I'm relatively new to Python and am trying to get some data prepped to train a RandomForest. For various reasons, we want the data to be discrete, so there are a few continuous variables that need to be discretized. I found qcut in pandas, which seems to do what I want - I can set a number of bins, and it will discretize the variable into that many bins, trying to keep the counts in each bin even.
However, the output of pandas.qcut is a list of Intervals, and the RandomForest classifier in scikit-learn needs a string. I found that I can convert an interval into a string by using .astype(str). Here's a quick example of what I'm doing:
import pandas as pd
from random import sample
vals = sample(range(0,100), 100)
cuts = pd.qcut(vals, q=5)
str_cuts = pd.qcut(vals, q=5).astype(str)
and then str_cuts is one of the variables passed into a random forest.
However, the intent of this system is to train a RandomForest, save it to a file, and then allow someone to load it at a later date and get a classification for a new test instance, that is not available at training time. And because the classifier was trained on discretized data, the new test instance will need to be discretized before it can be used. So what I want to be able to do is read in a new instance, apply the already-established discretization scheme to it, convert it to a string, and run it through the random forest. However, I'm getting hung up on the best way to 'apply the discretization scheme'.
Is there an easy way to handle this? I assume there's no straight-forward way to convert a string back into an Interval. I can get the list of all Interval values from the discretization (ex: cuts.unique()) and apply that at test-time, but that would require saving/loading a discretization dictionary alongside the random forest, which seems clunky, and I worry about running into issues trying to recreate a categorical variable (coming mostly from R, which is extremely particular about the format of categorical variables). Or is there another way around this that I'm not seeing?
Use the labelsargument in qcut and use pandas Categorical.
Either of those can help you create categories instead of interval for your variable. Then, you can use a form of encoding, for example Label Encoding or Ordinal Encoding to convert the categories (the factors if you're used to R) to numerical values which the Forest will be able to use.
Then the process goes :
cutting => categoricals => encoding
and you don't need to do it by hand anymore.
Lastly, some gradient boosted trees libraries have support for categorical variables though it's not a silver bullet and will depend on your goal. See catboost and lightgbm.
For future searchers, there are benefits to using transformers from scikit-learn instead of pandas. In this case, KBinsDiscretizer is the scikit equivalent of qcut.
It can be used in a pipeline, which will handle applying the previously-learned discretization to unseen data without the need for storing the discretization dictionary separately or round trip string conversion. Here's an example:
from sklearn.datasets import make_classification
from sklearn.ensemble import RandomForestClassifier
from sklearn.model_selection import train_test_split
from sklearn.pipeline import make_pipeline
from sklearn.preprocessing import KBinsDiscretizer
pipeline = make_pipeline(KBinsDiscretizer(n_bins=3, encode='ordinal', strategy='quantile'),
RandomForestClassifier())
X, y = make_classification()
X_train, X_test, y_train, y_test = train_test_split(X, y)
pipeline.fit(X_train, y_train)
predictions = pipeline.predict(X_test)
If you really need to convert back and forth between pandas IntervalIndex and string, you'll probably need to do some parsing as described in this answer: https://stackoverflow.com/a/65296110/3945991 and either use FunctionTransformer or write your own Transformer for pipeline integration.
While it may not be the cleanest-looking method, converting a string back into an interval is indeed possible:
import pandas as pd
str_intervals = [i.replace("(","").replace("]", "").split(", ") for i in str_cuts]
original_cuts = [pd.Interval(float(i), float(j)) for i, j in str_intervals]
Related
I have a dataframe with multiple x columns and one y column. I'd like to predict the linear relationship between y and multiple x variables.
so I am using smf.ols() model to predict the formula. I am wondering if I need to scale the data before fit the data using ols().
I checked ols website and it seems that they never talk about data scaling , for example, below website
https://www.statsmodels.org/devel/example_formulas.html
at the mean time, I used to take a course from datacamp and they don't mention about data scaling either, for example, below screenshot from datacamp course. You can see the regressed coefficient for each variable is not in the same order, like 3655 vs 83.
Here is what I did for my regression. I am wondering for my below example if we need to add scaling like
from sklearn.preprocessing import StandardScaler
scaler=StandardScaler()
scaler.fit(df_crossplot)
df_scaled=scaler.transform(df_crossplot)
then after that, I input df_scaled into the below function? do I have to do this above step? My hesitation is, if I scale it, then how to convert regressed formula back to a new formula based on original scale? Thanks for your help.
import statsmodels.formula.api as smf
def linear_regression_statsmodel(df_crossplot,crossplot_y,crossplot_x_list):
formula_crossplot=crossplot_y+'~'
for x in crossplot_x_list:
formula_crossplot=formula_crossplot+'+'+x
model_crossplot=smf.ols(formula=formula_crossplot,data=df_crossplot).fit()
df_crossplot['regressed']=model_crossplot.params[0]
regressed_x_string=f'{model_crossplot.params[0]:,.2f}'
for ix,x in enumerate(crossplot_x_list):
df_crossplot['regressed']=df_crossplot['regressed']+df_crossplot[x]*model_crossplot.params[ix+1]
if model_crossplot.params[ix+1]>0:
regressed_x_string=regressed_x_string+f'+{model_crossplot.params[ix+1]:,.2f}*{x}'
else: # no need + sign since we have already negative sign
regressed_x_string=regressed_x_string+f'{model_crossplot.params[ix+1]:,.2f}*{x}'
return df_crossplot,model_crossplot,regressed_x_string
So after a year of arduous work, my model is finally being implemented in my company's productive servers.
In this productive environment, my model is loaded in a Python script and a string is pulled from another server. I now have to parse this string and pass it to the model so it can make a prediction and return that output to the end user.
My current concern is efficiency. I am looking for a very fast way to convert the string to an array-like object that can be passed to my model.
Here's a replicable example:
# Load modules
from sklearn.datasets import load_breast_cancer
from sklearn.ensemble import GradientBoostingClassifier
# Load dummy data and target
X = load_breast_cancer()['data']
y = load_breast_cancer()['target']
# Initialize and fit classifier
clf = GradientBoostingClassifier(random_state=0)
clf.fit(X, y)
# [1] New string is received
string = '17.99|10.38|122.8|1001.0|0.1184|0.2776|0.3001|0.1471|0.2419|0.07871|1.095|0.9053|8.589|153.4|0.006399|0.04904|0.05373|0.01587|0.03003|0.006193|25.38|17.33|184.6|2019.0|0.1622|0.6656|0.7119|0.2654|0.4601|0.1189'
# [2] Convert string to array-like structure
import numpy as np
x = np.array(string.split('|')).astype(float)
# [3] Pass `x` to `clf` and predict probability
clf.predict_proba(x.reshape(-1, 30)).item(0)
> 0.9987537665581022
My question
Is there a more efficient way to parse a string and pass it to an sklearn model?
I think skipping the import numpy would speed things up. However, I'm open to any solution that can improve the runtime of steps [1], [2] and [3].
make sure that you indeed need double precision
and use
fromstring = np.fromstring
# ...
fromstring(string, 'f', -1, '|')
it will be 3-4x faster than
np.array(string.split('|')).astype(float)
I am trying to create an AI that reads my dataset and states whether an input outside the data is 1 or 0
My dataset has column for qualitative data and column for a boolean. Here is a sample from it:
Imports:
import pandas as pd
import numpy as np
import seaborn as sns
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.metrics import accuracy_score
from sklearn.metrics import classification_report
import re
import string
Open and cleaning dataset:
saisei_data = saisei_data.dropna(how='any',axis=0)
saisei_data = saisei_data.sample(frac=1)
X = saisei_data['Data']
y = saisei_data['Conscious']
saisei_data
Vectorisation:
from sklearn.feature_extraction.text import TfidfVectorizer
vectorization = TfidfVectorizer()
xv_train = vectorization.fit_transform(X_train)
xv_test = vectorization.fit_transform(X_test)
Example Algorithm - Logistic Regression:
LR = LogisticRegression()
LR.fit(xv_train,y_train)
pred_lr=LR.predict(xv_test) # Here is where I get an error
Everything works fine until I predict using the logistic regression algorithm.
