如何使用 Sklearn 在 Python 中对列表列表进行 L2 规范化

Question

s2 = [[0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194], [0.2, 0.4892574205256839, 0.2, 0.2, 0.383258146374831], [0.3193817886456925, 0.16666666666666666, 0.16666666666666666, 0.16666666666666666, 0.3193817886456925, 0.3193817886456925], [0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194]]

from sklearn.preprocessing import normalize
X = normalize(s2)

this is throwing error:

ValueError: setting an array element with a sequence.

How to L2 Normalize a list of lists in Python using Sklearn.

Answer 1

Since I don't have enough reputation to comment; hence posting it as an answer.

Let's quickly look at your datapoint.

I have converted the given datapoint into NumPy array. Since it doesn't have the same length, so it will look like.

>>> n2 = np.array([[0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194], [0.2, 0.4892574205256839, 0.2, 0.2, 0.383258146374831], [0.3193817886456925, 0.16666666666666666, 0.16666666666666666, 0.16666666666666666, 0.3193817886456925, 0.3193817886456925], [0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194]])
>>> n2
array([list([0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194]),
       list([0.2, 0.4892574205256839, 0.2, 0.2, 0.383258146374831]),
       list([0.3193817886456925, 0.16666666666666666, 0.16666666666666666, 0.16666666666666666, 0.3193817886456925, 0.3193817886456925]),
       list([0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194])],
      dtype=object)

And you can see here that converted values are not in Sequence of Values and to achieve this you need to keep the same length for the internal list ( looks like 0.16666666666666666 is copied multiple time in your array; if not then fix the length), it will look like

>>> n3 = np.array([[0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194], [0.2, 0.4892574205256839, 0.2, 0.2, 0.383258146374831], [0.3193817886456925, 0.16666666666666666, 0.16666666666666666, 0.16666666666666666, 0.319381788645692], [0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194]])
>>> n3
array([[0.2       , 0.2       , 0.2       , 0.30216512, 0.24462871],
       [0.2       , 0.48925742, 0.2       , 0.2       , 0.38325815],
       [0.31938179, 0.16666667, 0.16666667, 0.16666667, 0.31938179],
       [0.2       , 0.2       , 0.2       , 0.30216512, 0.24462871]])

As you can see now n3 has become a sequence of values.

and if you use normalize function, it simply works

>>> X = normalize(n3)
>>> X
array([[0.38408524, 0.38408524, 0.38408524, 0.58028582, 0.46979139],
       [0.28108867, 0.6876236 , 0.28108867, 0.28108867, 0.53864762],
       [0.59581303, 0.31091996, 0.31091996, 0.31091996, 0.59581303],
       [0.38408524, 0.38408524, 0.38408524, 0.58028582, 0.46979139]])

How to use NumPy array to avoid this issue, please have a look at this SO link ValueError: setting an array element with a sequence

Answer 2

Important: I removed one element from the 3rd list in order for all lists to have the same length.

I did that cause I really believe that it's a copy-paste error. If not, comment below and I will modify my answer.

import numpy as np

s2 = [[0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194], [0.2, 0.4892574205256839, 0.2, 0.2, 0.383258146374831], [0.3193817886456925, 0.16666666666666666, 0.16666666666666666, 0.3193817886456925, 0.3193817886456925], [0.2, 0.2, 0.2, 0.3021651247531982, 0.24462871026284194]]

X = normalize(np.array(s2))