Import error. Can't import kmeans_plusplus - python

I'm trying to replicate an example of a clustering model with scikit-learn:
import sklearn
sklearn.__version__
Returns:
'0.23.2'
And:
from sklearn.cluster import kmeans_plusplus
Returns the Error message:
ImportError: cannot import name 'kmeans_plusplus' from 'sklearn.cluster' (C:\Users\sddss\anaconda3\lib\site-packages\sklearn\cluster\__init__.py)

According to the documentation, kmeans_plusplus is
New in version 0.24.
so it is not available for the version 0.23.2 you are using.
Nevertheless, this should not be a real issue; the only difference between the "good old" K-Means already available in scikit-learn is the initialization of the cluster centers according to the kmeans++ algorithm; and this is already available in the standard KMeans. From the standard KMeans documentation regarding the init argument:
'k-means++' : selects initial cluster centers for k-mean clustering in a smart way to speed up convergence
So, what you need to do instead is simply to use the "vanilla" KMeans of scikit-learn with the argument init='kmeans++':
from sklearn.cluster import KMeans
kmeans = KMeans(n_clusters=n_clusters, init='kmeans++')

There is no kmeans_plusplus class or module for version 0.23.2. You need to import KMeans and set the init key word argument to kmeans++ to obtain the behaviour you want
from sklearn.cluster import KMeans
kmeans = KMeans(init='k-means++')

Related

Import Error: cannot import name 'tree' from 'sklearn.tree'

I am on my second day of re-taking Python for the gazillionth time!
I am doing a tutorial on ML in Python, using the following code:
import sklearn.tree
import pandas as pd
from sklearn.tree import DecisionTreeClassifier
from sklearn.tree import tree
music_data = pd.read_csv('music.csv')
x = music_data.drop(columns=['genre'])
y = music_data['genre']
model = DecisionTreeClassifier()
model.fit(x,y)
tree.export_graphviz(model, out_file='music-recommender.dot',
feature_names=['age','gender'],
class_names= sorted(y.unique()),
label='all',
rounded=True,
filled=True)
I keep getting the following error:
ImportError Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_13088/3820271611.py in <module>
2 import pandas as pd
3 from sklearn.tree import DecisionTreeClassifier
----> 4 from sklearn.tree import tree
5
6 music_data = pd.read_csv('music.csv')
ImportError: cannot import name 'tree' from 'sklearn.tree' (C:\Anaconda\lib\site-packages\sklearn\tree\__init__.py)
I've tried to find a solution online, but I don't think it's the version of Python/Anaconda because I literally just installed both. I also don't think it's the sklearn.tree since I was able to import DecisionClassifer.
As this answer indicates, you're looking at some older code; this is always a risk with programming. But there's another thing you need to know about your code.
First off, scikit-learn contains several modules, and almost everything you need from it is in one of those. In my experience, most people import things like this:
from sklearn.tree import DecisionTreeRegressor # A regressor class.
from sklearn.tree import plot_tree # A helpful function.
from sklearn.metrics import mean_squared_error # An evaluation function.
It looks like the tutorial wants something similar to plot_tree(). This new-ish function is much easier to use than the older Graphviz visualization. So unless you really need the DOT file for some reasons, you should be able to do this:
from sklearn.tree import plot_tree
sklearn.tree.plot_tree(model)
Bottom line: there will probably be more broken things in that material. So if I were you I'd either make a new environment with a version of sklearn matching whatever material you're using... or ditch that material and look for something newer.
from sklearn.tree import tree looks wrong. Did you mean from sklearn import tree ?
According to the official Scikit Learn Decision Trees Documentation you really do not need too much of importing.
It can be done simply as follows:
from sklearn import tree
import pandas as pd
music_data = pd.read_csv('music.csv')
X = music_data.drop(columns=['genre'])
y = music_data['genre']
model = tree.DecisionTreeClassifier()
model.fit(X,y)

Simple question I am not getting output as expected.(Linear regression)

I am new to programming. Currently, I am learning machine learning from this video.
This is related to linear regression
CODE:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn import linear_model
df=pd.read_csv('homeprices.csv')
reg = linear_model.LinearRegression()
Problem 1
reg.fit(df[['area']],df.price)
Expected output should be
LinearRegression(copy_X=True, fit_intercept=True, n_jobs=None,
normalize=False)
My output:
LinearRegression()
Problem 2
reg.predict(3300)
It's giving error when I use "()" but when I use 2D array "[[]]" It is giving me correct output, But I want to know why It is not giving me output(as shown in video) when I use the only parenthesis.
Problem 1 :
This is how the fitted model outputs are shown in the newest version of sklearn, i.e., 0.23. The parameters are the same, but they are not shown in the output.
You can use reg.get_params() to view the parameters.
Problem 2 :
Newer versions of Scikit-learn require 2D inputs for the predict function and we can make 3300 2D by [[3300]]:
reg.predict( [[3300]] )
Problem1:
it depends on the default parameters which you might have changed it before or any other reason which has changed it, but you can easily set your desired parameters while you are initializing the Linear classifier in this way:
reg = linear_model.LinearRegression(copy_X=True, fit_intercept=True, n_jobs=None, normalize=False)
Problem 2:
reg.predict(3300) it's not correct to pass the parameter to Pandas in that way and you can see that the instructor has also corrected his answer to the reg.predict([3300]) in the description of the youtube Post
try this but you should define your variable and fit them to get desired output
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression()
df=pd.read_csv('homeprices.csv')
reg =LinearRegression()