The Error:
ValueError: X has 112 features per sample; expecting 23
This seems to change to similar errors such as:
ValueError: X has 92 features per sample; expecting 45
I am new to machine learning so I don't really know what I'm doing when it comes to using the algorithms, however I tried printing the xv_test variable and here is a sample of the output (also changes often):
Any ideas?
That is because you erroneously apply .fit_transform() to your test data; and, in this case, you are lucky enough that the process produces a programming error, thus alerting you that you are doing something methodologically wrong (which is not always the case).
We never apply either .fit() or .fit_transform() to unseen (test) data. The fitting should be done only once with the training data, like you have done here:
xv_train = vectorization.fit_transform(X_train)
For subsequent transformations of unseen (test) data, we use only .transform(). So, your next line should be
xv_test = vectorization.transform(X_test)
That way, the features in the test set will be the same with the ones in the training set, as it should be in the first place.
Notice the difference between the two methods in the docs (emphasis mine):
fit_transform:
Learn vocabulary and idf, return document-term matrix.
transform:
Transform documents to document-term matrix.
Uses the vocabulary and document frequencies (df) learned by fit (or fit_transform).
and recall that we don't ever use the test set to learn anything.
So, simple general mnemonic rule, applicable practically everywhere:
The terms "fit" and "test data" are always (always...) incompatible; mixing them will create havoc.
I have created a model for text classification using python. I have CountVectorizer and it results in a document term matrix of 2034 rows and 4063 columns ( unique words ). I saved the model I used for new test data. My new test data
test_data = ['Love', 'python', 'every','time']
But the problem is I converted the above test data tokens into a feature vector, but it differs in shape. Because the model expect a 4063 vector. I know how to solve it by taking vocabulary of CountVectorizer and search for each token in test data and putting it in that index. But is there any easy way to handle this problem in scikit-learn itself.
You should not fit a new CountVectorizer on the test data, you should use the one you fit on the training data and call transfrom(test_data) on it.
You have two ways to solve this
1. you can use the same CountVectorizer that you used for your train features like this
cv = CountVectorizer(parameters desired)
X_train = cv.fit_transform(train_data)
X_test = cv.transform(test_data)
2. You can also creat another CountVectorizer, if you really want to(but not advisable since you would be wasting space and you'd still want to use the same parameters for your CV), and use the same feature.
cv_train = CountVectorizer(parameters desired)
X_train = cv_train.fit_transform(train_data)
cv_test = CountVectorizer(vocabulary=cv_train.get_feature_names(),desired params)
X_test = cv_test.fit_transform(test_data)
try to use:
test_features = inverse_transform(test_data)
this should return you what you wish for.
I added .toarray() to the wole command in order to see the results as a matrix.
so you should write:
X_test_analyst = Pipeline.named_steps['count_vectorizer'].transform(X_test).toarray()
I'm mega late for this discussion, but I just want to leave something for people come from the search engine.
Sorry for my bad English.
;)
As mention by #Andreas Mueller, you shouldn't create a new CountVectorizer with your new data(set), u can imagine what count vectorizer do is make a 2d array(or think as a excel table), every column is a unique word, every row representing a document(or sentence), and the value (i,j) means in i^th sentence, the frequency of j^th word.
If you make a new CountVectorizer using your new data, the unique word probably(if not must) be different. When u make model.predict using this data, it will report some sort of error telling u the dim are not correct.
What I did in my code is the following:
If you train your model in different .py / .ipynb file, you can use import pickle followed by dump function for your fitted count vectorizer. You can follow the detail in this post.
