DataFrame：XA和XB，例如每个都有3行和2列：

import pandas as pd

XA = pd.DataFrame({
    'x1': [1, 2, 3],
    'x2': [4, 5, 6]
})

XB = pd.DataFrame({
    'x1': [8, 7, 6],
    'x2': [5, 4, 3]
})

对于XA中的每条记录，我想找到XB中最近的记录（例如，基于欧几里得距离）以及相应的距离。例如，这可能返回在DataFrame上建立索引的id_A，并带有id_B和distance的列。

如何最有效地做到这一点？

Answer 1

一种方法是计算完整距离矩阵，然后melt并使用nsmallest进行汇总，这将返回索引以及值：

from scipy.spatial.distance import cdist

def nearest_record(XA, XB):
    """Get the nearest record in XA for each record in XB.

    Args:
        XA: DataFrame. Each record is matched against the nearest in XB.
        XB: DataFrame.

    Returns:
        DataFrame with columns for id_A (from XA), id_B (from XB), and dist.
        Each id_A maps to a single id_B, which is the nearest record from XB.
    """
    dist = pd.DataFrame(cdist(XA, XB)).reset_index().melt('index')
    dist.columns = ['id_A', 'id_B', 'dist']
    # id_B is sometimes returned as an object.
    dist['id_B'] = dist.id_B.astype(int)
    dist.reset_index(drop=True, inplace=True)
    nearest = dist.groupby('id_A').dist.nsmallest(1).reset_index()
    return nearest.set_index('level_1').join(dist.id_B).reset_index(drop=True)

这表明id_B 2是最接近XA中三个记录的每个记录：

nearest_record(XA, XB)

 id_A       dist id_B
0   0   5.099020    2
1   1   4.472136    2
2   2   4.242641    2

但是，由于这涉及到计算完整距离矩阵，因此当XA和XB较大时，它将变慢或失败。一种计算每一行最近的替代方法可能会更快。

Answer 2

修改this answer以避开完整距离矩阵，您可以在XA（nearest_record1()）中找到每一行的最近记录和距离，然后调用apply来运行在每行（nearest_record()上浏览。这样test中的运行时间减少了约85％。

from scipy.spatial.distance import cdist

def nearest_record1(XA1, XB):
    """Get the nearest record between XA1 and XB.

    Args:
        XA: Series.
        XB: DataFrame.

    Returns:
        DataFrame with columns for id_B (from XB) and dist.
    """
    dist = cdist(XA1.values.reshape(1, -1), XB)[0]
    return pd.Series({'dist': np.amin(dist), 'id_B': np.argmin(dist)})

def nearest_record(XA, XB):
    """Get the nearest record in XA for each record in XB.

    Args:
        XA: DataFrame. Each record is matched against the nearest in XB.
        XB: DataFrame.

    Returns:
        DataFrame with columns for id_A (from XA), id_B (from XB), and dist.
        Each id_A maps to a single id_B, which is the nearest record from XB.
    """
    res = XA.apply(lambda x: nearest_record1(x, XB), axis=1)
    res['id_A'] = XA.index
    # id_B is sometimes returned as an object.
    res['id_B'] = res.id_B.astype(int)
    # Reorder columns.
    return res[['id_A', 'id_B', 'dist']]

这还会返回正确的结果：

nearest_record(XA, XB)
    id_A    id_B        dist
0      0       2    5.099020
1      1       2    4.472136
2      2       2    4.242641

最近记录和两个DataFrame中每个记录之间的对应距离

2 个答案: