Welcome to the first project of the Machine Learning Engineer Nanodegree! In this notebook, some template code has already been provided for you, and you will need to implement additional functionality to successfully complete this project. You will not need to modify the included code beyond what is requested. Sections that begin with 'Implementation' in the header indicate that the following block of code will require additional functionality which you must provide. Instructions will be provided for each section and the specifics of the implementation are marked in the code block with a 'TODO' statement. Please be sure to read the instructions carefully!
In addition to implementing code, there will be questions that you must answer which relate to the project and your implementation. Each section where you will answer a question is preceded by a 'Question X' header. Carefully read each question and provide thorough answers in the following text boxes that begin with 'Answer:'. Your project submission will be evaluated based on your answers to each of the questions and the implementation you provide.
Note: Code and Markdown cells can be executed using the Shift + Enter keyboard shortcut. In addition, Markdown cells can be edited by typically double-clicking the cell to enter edit mode.
In this project, you will evaluate the performance and predictive power of a model that has been trained and tested on data collected from homes in suburbs of Boston, Massachusetts. A model trained on this data that is seen as a good fit could then be used to make certain predictions about a home — in particular, its monetary value. This model would prove to be invaluable for someone like a real estate agent who could make use of such information on a daily basis.
The dataset for this project originates from the UCI Machine Learning Repository. The Boston housing data was collected in 1978 and each of the 506 entries represent aggregated data about 14 features for homes from various suburbs in Boston, Massachusetts. For the purposes of this project, the following preprocessing steps have been made to the dataset:
'MEDV' value of 50.0. These data points likely contain missing or censored values and have been removed.'RM' value of 8.78. This data point can be considered an outlier and has been removed.'RM', 'LSTAT', 'PTRATIO', and 'MEDV' are essential. The remaining non-relevant features have been excluded.'MEDV' has been multiplicatively scaled to account for 35 years of market inflation.Run the code cell below to load the Boston housing dataset, along with a few of the necessary Python libraries required for this project. You will know the dataset loaded successfully if the size of the dataset is reported.
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# Import libraries necessary for this project
import numpy as np
import pandas as pd
import visuals as vs # Supplementary code
from sklearn.cross_validation import ShuffleSplit
# Pretty display for notebooks
%matplotlib inline
# Load the Boston housing dataset
data = pd.read_csv('housing.csv')
prices = data['MEDV']
features = data.drop('MEDV', axis = 1)
# Success
print "Boston housing dataset has {} data points with {} variables each.".format(*data.shape)
In this first section of this project, you will make a cursory investigation about the Boston housing data and provide your observations. Familiarizing yourself with the data through an explorative process is a fundamental practice to help you better understand and justify your results.
Since the main goal of this project is to construct a working model which has the capability of predicting the value of houses, we will need to separate the dataset into features and the target variable. The features, 'RM', 'LSTAT', and 'PTRATIO', give us quantitative information about each data point. The target variable, 'MEDV', will be the variable we seek to predict. These are stored in features and prices, respectively.
For your very first coding implementation, you will calculate descriptive statistics about the Boston housing prices. Since numpy has already been imported for you, use this library to perform the necessary calculations. These statistics will be extremely important later on to analyze various prediction results from the constructed model.
In the code cell below, you will need to implement the following:
'MEDV', which is stored in prices.
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print prices.head()
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print prices.describe()
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print np.mean(prices)
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# TODO: Minimum price of the data
minimum_price = np.min(prices)
# TODO: Maximum price of the data
maximum_price = np.max(prices)
# TODO: Mean price of the data
mean_price = np.mean(prices)
# TODO: Median price of the data
median_price = np.median(prices)
# TODO: Standard deviation of prices of the data
std_price = np.std(prices)
# Show the calculated statistics
print "Statistics for Boston housing dataset:\n"
print "Minimum price: ${:,.2f}".format(minimum_price)
print "Maximum price: ${:,.2f}".format(maximum_price)
print "Mean price: ${:,.2f}".format(mean_price)
print "Median price ${:,.2f}".format(median_price)
print "Standard deviation of prices: ${:,.2f}".format(std_price)
As a reminder, we are using three features from the Boston housing dataset: 'RM', 'LSTAT', and 'PTRATIO'. For each data point (neighborhood):
'RM' is the average number of rooms among homes in the neighborhood.'LSTAT' is the percentage of homeowners in the neighborhood considered "lower class" (working poor).'PTRATIO' is the ratio of students to teachers in primary and secondary schools in the neighborhood.Using your intuition, for each of the three features above, do you think that an increase in the value of that feature would lead to an increase in the value of 'MEDV' or a decrease in the value of 'MEDV'? Justify your answer for each.
Hint: Would you expect a home that has an 'RM' value of 6 be worth more or less than a home that has an 'RM' value of 7?
Answer: I would think that an increase in the average number of rooms in a home(RM) would lead to an increase in MEDV(prices).Justification the larger the home generally the more rooms although mansions to not fit this line of thought-square footage might be a better predictor; I would predict that an increase in the value of LSTAT would decrease MEDV-jusitification beacuse the lower the salary the less expendible income to take care of yard and home maintenence and the lower the slary the less free hours to work on the home upkeep, (although I personally do not believe this bias is true) and lastly an increase in PTRATIO would lead to a decrease in MEDV-justification- the higher the number of students in the class the less personal attention so parent would want a lower student teacher ratio. A school area that has a loweer student to teacher ratio would be more desirable and higher student to teacher ration would be less desirable
In this second section of the project, you will develop the tools and techniques necessary for a model to make a prediction. Being able to make accurate evaluations of each model's performance through the use of these tools and techniques helps to greatly reinforce the confidence in your predictions.
It is difficult to measure the quality of a given model without quantifying its performance over training and testing. This is typically done using some type of performance metric, whether it is through calculating some type of error, the goodness of fit, or some other useful measurement. For this project, you will be calculating the coefficient of determination, R2, to quantify your model's performance. The coefficient of determination for a model is a useful statistic in regression analysis, as it often describes how "good" that model is at making predictions.
The values for R2 range from 0 to 1, which captures the percentage of squared correlation between the predicted and actual values of the target variable. A model with an R2 of 0 always fails to predict the target variable, whereas a model with an R2 of 1 perfectly predicts the target variable. Any value between 0 and 1 indicates what percentage of the target variable, using this model, can be explained by the features. A model can be given a negative R2 as well, which indicates that the model is no better than one that naively predicts the mean of the target variable.
For the performance_metric function in the code cell below, you will need to implement the following:
r2_score from sklearn.metrics to perform a performance calculation between y_true and y_predict.score variable.
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# TODO: Import 'r2_score'
from sklearn.metrics import r2_score
def performance_metric(y_true, y_predict ):
""" Calculates and returns the performance score between
true and predicted values based on the metric chosen. """
# TODO: Calculate the performance score between 'y_true' and 'y_predict'
score = r2_score(y_true, y_predict)
# Return the score
return score
Assume that a dataset contains five data points and a model made the following predictions for the target variable:
| True Value | Prediction |
|---|---|
| 3.0 | 2.5 |
| -0.5 | 0.0 |
| 2.0 | 2.1 |
| 7.0 | 7.8 |
| 4.2 | 5.3 |
Would you consider this model to have successfully captured the variation of the target variable? Why or why not?
Run the code cell below to use the performance_metric function and calculate this model's coefficient of determination.
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# Calculate the performance of this model
score = performance_metric([3, -0.5, 2, 7, 4.2], [2.5, 0.0, 2.1, 7.8, 5.3])
print "Model has a coefficient of determination, R^2, of {:.3f}.".format(score)
Answer:I think the model has reasonably captured the variation in the data because it has fit a rsquared formula which delivers a 0.923 out of 1 or 92% correlation between the actual and predicted values. Possibility there is room to improve this
Your next implementation requires that you take the Boston housing dataset and split the data into training and testing subsets. Typically, the data is also shuffled into a random order when creating the training and testing subsets to remove any bias in the ordering of the dataset.
For the code cell below, you will need to implement the following:
train_test_split from sklearn.cross_validation to shuffle and split the features and prices data into training and testing sets.random_state for train_test_split to a value of your choice. This ensures results are consistent.X_train, X_test, y_train, and y_test.
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from sklearn import cross_validation
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from sklearn.cross_validation import train_test_split
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# TODO: Import 'train_test_split'
# TODO: Shuffle and split the data into training and testing subsets
def shuffle_slpit_data(X, y):
X_train, X_test, y_train, y_test = train_test_split(features, prices, test_size=0.2, random_state=1)
return X_train, y_train, X_test, y_test
# Success
print "Angie, your data has been split"
Answer: Benefit: Splitting the data into multiple sets allows the alorgorithm to be tested against a set it has not seen before. It tests the robustness of the algorithm against unseen data such as would be encountered in the real world.What could go wrong if you do not have a way to test your model is that the model can learn and perform very well using all the data and the only way that you would know that it actually performs poorly on unseen data is when you go into production against unseen data and it performs poorly. If you had split and tested against the unseen held back data then you would have seen this during model creation.
In this third section of the project, you'll take a look at several models' learning and testing performances on various subsets of training data. Additionally, you'll investigate one particular algorithm with an increasing 'max_depth' parameter on the full training set to observe how model complexity affects performance. Graphing your model's performance based on varying criteria can be beneficial in the analysis process, such as visualizing behavior that may not have been apparent from the results alone.
The following code cell produces four graphs for a decision tree model with different maximum depths. Each graph visualizes the learning curves of the model for both training and testing as the size of the training set is increased. Note that the shaded region of a learning curve denotes the uncertainty of that curve (measured as the standard deviation). The model is scored on both the training and testing sets using R2, the coefficient of determination.
Run the code cell below and use these graphs to answer the following question.
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# Produce learning curves for varying training set sizes and maximum depths
vs.ModelLearning(features, prices)
Choose one of the graphs above and state the maximum depth for the model. What happens to the score of the training curve as more training points are added? What about the testing curve? Would having more training points benefit the model?
Hint: Are the learning curves converging to particular scores?
Answer: max_depth = 3 model. As the training points increase the scores goes from one to near 0.8. The testing score increases from just above 0.6 to just below 0.8. It appears having more training points would not benefit max depth 1,3 or 10. In the Max depth 6 graph the final data point score appears to possibly be less than the point before, so this depth could possible perform worse as more training points are added. In max_depth 3 graph it appears the learning curves might be converging towards the score of 0.8. The other graphs also appear to be converging except max_depth = 10 which appears to be more parallel.
The following code cell produces a graph for a decision tree model that has been trained and validated on the training data using different maximum depths. The graph produces two complexity curves — one for training and one for validation. Similar to the learning curves, the shaded regions of both the complexity curves denote the uncertainty in those curves, and the model is scored on both the training and validation sets using the performance_metric function.
Run the code cell below and use this graph to answer the following two questions.
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vs.ModelComplexity(X_train, y_train)
When the model is trained with a maximum depth of 1, does the model suffer from high bias or from high variance? How about when the model is trained with a maximum depth of 10? What visual cues in the graph justify your conclusions?
Hint: How do you know when a model is suffering from high bias or high variance?
Answer: I had to lok up the definitions of high bias and high variance from http://www.astroml.org/sklearn_tutorial/practical.html to help answer this question. High bias is underfitting the data. Indicated when both the training and cross-validation errors are very high which means the r2 is low. If this is the case, I can add more features,use a more sophisticated model, use fewer samples or decrease any regularization (penalty terms) that may be in the algorithm. They noted that adding more training data will not help matters if both lines have converged to a relatively high error. High Variance is overfitting. Indicated when the training error is much less than the cross-validation error. If this is the case, adding more training data may not help matters, the training error will climb and the cross validation error will decrease until they begin to converge and both lines tend to converge to a relatively high error. To fix high variance I can use fewer features,use more training samples and/or increase regularization (add penalty terms). Looking at the graphs here it appears that when max depth is 1 both the training error is high and the validation error is high (low r2) which indicates high bias (under fitting). When the max depth is 10 the training error is low (high r2) but the validation error is high (lowr2) which indicates high variance. The visual clues are the score points in relationship to the max depth. The shaded uncertainty appears relatively consistant decreasing on the traing score at max depth 10 which makes sense if it is overfitting the data
Answer: I would guess max depth 4 based upon the graph. It appears that overfitting (high variance) is begining at max depth 6 and the error does not appear improved at max depth 5 as compared to 4
Answer: To answer this question I went to http://scikit-learn.org/stable/modules/generated/sklearn.grid_search.GridSearchCV.html and http://scikit-learn.org/stable/modules/generated/sklearn.metrics.make_scorer.html#sklearn.metrics.make_scorer The grid search technique searches over the parameter values to return estimators. Because it can help determine the best parameters for the model it can be useful for parameter tuning
Answer:I had to research to answer this question. I used https://www.cs.cmu.edu/~schneide/tut5/node42.html to find the definition of the technique: "K-fold cross validation is one way to improve over the holdout method. The data set is divided into k subsets, and the holdout method is repeated k times. Each time, one of the k subsets is used as the test set and the other k-1 subsets are put together to form a training set" It seems to benefit grid search because each data point is in the test set exactly once (and in the training set k-1 times.) which then implies that using grid search without a cross validation set you could overfit or increase the variance
Your final implementation requires that you bring everything together and train a model using the decision tree algorithm. To ensure that you are producing an optimized model, you will train the model using the grid search technique to optimize the 'max_depth' parameter for the decision tree. The 'max_depth' parameter can be thought of as how many questions the decision tree algorithm is allowed to ask about the data before making a prediction. Decision trees are part of a class of algorithms called supervised learning algorithms.
For the fit_model function in the code cell below, you will need to implement the following:
DecisionTreeRegressor from sklearn.tree to create a decision tree regressor object.'regressor' variable.'max_depth' with the values from 1 to 10, and assign this to the 'params' variable.make_scorer from sklearn.metrics to create a scoring function object.performance_metric function as a parameter to the object.'scoring_fnc' variable.GridSearchCV from sklearn.grid_search to create a grid search object.'regressor', 'params', 'scoring_fnc', and 'cv_sets' as parameters to the object. GridSearchCV object to the 'grid' variable.
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from sklearn import grid_search
from sklearn import tree
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from sklearn.metrics import mean_squared_error, make_scorer
from sklearn.grid_search import GridSearchCV
from sklearn.tree import DecisionTreeRegressor
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# TODO: Import 'make_scorer', 'DecisionTreeRegressor', and 'GridSearchCV'
def fit_model(X, y):
""" Performs grid search over the 'max_depth' parameter for a
decision tree regressor trained on the input data [X, y]. """
# Create cross-validation sets from the training data
cv_sets = ShuffleSplit(X.shape[0], n_iter = 10, test_size = 0.20, random_state = 0)
# TODO: Create a decision tree regressor object
regressor = DecisionTreeRegressor()
# TODO: Create a dictionary for the parameter 'max_depth' with a range from 1 to 10
params = {'max_depth': (1,2,3,4,5,6,7,8,9,10)}
# TODO: Transform 'performance_metric' into a scoring function using 'make_scorer'
scoring_function = make_scorer(score_func = mean_squared_error, greater_is_better = False)
# TODO: Create the grid search object
grid = GridSearchCV(estimator = regressor,param_grid = params,scoring = scoring_function,cv = cv_sets)
# Fit the grid search object to the data to compute the optimal model
grid = grid.fit(X, y)
# Return the optimal model after fitting the data
return grid.best_estimator_
try:
grid = fit_model(features, prices)
print "Yea, fit a model!", grid
except:
print "Something went wrong with fitting a model."
Once a model has been trained on a given set of data, it can now be used to make predictions on new sets of input data. In the case of a decision tree regressor, the model has learned what the best questions to ask about the input data are, and can respond with a prediction for the target variable. You can use these predictions to gain information about data where the value of the target variable is unknown — such as data the model was not trained on.
