Consider the following machine translation problem. Let s be a source sentence and t be a target sentence. Both sentences are conceptually represented as lists of indices, where the indices correspond to the position of the words in the associated dictionaries. Example:
s = [34, 68, 91, 20]
t = [29, 0, 43]
Note that s and t don't necessarily have the same length. Now let S and T be sets of such instances. In other words, they are a parallel corpus. Example:
S = [[34, 68, 91, 20], [4, 7, 1]]
T = [[29, 0, 43], [190, 37, 25, 60]]
Note that not all s's in S have the same length. That is, sentences have variable numbers of words.
I am implementing a machine translation system in Theano, and the first design decision is what kind of data structures to use for S and T. From one of the answers posted on Matrices with different row lengths in numpy , I learnt that typed lists are a good solution for storing variable length tensors.
However, I realise that they complicate my code a lot. Let me give you one example. Say that we have two typed lists y and p_y_given_x and aim to calculate the negative loss likelihood. If they were regular tensors, a simple statement like this would suffice:
loss = t.mean(t.nnet.categorical_crossentropy(p_y_given_x, y))
But categorical_crossentropy can only be applied to tensors, so in case of typed lists I have to iterate over them and apply the function separately to each element:
_loss, _ = theano.scan(fn=lambda i, p, y: t.nnet.categorical_crossentropy(p[i], y[i]),
non_sequences=[p_y_given_x, y],
sequences=[t.arange(y.__len__(), dtype='int64')])
loss = t.mean(_loss)
On top of making my code more and more messy, these problems propagate. For instance, if I want to calculate the gradient of the loss, the following doesn't work anymore:
grad_params = t.grad(loss, params)
I don't know exactly why it doesn't work. I'm sure it has to do with the type of loss, but I am not interested in investigating any further how I could make it work. The mess is growing exponentially, and what I would like is to know whether I am using typed lists in the wrong way, or if it is time to give up on them because they are not well enough supported yet.
Typed list isn't used by anybody yet. But the idea for having them is that you iterate on them with scan for each sentence. Then you do everything you need in 1 scan. You don't do 1 scan for each operation.
So the scan is only used to do the iteration on each example in the minibatch, and the inside of scan is all what is done on one example.
We haven't tested typed list with grad yet. It is possible that it is missing some implementations.
Related
I have a dictionary given in the form {(i1,r1,m1):w, (i2,r2,m1):w, (i1,r1,m2):w ...} where i is the activity, r the type of resource, m the mode, and w is the resources of type r needed of activity i in mode m.
Now I would like to choose for every activity the mode, that requires the least resources (w). If possible, at the end in a list in the form [(i,m),...] for every i.
My tutor suggested to work with np.argmin(), but for this I have to convert the dictionary into an array. So I tried to convert the dictionary into an array:
w_list = list(w.items())
w_array = np.array(w_list)
print(w_array)
array([[(0, 1, 1), 0],
[(0, 2, 1), 0],
[(1, 1, 1), 9],
[(1, 2, 1), 0], ...
However, this array arrangement cannot be used for np.argmin.
Does anyone have any other idea how I can get the desired list mentioned above?
Here's one trivial non-numpy solution - simply create a new dictionary, and fill it with the mode and lowest cost per activity by iterating over the original dict:
w = {(i1,r1,m1): w1, (i2,r2,m1): w2, (i1,r1,m2): w3} #your original dict
result = {}
for (activity, _, mode), requiredResources in w.items():
if activity not in result or result[activity][1] > requiredResources:
result[activity] = mode, requiredResources
Now result holds a mapping from i to a tuple of m and w for the lowest w. In case of ambiguous entries for some i, the first entry in the iteration order will win (and as dicts are unordered, the iteration order is an implementation detail and dependend on things such as the specific keys and the dict size).
If you want to turn this into a list of i and m tuples, simply use a list comprehension:
resultList = [(k, v[0]) for k, v in result.items()]
An observation on the side: when confronted with any python problem, some people instantly recommend using numpy or similar libraries. IMO this is simply an expression of their own inexperience or ignorance - in many cases numpy is not just unneccessary, but actively detrimental if you don't know what you're doing.
If you're intending to seriously work with python, you would do well to first master the basics of the language (functions, classes, lists, dictionaries, loops, comprehensions, basic variable scoping rules), and get a rough overview of the vast python standard library - at least enough to know how to look up if something you need is readily available in some built-in module. Then next time when you need some functionality, you will be better equipped for deciding if this is something you can easily implement yourself (potentially with help from the standard lib), or if it makes sense to use functionality from external libraries such as numpy.
I'm trying to use Hyperopt on a regression model such that one of its hyperparameters is defined per variable and needs to be passed as a list. For example, if I have a regression with 3 independent variables (excluding constant), I would pass hyperparameter = [x, y, z] (where x, y, z are floats).
The values of this hyperparameter have the same bounds regardless of which variable they are applied to. If this hyperparameter was applied to all variables, I could simply use hp.uniform('hyperparameter', a, b). What I want the search space to be instead is a cartesian product of hp.uniform('hyperparameter', a, b) of length n, where n is the number of variables in a regression (so, basically, itertools.product(hp.uniform('hyperparameter', a, b), repeat = n))
I'd like to know whether this is possible within Hyperopt. If not, any suggestions for an optimizer where this is possible are welcome.
As noted in my comment, I am not 100% sure what you are looking for, but here is an example of using hyperopt to optimize 3 variables combination:
import random
# define an objective function
def objective(args):
v1 = args['v1']
v2 = args['v2']
v3 = args['v3']
result = random.uniform(v2,v3)/v1
return result
# define a search space
from hyperopt import hp
space = {
'v1': hp.uniform('v1', 0.5,1.5),
'v2': hp.uniform('v2', 0.5,1.5),
'v3': hp.uniform('v3', 0.5,1.5),
}
# minimize the objective over the space
from hyperopt import fmin, tpe, space_eval
best = fmin(objective, space, algo=tpe.suggest, max_evals=100)
print(best)
they all have the same search space in this case (as I understand this was your problem definition). Hyperopt aims to minimize the objective function, so running this will end up with v2 and v3 near the minimum value, and v1 near the maximum value. Since this most generally minimizes the result of the objective function.
You could use this function to create the space:
def get_spaces(a, b, num_spaces=9):
return_set = {}
for set_num in range(9):
name = str(set_num)
return_set = {
**return_set,
**{name: hp.uniform(name, a, b)}
}
return return_set
I would first define my pre-combinatorial space as a dict. The keys are names. The values are a tuple.
from hyperopt import hp
space = {'foo': (hp.choice, (False, True)), 'bar': (hp.quniform, 1, 10, 1)}
Next, produce the required combinatorial variants using loops or itertools. Each name is kept unique using a suffix or prefix.
types = (1, 2)
space = {f'{name}_{type_}': args for type_ in types for name, args in space.items()}
>>> space
{'foo_1': (<function hyperopt.pyll_utils.hp_choice(label, options)>,
(False, True)),
'bar_1': (<function hyperopt.pyll_utils.hp_quniform(label, *args, **kwargs)>,
1, 10, 1),
'foo_2': (<function hyperopt.pyll_utils.hp_choice(label, options)>,
(False, True)),
'bar_2': (<function hyperopt.pyll_utils.hp_quniform(label, *args, **kwargs)>,
1, 10, 1)}
Finally, initialize and create the actual hyperopt space:
space = {name: fn(name, *args) for name, (fn, *args) in space.items()}
values = tuple(space.values())
>>> space
{'foo_1': <hyperopt.pyll.base.Apply at 0x7f291f45d4b0>,
'bar_1': <hyperopt.pyll.base.Apply at 0x7f291f45d150>,
'foo_2': <hyperopt.pyll.base.Apply at 0x7f291f45d420>,
'bar_2': <hyperopt.pyll.base.Apply at 0x7f291f45d660>}
This was done with hyperopt 0.2.7. As a disclaimer, I strongly advise against using hyperopt because in my experience it has significantly poor performance relative to other optimizers.
Hi so I implemented this solution with optuna. The advantage of optuna is that it will create a hyperspace for all individual values, but optimizes this values in a more intelligent way and just uses one hyperparameter optimization. For example I optimized a neural network with the Batch-SIze, Learning-rate and Dropout-Rate:
The search space is much larger than the actual values being used. This safes a lot of time instead of an grid search.
The Pseudo-Code of the implementation is:
def function(trial): #trials is the parameter of optuna, which selects the next hyperparameter
distribution = [0 , 1]
a = trials.uniform("a": distribution) #this is a uniform distribution
b = trials.uniform("a": distribution)
return (a*b)-b
#This above is the function which optuna tries to optimze/minimze
For more detailed source-Code visit Optuna. It saved a lot of time for me and it was a really good result.
I'm training a CNN for image classification. The same object (with the same label then) is present in the test set twice (like two view-point). I'd like to take advantage of this when predicting the class.
Right now the final layer is a Linear layer (PyTorch) and I'm using cross-entropy as loss function. I was wondering what is the best way to take the most confident prediction for each object. Should I first compute the LogSoftMax and take the class with the highest probability (among both arrays of predictions), or should I take the logits directly?
Since LogSoftMax preserves order, the largest logit will always correspond to the highest confidence. Therefore there's no need to perform the operation if all you're interested in is finding the index of most confident class.
Probably the easiest way to get the index of the most confident class is by using torch.argmax.
e.g.
batch_size = 5
num_logits = 10
y = torch.randn(batch_size, num_logits)
preds = torch.argmax(y, dim=1)
which in this case results in
>>> print(preds)
tensor([9, 7, 2, 4, 6])
I have read an article on data leakage. In a hackathon there are two sets of data, train data on which participants train their algorithm and test set on which performance is measured.
Data leakage helps in getting a perfect score in test data, with out viewing train data by exploiting the leak.
I have read the article, but I am missing the crux how the leakage is exploited.
Steps as shown in article are following:
Let's load the test data.
Note, that we don't have any training data here, just test data. Moreover, we will not even use any features of test objects. All we need to solve this task is the file with the indices for the pairs, that we need to compare.
Let's load the data with test indices.
test = pd.read_csv('../test_pairs.csv')
test.head(10)
pairId FirstId SecondId
0 0 1427 8053
1 1 17044 7681
2 2 19237 20966
3 3 8005 20765
4 4 16837 599
5 5 3657 12504
6 6 2836 7582
7 7 6136 6111
8 8 23295 9817
9 9 6621 7672
test.shape[0]
368550
For example, we can think that there is a test dataset of images, and each image is assigned a unique Id from 0 to N−1 (N -- is the number of images). In the dataframe from above FirstId and SecondId point to these Id's and define pairs, that we should compare: e.g. do both images in the pair belong to the same class or not. So, for example for the first row: if images with Id=1427 and Id=8053 belong to the same class, we should predict 1, and 0 otherwise.
But in our case we don't really care about the images, and how exactly we compare the images (as long as comparator is binary).
print(test['FirstId'].nunique())
print(test['SecondId'].nunique())
26325
26310
So the number of pairs we are given to classify is very very small compared to the total number of pairs.
To exploit the leak we need to assume (or prove), that the total number of positive pairs is small, compared to the total number of pairs. For example: think about an image dataset with 1000 classes, N images per class. Then if the task was to tell whether a pair of images belongs to the same class or not, we would have 1000*N*(N−1)/2 positive pairs, while total number of pairs was 1000*N(1000N−1)/2.
Another example: in Quora competitition the task was to classify whether a pair of qustions are duplicates of each other or not. Of course, total number of question pairs is very huge, while number of duplicates (positive pairs) is much much smaller.
Finally, let's get a fraction of pairs of class 1. We just need to submit a constant prediction "all ones" and check the returned accuracy. Create a dataframe with columns pairId and Prediction, fill it and export it to .csv file. Then submit
test['Prediction'] = np.ones(test.shape[0])
sub=pd.DataFrame(test[['pairId','Prediction']])
sub.to_csv('sub.csv',index=False)
All ones have accuracy score is 0.500000.
So, we assumed the total number of pairs is much higher than the number of positive pairs, but it is not the case for the test set. It means that the test set is constructed not by sampling random pairs, but with a specific sampling algorithm. Pairs of class 1 are oversampled.
Now think, how we can exploit this fact? What is the leak here? If you get it now, you may try to get to the final answer yourself, othewise you can follow the instructions below.
Building a magic feature
In this section we will build a magic feature, that will solve the problem almost perfectly. The instructions will lead you to the correct solution, but please, try to explain the purpose of the steps we do to yourself -- it is very important.
Incidence matrix
First, we need to build an incidence matrix. You can think of pairs (FirstId, SecondId) as of edges in an undirected graph.
The incidence matrix is a matrix of size (maxId + 1, maxId + 1), where each row (column) i corresponds i-th Id. In this matrix we put the value 1to the position [i, j], if and only if a pair (i, j) or (j, i) is present in a given set of pais (FirstId, SecondId). All the other elements in the incidence matrix are zeros.
Important! The incidence matrices are typically very very sparse (small number of non-zero values). At the same time incidence matrices are usually huge in terms of total number of elements, and it is impossible to store them in memory in dense format. But due to their sparsity incidence matrices can be easily represented as sparse matrices. If you are not familiar with sparse matrices, please see wiki and scipy.sparse reference. Please, use any of scipy.sparseconstructors to build incidence matrix.
For example, you can use this constructor: scipy.sparse.coo_matrix((data, (i, j))). We highly recommend to learn to use different scipy.sparseconstuctors, and matrices types, but if you feel you don't want to use them, you can always build this matrix with a simple for loop. You will need first to create a matrix using scipy.sparse.coo_matrix((M, N), [dtype]) with an appropriate shape (M, N) and then iterate through (FirstId, SecondId) pairs and fill corresponding elements in matrix with ones.
Note, that the matrix should be symmetric and consist only of zeros and ones. It is a way to check yourself.
import networkx as nx
import numpy as np
import pandas as pd
import scipy.sparse
import matplotlib.pyplot as plt
test = pd.read_csv('../test_pairs.csv')
x = test[['FirstId','SecondId']].rename(columns={'FirstId':'col1', 'SecondId':'col2'})
y = test[['SecondId','FirstId']].rename(columns={'SecondId':'col1', 'FirstId':'col2'})
comb = pd.concat([x,y],ignore_index=True).drop_duplicates(keep='first')
comb.head()
col1 col2
0 1427 8053
1 17044 7681
2 19237 20966
3 8005 20765
4 16837 599
data = np.ones(comb.col1.shape, dtype=int)
inc_mat = scipy.sparse.coo_matrix((data,(comb.col1,comb.col2)), shape=(comb.col1.max() + 1, comb.col1.max() + 1))
rows_FirstId = inc_mat[test.FirstId.values,:]
rows_SecondId = inc_mat[test.SecondId.values,:]
f = rows_FirstId.multiply(rows_SecondId)
f = np.asarray(f.sum(axis=1))
f.shape
(368550, 1)
f = f.sum(axis=1)
f = np.squeeze(np.asarray(f))
print (f.shape)
Now build the magic feature
Why did we build the incidence matrix? We can think of the rows in this matix as of representations for the objects. i-th row is a representation for an object with Id = i. Then, to measure similarity between two objects we can measure similarity between their representations. And we will see, that such representations are very good.
Now select the rows from the incidence matrix, that correspond to test.FirstId's, and test.SecondId's.
So do not forget to convert pd.series to np.array
These lines should normally run very quickly
rows_FirstId = inc_mat[test.FirstId.values,:]
rows_SecondId = inc_mat[test.SecondId.values,:]
Our magic feature will be the dot product between representations of a pair of objects. Dot product can be regarded as similarity measure -- for our non-negative representations the dot product is close to 0 when the representations are different, and is huge, when representations are similar.
Now compute dot product between corresponding rows in rows_FirstId and rows_SecondId matrices.
From magic feature to binary predictions
But how do we convert this feature into binary predictions? We do not have a train set to learn a model, but we have a piece of information about test set: the baseline accuracy score that you got, when submitting constant. And we also have a very strong considerations about the data generative process, so probably we will be fine even without a training set.
We may try to choose a thresold, and set the predictions to 1, if the feature value f is higer than the threshold, and 0 otherwise. What threshold would you choose?
How do we find a right threshold? Let's first examine this feature: print frequencies (or counts) of each value in the feature f.
For example use np.unique function, check for flags
Function to count frequency of each element
from scipy.stats import itemfreq
itemfreq(f)
array([[ 14, 183279],
[ 15, 852],
[ 19, 546],
[ 20, 183799],
[ 21, 6],
[ 28, 54],
[ 35, 14]])
Do you see how this feature clusters the pairs? Maybe you can guess a good threshold by looking at the values?
In fact, in other situations it can be not that obvious, but in general to pick a threshold you only need to remember the score of your baseline submission and use this information.
Choose a threshold below:
pred = f > 14 # SET THRESHOLD HERE
pred
array([ True, False, True, ..., False, False, False], dtype=bool)
submission = test.loc[:,['pairId']]
submission['Prediction'] = pred.astype(int)
submission.to_csv('submission.csv', index=False)
I want to understand the idea behind this. How we are exploiting the leak from the test data only.
There's a hint in the article. The number of positive pairs should be 1000*N*(N−1)/2, while the number of all pairs is 1000*N(1000N−1)/2. Of course, the number of all pairs is much, much larger if the test set was sampled at random.
As the author mentions, after you evaluate your constant prediction of 1s on the test set, you can tell that the sampling was not done at random. The accuracy you obtain is 50%. Had the sampling been done correctly, this value should've been much lower.
Thus, they construct the incidence matrix and calculate the dot product (the measure of similarity) between the representations of our ID features. They then reuse the information about the accuracy obtained with constant predictions (at 50%) to obtain the corresponding threshold (f > 14). It's set to be greater than 14 because that constitutes roughly half of our test set, which in turn maps back to the 50% accuracy.
The "magic" value didn't have to be greater than 14. It could have been equal to 14. You could have adjusted this value after some leader board probing (as long as you're capturing half of the test set).
It was observed that the test data was not sampled properly; same-class pairs were oversampled. Thus there is a much higher probability of each pair in the training set to have target=1 than any random pair. This led to the belief that one could construct a similarity measure based only on the pairs that are present in the test, i.e., whether a pair made it to the test is itself a strong indicator of similarity.
Using this insight one can calculate an incidence matrix and represent each id j as a binary array (the i-th element representing the presence of i-j pair in test, and thus representing the strong probability of similarity between them). This is a pretty accurate measure, allowing one to find the "similarity" between two rows just by taking their dot product.
The cutoff arrived at is purely by the knowledge of target-distribution found by leaderboard probing.
When we posted a homework assignment about PCA we told the course participants to pick any way of calculating the eigenvectors they found. They found multiple ways: eig, eigh (our favorite was svd). In a later task we told them to use the PCAs from scikit-learn - and were surprised that the results differed a lot more than we expected.
I toyed around a bit and we posted an explanation to the participants that either solution was correct and probably just suffered from numerical instabilities in the algorithms. However, recently I picked that file up again during a discussion with a co-worker and we quickly figured out that there's an interesting subtle change to make to get all results to be almost equivalent: Transpose the eigenvectors obtained from the SVD (and thus from the PCAs).
A bit of code to show this:
def pca_eig(data):
"""Uses numpy.linalg.eig to calculate the PCA."""
data = data.T # data
val, vec = np.linalg.eig(data)
return val, vec
versus
def pca_svd(data):
"""Uses numpy.linalg.svd to calculate the PCA."""
u, s, v = np.linalg.svd(data)
return s ** 2, v
Does not yield the same result. Changing the return of pca_svd to s ** 2, v.T, however, works! It makes perfect sense following the definition by wikipedia: The SVD of X follows X=UΣWT where
the right singular vectors W of X are equivalent to the eigenvectors of XTX
So to get the eigenvectors we need to transposed the output v of np.linalg.eig(...).
Unless there is something else going on? Anyway, the PCA and IncrementalPCA both show wrong results (or eig is wrong? I mean, transposing that yields the same equality), and looking at the code for PCA reveals that they are doing it as I did it initially:
U, S, V = linalg.svd(X, full_matrices=False)
# flip eigenvectors' sign to enforce deterministic output
U, V = svd_flip(U, V)
components_ = V
I created a little gist demonstrating the differences (nbviewer), the first with PCA and IncPCA as they are (also no transposition of the SVD), the second with transposed eigenvectors:
Comparison without transposition of SVD/PCAs (normalized data)
Comparison with transposition of SVD/PCAs (normalized data)
As one can clearly see, in the upper image the results are not really great, while the lower image only differs in some signs, thus mirroring the results here and there.
Is this really wrong and a bug in scikit-learn? More likely I am using the math wrong – but what is right? Can you please help me?
If you look at the documentation, it's pretty clear from the shape that the eigenvectors are in the rows, not the columns.
The point of the sklearn PCA is that you can use the transform method to do the correct transformation.