Getting deprecation warning in Sklearn over 1d array, despite not having a 1D array

Question

I am trying to use SKLearn to run an SVM model. I am just trying it out now with some sample data. Here is the data and the code:

import numpy as np
from sklearn import svm
import random as random

A = np.array([[random.randint(0, 20) for i in range(2)] for i in range(10)])
lab = [0, 1, 0, 1, 0, 1, 0, 1, 0, 1]

clf = svm.SVC(kernel='linear', C=1.0)
clf.fit(A, lab)

FYI, when I run

import sklearn
sklearn.__version__

It outputs 0.17.

Now, when I run print(clf.predict([1, 1])), I get the following warning:

C:\Users\me\AppData\Local\Continuum\Anaconda2\lib\site-packages\sklearn\ut
ils\validation.py:386: DeprecationWarning: Passing 1d arrays as data is deprecat
ed in 0.17 and willraise ValueError in 0.19. Reshape your data either using X.re
shape(-1, 1) if your data has a single feature or X.reshape(1, -1) if it contain
s a single sample.
  DeprecationWarning)

It does give me a prediction, which is great. However, I find this weird for a few reasons.

I don't have a 1d array. If you print A, you get

array([[ 9, 12],
       [ 2, 16],
       [14, 14],
       [ 4,  2],
       [ 8,  4],
       [12,  3],
       [ 0,  0],
       [ 3, 13],
       [15, 17],
       [15, 16]])

Which appears to me to be 2 dimensional. But okay, let's just say that what I have is in fact a 1D array. Let's try to change it using reshape, as suggested by the error.

Same code as above, but now we have

A = np.array([[random.randint(0, 20) for i in range(2)] for i in range(10)]).reshape(-1,1)

But then this outputs an array of length 20, which makes no sense and is not what I want. I also tried it with reshape(1, -1) but then this gives me a single observation / list with 20 items in it.

How can I reshape my data in numpy arrays so that I don't get this warning?

I looked at two answers on SO, and neither worked for me. Question 1 and Question 2. It seems that Q1 was actually 1D data and was solved using reshape, which I tried and failed at. Q2 has an answer about how to track warnings and errors, which isn't what I want. The other answer is again an instance of a 1D array.

juanpa.arrivillaga · Accepted Answer · 2016-06-13T20:19:22.397

The error is coming from the predict method. Numpy will interpret [1,1] as a 1d array. So this should avoid the warning:

clf.predict(np.array([[1,1]]))

Notice that:

In [14]: p1 = np.array([1,1])

In [15]: p1.shape
Out[15]: (2,)

In [16]: p2 = np.array([[1,1]])

In [17]: p2.shape
Out[17]: (1, 2)

Also, note that you can't use an array of shape (2,1)

In [21]: p3 = np.array([[1],[1]])

In [22]: p3.shape
Out[22]: (2, 1)

In [23]: clf.predict(p3)
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-23-e4070c037d78> in <module>()
----> 1 clf.predict(p3)

/home/juan/anaconda3/lib/python3.5/site-packages/sklearn/svm/base.py in predict(self, X)
    566             Class labels for samples in X.
    567         """
--> 568         y = super(BaseSVC, self).predict(X)
    569         return self.classes_.take(np.asarray(y, dtype=np.intp))
    570 

/home/juan/anaconda3/lib/python3.5/site-packages/sklearn/svm/base.py in predict(self, X)
    303         y_pred : array, shape (n_samples,)
    304         """
--> 305         X = self._validate_for_predict(X)
    306         predict = self._sparse_predict if self._sparse else self._dense_predict
    307         return predict(X)

/home/juan/anaconda3/lib/python3.5/site-packages/sklearn/svm/base.py in _validate_for_predict(self, X)
    472             raise ValueError("X.shape[1] = %d should be equal to %d, "
    473                              "the number of features at training time" %
--> 474                              (n_features, self.shape_fit_[1]))
    475         return X
    476 

ValueError: X.shape[1] = 1 should be equal to 2, the number of features at training time

When your array is stored in a variable let's say `var`, you can also do `np.array([var])` — Yohan Obadia, Jan 25 '17 at 13:22
Does this slow down prediction? i.e. is predict(var) faster than predict(np.array([var])) ? — user3494047, Sep 21 '17 at 15:06
@user3494047 for a single observation I can't imagine it would make an appreciable difference. If you were running this on many observations with lots of features, i.e. you `var` was a shape like `100000, 5000` then yes, a `np.ndarray` will be faster. My suspicion is that sklearn turns the inputs into `np.ndarrays` *antway*. — juanpa.arrivillaga, Sep 21 '17 at 15:42

score 5 · Answer 2 · answered Apr 30 '17 at 10:32

5

Instead of running

print(clf.predict([1, 1]))

Run

print(clf.predict([[1,1]])

answered Apr 30 '17 at 10:32

Bharath M Shetty

30,075
6
57
108

score 0 · Answer 3 · answered Jun 15 '16 at 00:47

the sample for the prediction intuitively could be:

[1,9]

But you can reshape the vector as the previous answer. Or just do the following:

[[1,9]]

EXAMPLE

import numpy as np from sklearn import svm import random as random

A = np.array([[random.randint(0, 20) for i in range(2)] for i in range(10)]) lab = [0, 1, 0, 1, 0, 1, 0, 1, 0, 1]

clf = svm.SVC(kernel='linear', C=1.0) clf.fit(A, lab)

print clf.predict([[1,9]])

Getting deprecation warning in Sklearn over 1d array, despite not having a 1D array

3 Answers3

EXAMPLE

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