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# Fit the training data to the model using grid search
reg = fit_model(X_train, y_train)
# Produce the value for 'max_depth'
print "Parameter 'max_depth' is {} for the optimal model.".format(reg.get_params()['max_depth'])
Answer: It choose max depth 5 and I quessed that max depth would be 4 for the optimal model
Imagine that you were a real estate agent in the Boston area looking to use this model to help price homes owned by your clients that they wish to sell. You have collected the following information from three of your clients:
| Feature | Client 1 | Client 2 | Client 3 |
|---|---|---|---|
| Total number of rooms in home | 5 rooms | 4 rooms | 8 rooms |
| Neighborhood poverty level (as %) | 17% | 32% | 3% |
| Student-teacher ratio of nearby schools | 15-to-1 | 22-to-1 | 12-to-1 |
What price would you recommend each client sell his/her home at? Do these prices seem reasonable given the values for the respective features?
Hint: Use the statistics you calculated in the Data Exploration section to help justify your response.
Run the code block below to have your optimized model make predictions for each client's home.
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# Produce a matrix for client data
client_data = [[5, 17, 15], # Client 1
[4, 32, 22], # Client 2
[8, 3, 12]] # Client 3
# Show predictions
for i, price in enumerate(reg.predict(client_data)):
print "Predicted selling price for Client {}'s home: ${:,.2f}".format(i+1, price)
Answer: The statistics show that the: Minimum price: 105,000.00 Maximum price: 1,024,800.00 Mean price: 454,342.94 Median price 438,900.00 Standard deviation of prices: 165,171.13 so the values appear within reason based upon the statistics of the sales prices the poverty levels and the student teacher ratios. Although client 2 home has an extremely low valuation it would be interesting to see if the current comps maintain that low level over time. To elaborate as requested: client 1 has 5 rooms, a median poverty level and a median student teacher povery level as compared to the other 2 so a predicted selling price of 419,000 appears within the median of the 3 estimates(consistant with the median statistical home price), as is the case for home 2 which has fewer rooms than client 1 but the poverty level is the highest of the 3 and the student teacher ratio is also the highest of the 3 so the lowest valuation matches the the expectation of closest to the minimum price of the 3; also client 3 has 8 rooms, 3% poverty level and the best student teacher ratio which should place it closest to the maximum price of the three. In summation, the features of the clients homes(room.poverty level,student teacher level) which are relatively consistant with anticipated poor valuations and are consistant in their relative comparison to the statistical results also based upon rooms,poverty level and student/teacher ratio. The model appears consistant with the data and features presented. Although one cannot say of this result that these three features are the best predictor of home prices in totality
An optimal model is not necessarily a robust model. Sometimes, a model is either too complex or too simple to sufficiently generalize to new data. Sometimes, a model could use a learning algorithm that is not appropriate for the structure of the data given. Other times, the data itself could be too noisy or contain too few samples to allow a model to adequately capture the target variable — i.e., the model is underfitted. Run the code cell below to run the fit_model function ten times with different training and testing sets to see how the prediction for a specific client changes with the data it's trained on.
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vs.PredictTrials(features, prices, fit_model, client_data)
In a few sentences, discuss whether the constructed model should or should not be used in a real-world setting.
Hint: Some questions to answering:
Answer: I don't think the constructed model can be used today in a real world setting. Pricing data from the 1978 would be irrelevant for a pricing prediction today unless the model consistantly updated and trained to keep up with pricing changes dynamically. The features in the data are realatively sufficient to descibe a home but their are variables (features that can make a sugnificant difference in a home price such as a pool or tennis court/putting green/proximity to ocean/view etc. The model appears to have a 69,044.61 difference in price range in its predictions. That is very significant and it would be better to have a model that predicts within a maximum 10,000 range for the real estate market. The data collected in a urban city like boston would have difficulty being applicable in a rural city or in a city like New York where 1 bedroom can be 1 million Or Los Angeles where 2 rooms could be a 5,00 square foot loft. For realestate, I think the model should betrained on data from that individual city if possible to increase accuracy and robustness of the model