If you train your model in same .py/.ipynb file, you can directly follow what #Andreas Mueller said.
code:
import pickle
pk.dump(vectorizer,open(r'/relative path','wb'))
pk.dump(pca,open(r'/relative path','wb'))
# ...
# When you want to use:
import pickle
vectoriser = pk.load(open(r'/relative path','rb'))
pea = pk.load(open(r'/relative path','rb'))
#...
Side note:
If I remember correctly, you can also export class or other things using pickle, but when you did so, make sure the class is already defined when you load the object. Not sure if this matters in this case, but I still import PCA and CountVectorizer before I did the pk.load function.
I'm just a beginner in coding so please test my code before use it in your project.
I am trying to apply a univariate feature selection method using the Python module scikit-learn to a regression (i.e. continuous valued response values) dataset in svmlight format.
I am working with scikit-learn version 0.11.
I have tried two approaches - the first of which failed and the second of which worked for my toy dataset but I believe would give meaningless results for a real dataset.
I would like advice regarding an appropriate univariate feature selection approach I could apply to select the top N features for a regression dataset. I would either like (a) to work out how to make the f_regression function work or (b) to hear alternative suggestions.
The two approaches mentioned above:
I tried using sklearn.feature_selection.f_regression(X,Y).
This failed with the following error message:
"TypeError: copy() takes exactly 1 argument (2 given)"
I tried using chi2(X,Y). This "worked" but I suspect this is because the two response values 0.1 and 1.8 in my toy dataset were being treated as class labels? Presumably, this would not yield a meaningful chi-squared statistic for a real dataset for which there would be a large number of possible response values and the number in each cell [with a particular response value and value for the attribute being tested] would be low?
Please find my toy dataset pasted into the end of this message.
The following code snippet should give the results I describe above.
from sklearn.datasets import load_svmlight_file
X_train_data, Y_train_data = load_svmlight_file(svmlight_format_train_file) #i.e. change this to the name of my toy dataset file
from sklearn.feature_selection import SelectKBest
featureSelector = SelectKBest(score_func="one of the two functions I refer to above",k=2) #sorry, I hope this message is clear
featureSelector.fit(X_train_data,Y_train_data)
print [1+zero_based_index for zero_based_index in list(featureSelector.get_support(indices=True))] #This should print the indices of the top 2 features
Thanks in advance.
Richard
Contents of my contrived svmlight file - with additional blank lines inserted for clarity:
1.8 1:1.000000 2:1.000000 4:1.000000 6:1.000000#mA
1.8 1:1.000000 2:1.000000#mB
0.1 5:1.000000#mC
1.8 1:1.000000 2:1.000000#mD
0.1 3:1.000000 4:1.000000#mE
0.1 3:1.000000#mF
1.8 2:1.000000 4:1.000000 5:1.000000 6:1.000000#mG
1.8 2:1.000000#mH
As larsmans noted, chi2 cannot be used for feature selection with regression data.
Upon updating to scikit-learn version 0.13, the following code selected the top two features (according to the f_regression test) for the toy dataset described above.
def f_regression(X,Y):
import sklearn
return sklearn.feature_selection.f_regression(X,Y,center=False) #center=True (the default) would not work ("ValueError: center=True only allowed for dense data") but should presumably work in general
from sklearn.datasets import load_svmlight_file
X_train_data, Y_train_data = load_svmlight_file(svmlight_format_train_file) #i.e. change this to the name of my toy dataset file
from sklearn.feature_selection import SelectKBest
featureSelector = SelectKBest(score_func=f_regression,k=2)
featureSelector.fit(X_train_data,Y_train_data)
print [1+zero_based_index for zero_based_index in list(featureSelector.get_support(indices=True))]
You could also try to do feature selection by L1/Lasso regularization. The class specifically designed for this is RandomizedLasso which will train LassoRegression on multiple subsamples of your data and select features that are selected most frequently by these models. You can also just use Lasso, LassoLars or SGDClassifier to do same thing without the benefit of resampling but faster.