10

In python 3 the print function must have parenthesis, so print(clf.predict([[150, 0]])) will work


5

There can be other reasons related to the model but the most simple explanation is that the data contains contradicting patterns: if the same features correspond to different target values, there is no way for any model to achieve perfect performance. Let me illustrate with a toy example: x y 0 4 0 4 1 3 1 4 1 3 1 3 2 8 2 7 2 7 2 6 ...


3

I had the same problem. It seems to be a bug in keras that occurs with nested arrays as parameters for the grid search. I was able to solve it by nesting tuples instead of arrays. So try to change layers = [[50],[50, 20], [50, 30, 15], [70,45,15,5]] to layers = [(50,),(50, 20), (50, 30, 15), (70,45,15,5)] (add a ,(comma) when only 1 value is available in ...


2

Use the below instructions as it was worked for me from sklearn.metrics import accuracy_score predictions = nordan_tree.predict(X_test) print(accuracy_score(y_test, predictions))


2

You have not defined the variable predictions anywhere. You will need to get them from your classifier somehow. You have fit your nordan_tree on your training data, now you can use the fitted nordan_tree to generate the predictions, for example like this: predictions = nordan_tree.predict(X_test) Then your line of: print(accuracy_score(y_test, predictions)...


2

The answer to this severely depends on what you mean by 'noisy' data. Are the labels noisy i.e. wrong? Or are the features noisy? Or both? If only the features are noisy, definitely use the noisy data and probably also the clean data. If only the labels, definitely do not use the noisy data. If both, is it possible to correct the labels? At the very least, ...


2

Can you post a model summary using: model.summary() Also, elaborate on how exactly the Y_train dataset works with the X_train? It's not clear how the 25 time steps from X_train data correspond to the Y_train 5 outputs.


2

Your max iteration values are strings. max_iter': ['200', '1000', '5000', '10000'] Try max_iter': [200, 1000, 5000, 10000] }


2

TL;DR Use fuzzy string matching to account for spelling mistakes. Solving Jennifer for Alice will require you to know where to look in your DB for these cases, or talk to make a better Excel file to force people to only input first names (e.g. make a an entry to a cell restricted to a given list). Fuzzy String Matching In R, you can use adist or the ...


2

I agree with @ssdecontrol that a minimal reproducible example would be the most helpful. However, looking at your code (pay attention to the sequence Error: ... Warning: ...), I believe that the issue you are experiencing is due to an inappropriate setting of R's global warn option. It appears that your current setting is likely 2, which refers to converting ...


1

See also this stackoverflow answer, if you just want the unique values you can use pandas.Series.unique() or pandas.DataFrame.drop_duplicates(). If you need the python set object you can use set(df['colname']).


1

RSS = 0 implies that the model is a perfect fit since there was no residual. The limit of the log of 0 is $-\infty$, and since lower AICs are better, and this model is perfect,it makes perfect sense that the AIC should be a negative number such that no number can be lower.


1

This will give you list of companies then you need an other loop/or list comprehension to extract from the children. for para in content1: comp_name = para.find_all('a') all_comps = [one_para.get("href") for one_para in para] print(all_comps )


1

Use list comprehensions: comp_name = [link['href'] for link in para.find_all('a')]


1

The error comes from attempting to fit a classifier (logistic regression) on a regression problem. If you are trying to predict prices (continuous outcome), you should use linear regression.


1

You could solve the typos problem with regex. With regard to knowing who does and does not go by their middle name you are going to have to tell Alice that she is in the database as Jennifer. For the regex: this is a paraphrased excerpt from the book Regular Expressions Cookbook by Jan Goyvaerts and Steven Levithan (2012). I recommend this book. To match ...


1

In general you should not attemt to make your model look better than it actually is. Always stick to the truth. Some ideas: Outliers: It is not okay to simply drop the outliers. If you want to make a prediction or do inference, you have to ask yourself how relevant such outliers are in practice. Where do they come from (are data reliable or is it an error)? ...


1

I think you need this one too: from tensorflow import set_random_seed set_random_seed(123)


1

Random guessing of value of whcih column? For example random guessing the gender of each row is: probability of it is male but guess female + vice versa which is $\frac{7}{10}\times\frac{1}{2} + \frac{3}{10}\times \frac{1}{2} = \frac{1}{2}$ and it means the error of random guessing the value of Gender is $50\%$. Also you can compute for the others by the ...


1

I got a similar error,but not the RDD Memory calibration, the problem was infact with the installation , had been upgrading part of the libraries , there was no proper handshake for some internal libraries which was pushing the Python EOF error even after tweaking the memory. Created a Virtual environment and ran Pyspark there worked as expected. Just to ...


1

Adding white noise is simple enough and should work, alternatively you could permute the values of your variable, which is another commonly used method.


1

A few mistakes that I noticed in your code: a, b = tf.placeholder(tf.float32, shape=[None, len(X)]), tf.placeholder(tf.float32, shape=[None, 8]). Python does not support assignments of a, b= c, d form. I am positively sure you are feeding wrong dimensions for X in the placeholder. I do not think creating a placeholder within a session is a good idea, since ...


1

You're getting a placeholder exception because you are not feeding a concrete value to place_holder. It looks like a is not holding a concrete value when passed to feed_dict.


1

I think this paper just compares algorithms: https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.294.9414&rep=rep1&type=pdf If you want something specific, here's the white paper for SLIPPER: https://www.aaai.org/Papers/AAAI/1999/AAAI99-049.pdf Hope that helps.


1

One can use the matplotlib.axes.Axes.errorbar class. From this example in the documentation: # example error bar values that vary with x-position error = 0.1 + 0.2 * x fig, (ax0, ax1) = plt.subplots(nrows=2, sharex=True) ax0.errorbar(x, y, yerr=error, fmt='-o') ax0.set_title('variable, symmetric error') Acknowledgments: Thanks to Emre!


1

Typical things to look for when training neural networks: Learning Curve: a plot of the error on the validation set X size of training set. The error must decrease as the network is trained, so if you see fluctuations or an increase something is wrong. Learning Rate: the learning rate influences the convergence of gradient descent. A small learning rate ...


1

from sklearn import tree features = [[2200,1], [1500,1], [1800,1], [900,2], [1000,2]] labels = ['SUV', 'SUV', 'SUV', 'hatchback', 'hatchback'] clf = tree.DecisionTreeClassifier() clf = clf.fit(features, labels) print(clf.predict([[1350, 1]]))


1

I found an answer to my question which in my opinion is correct (or at least has a point). Nevertheless it would be nice to have a validation from someone that might know more on the topic. Since the average is calculated with a trapezoidal (which in reality is a summation) then the standard deviation can also be calculated by uncertainty propagation ...


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