plot calibration curve for machine learning

I have the code below and this code work only with the binary class so how can I use with three classes.
from sklearn.tree import DecisionTreeClassifier
import pandas as pd
from sklearn.model_selection import train_test_split
import matplotlib.pyplot as plt
import scikitplot as skp
orgnal_data = pd.read_excel("movie.xls")
# Program extracting first column
text = orgnal_data.iloc[:,0]
lable = orgnal_data.iloc[:,1]
x_train,x_test,y_train,y_test=train_test_split(fe,lable,test_size=0.30,random_state=40)
DT = DecisionTreeClassifier()
DT_y = DT.fit(x_train,y_train).predict(x_test)
clf_names = ['Decision Tree']
skp.metrics.plot_calibration_curve(y_test,DT_y,clf_names)
plt.show()
Since you use scikit-plot module, there is no function for a multiclass problem.
Read the source code here:
This function currently only works for binary classification.
So you can either 1) modify the source code or 2) open a github issue and request a function for multiclass problems.
EDIT 1:
Using scikit-learn you have some ML models that can handle multiclass problems. For example for the LinearSVC function here, the multiclass support is handled according to a one-vs-the-rest scheme.
So you can actually have models like this and then use the plot_calibration_curve function for each case (one VS rest) separately.

How to use the imbalanced library with sklearn pipeline?

I am trying to solve a text classification problem. I want to create baseline model using MultinomialNB
my data is highly imbalnced for few categories, hence decided to use the imbalanced library with sklearn pipeline and referring the tutorial.
The model is failing and giving error after introducing the two stages in pipeline as suggested in docs.
from imblearn.pipeline import make_pipeline
from sklearn.feature_extraction.text import CountVectorizer
from sklearn.feature_extraction.text import TfidfTransformer
from sklearn.naive_bayes import MultinomialNB
from imblearn.under_sampling import (EditedNearestNeighbours,
RepeatedEditedNearestNeighbours)
# Create the samplers
enn = EditedNearestNeighbours()
renn = RepeatedEditedNearestNeighbours()
pipe = make_pipeline_imb([('vect', CountVectorizer(max_features=100000,\
ngram_range= (1, 2),tokenizer=tokenize_and_stem)),\
('tfidf', TfidfTransformer(use_idf= True)),\
('enn', EditedNearestNeighbours()),\
('renn', RepeatedEditedNearestNeighbours()),\
('clf-gnb', MultinomialNB()),])
Error:
TypeError: Last step of Pipeline should implement fit. '[('vect', CountVectorizer(analyzer='word', binary=False, decode_error='strict',
Can someone please help here. I am also open to use different way of (Boosting/SMOTE) implementation as well ?
It seems that the pipeline from ìmblearn doesn't support naming like the one in sklearn. From imblearn documentation :
*steps : list of estimators.
You should modify your code to :
pipe = make_pipeline_imb( CountVectorizer(max_features=100000,\
ngram_range= (1, 2),tokenizer=tokenize_and_stem),\
TfidfTransformer(use_idf= True),\
EditedNearestNeighbours(),\
RepeatedEditedNearestNeighbours(),\
MultinomialNB())

How can I use libsvm on scikit learn?

I want to use libsvm as a classifier for predicition. I have used the following code:
import numpy as np
import sklearn
from sklearn.svm import libsvm
X = np.array([[0,1.22,45,2.111,9.344,0], [0,1.5,25,5,1,0]])
y = np.array([0.0,1.0])
clf=sklearn.svm.libsvm
clf.fit(X,y)
print(clf.predict([1,1.12,42,4.223,2.33,0]))
I got following error:
File "sklearn/svm/libsvm.pyx", line 270, in sklearn.svm.libsvm.predict (sklearn/svm/libsvm.c:3917)
TypeError: predict() takes at least 6 positional arguments (1 given)
Is this the correct way? How can I resolve the error?
Basically use sklearn.svm.SVC, since as it is stated in the documentation of sklearn, SVC is based on libsvm:
class SVC(BaseSVC):
C-Support Vector Classification.
The implementation is based on libsvm. The fit time complexity
is more than quadratic with the number of samples which makes it hard
to scale to dataset with more than a couple of 10000 samples.

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