与 pandas 成对的欧几里得距离忽略NaNs [英] Pairwise Euclidean distance with pandas ignoring NaNs

问题描述

我从字典开始，这是我的数据已被格式化的方式：

I start with a dictionary, which is the way my data was already formatted:

import pandas as pd
dict2 = {'A': {'a':1.0, 'b':2.0, 'd':4.0}, 'B':{'a':2.0, 'c':2.0, 'd':5.0}, 
'C':{'b':1.0,'c':2.0, 'd':4.0}}

然后将其转换为熊猫数据框：

I then convert it to a pandas dataframe:

df = pd.DataFrame(dict2)
print(df)
     A    B    C
a  1.0  2.0  NaN
b  2.0  NaN  1.0
c  NaN  2.0  2.0
d  4.0  5.0  4.0

当然，通过执行以下操作，我可以一次得到一个差异：

Of course, I can get the difference one at a time by doing this:

df['A'] - df['B']
Out[643]: 
a   -1.0
b    NaN
c    NaN
d   -1.0
dtype: float64

我想出了如何遍历和计算AA ，AB，AC：

I figured out how to loop through and calculate A-A, A-B, A-C:

for column in df:
print(df['A'] - df[column])

a    0.0
b    0.0
c    NaN
d    0.0
Name: A, dtype: float64
a   -1.0
b    NaN
c    NaN
d   -1.0
dtype: float64
a    NaN
b    1.0
c    NaN
d    0.0
dtype: float64

我想要做的是遍历各列以便计算| AB |，| AC |和| BC |并将结果存储在另一个字典中。

What I would like to do is iterate through the columns so as to calculate |A-B|, |A-C|, and |B-C| and store the results in another dictionary.

我要执行此操作，以便稍后计算所有列组合之间的欧几里得距离。如果有更简单的方法可以做到这一点，我也希望看到它。谢谢。

I want to do this so as to calculate the Euclidean distance between all combinations of columns later on. If there is an easier way to do this I would like to see it as well. Thank you.

推荐答案

您可以使用numpy广播来计算向量化的欧几里得距离（L2-范数），而忽略使用<$ c的NaN $ c> np.nansum 。

You can use numpy broadcasting to compute vectorised Euclidean distance (L2-norm), ignoring NaNs using np.nansum.

i = df.values.T
j = np.nansum((i - i[:, None]) ** 2, axis=2) ** .5

如果您想要一个表示距离矩阵的DataFrame ，如下所示：

If you want a DataFrame representing a distance matrix, here's what that would look like:

df = (lambda v, c: pd.DataFrame(v, c, c))(j, df.columns)
df
          A         B    C
A  0.000000  1.414214  1.0
B  1.414214  0.000000  1.0
C  1.000000  1.000000  0.0

df [i，j] 表示第i个和第j ^th 列在原始DataFrame中。

df[i, j] represents the distance between the i^th and j^th column in the original DataFrame.

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