Is there any way I can extract the phase from the Lomb Scargle periodogram? I'm using the LombScargle implementation from gatspy.
from gatspy.periodic import LombScargleFast
model = LombScargleFast().fit(t, y)
periods, power = model.periodogram_auto()
frequencies = 1 / periods
fig, ax = plt.subplots()
ax.plot(frequencies, power)
plt.show()
Power gives me an absolute value. Any way I can extract the phase for each frequency as I can for a discrete fourier transform.
The Lomb-Scargle method produces a periodogram, i.e., powers at each frequency. This is in order to be able to be performant, compared to directly least-squares fitting a sinusoidal model. I don't know about gatspy, but astropy does allow you to compute the best phase for a specific frequency of interest, see http://docs.astropy.org/en/stable/stats/lombscargle.html#the-lomb-scargle-model . I imagine doing this for many frequencies is many times slower than computing the periodogram.
-EDIT-
The docs outline moved to:
https://docs.astropy.org/en/stable/timeseries/lombscargle.html
let's consider that you're looking for a specific frequency fo. Then the corresponding period can be given by P = 1/fo.
We can define a function, as in below:
def phase_plot(t,period):
#t is the array of timesteps
phases = (time/period)%1.
this will give you all the phases for that particular frequency of interest.
Related
I have snippets of audio that are almost the same that I want to group together (samples 5 and 3 below). There are other portions that are similar, but differ (3 and 4, there is a double drum hit at the end for 3) and completely different ones (sample 8).
How can I group together samples that are almost the same? I tried taking the difference (attempting to minimize it), but that does not work since they are not aligned. I also tried to take audio features like pitch distribution, but since the sounds are similar in pitch those don't get separated well.
The files are available here: https://drive.google.com/drive/folders/14UQQDfIBUNRO_1Pv8bkPf9noi86M7lKd
Here's something that appears to work for the data you are using but may (likely does) have weaknesses when it comes to other data or other sorts of data. But maybe it will be helpful nonetheless.
The basic idea of this solution is to compute the MFCCs of each of the samples to get feature vectors and then find a distance (here just using basic Euclidean distance) between those feature sets with the assumption (which seems to be true for your data) that the least similar samples will have a large distance and the closest will have the least. Here's the code:
import librosa
import scipy
import matplotlib.pyplot as plt
sample3, rate = librosa.load('sample3.wav', sr=None)
sample4, rate = librosa.load('sample4.wav', sr=None)
sample5, rate = librosa.load('sample5.wav', sr=None)
sample8, rate = librosa.load('sample8.wav', sr=None)
# cut the longer sounds to same length as the shortest
len5 = len(sample5)
sample3 = sample3[:len5]
sample4 = sample4[:len5]
sample8 = sample8[:len5]
mf3 = librosa.feature.mfcc(sample3, sr=rate)
mf4 = librosa.feature.mfcc(sample4, sr=rate)
mf5 = librosa.feature.mfcc(sample5, sr=rate)
mf8 = librosa.feature.mfcc(sample8, sr=rate)
# average across the frames. dubious?
amf3 = mf3.mean(axis=0)
amf4 = mf4.mean(axis=0)
amf5 = mf5.mean(axis=0)
amf8 = mf8.mean(axis=0)
f_list = [amf3, amf4, amf5, amf8]
results = []
for i, features_a in enumerate(f_list):
results.append([])
for features_b in f_list:
result = scipy.spatial.distance.euclidean(features_a,
features_b)
results[i].append(result)
plt.ion()
fig, ax = plt.subplots()
ax.imshow(results, cmap='gray_r', interpolation='nearest')
spots = [0, 1, 2, 3]
labels = ['s3', 's4', 's5', 's8']
ax.set_xticks(spots)
ax.set_xticklabels(labels)
ax.set_yticks(spots)
ax.set_yticklabels(labels)
The code plots a heatmap of the distances across all the samples. The code is lazy so it both re-computes the elements that are symmetric across the diagonal, which are the same, and the diagonal itself (which should be zero distance) but those are sort of sanity checks as it is nice to see white down the diagonal and that the matrix is symmetric.
The real information is that clip 8 is black against all the other clips (i.e. furthest from them) and clip 3 and clip 5 are the least distant from one another.
This basic idea could be done with a feature vector generated in a different sort of way (e.g. instead of MFCCs, you could use the embeddings from something like YAMNet) or with a different way of finding a distance between the feature vectors.
For the grouping part of what you want to do, you could experimentally work out a threshold on the distance metric below which you would consider a clip to be in the same group as another. With more clips, you could compute all these distances and then hand that distance matrix over to a clustering algorithm (like HDBSCAN) to cluster the clips.
I am doing a project on multiclass semantic segmentation. I have formulated a model that outputs pretty descent segmented images by decreasing the loss value. However, I cannot evaluate the model performance in metrics, such as meanIoU or Dice coefficient.
In case of binary semantic segmentation it was easy just to set the threshold of 0.5, to classify the outputs as an object or background, but it does not work in the case of multiclass semantic segmentation. Could you please tell me how to obtain model performance on the aforementioned metrics? Any help will be highly appreciated!
By the way, I am using PyTorch framework and CamVid dataset.
If anyone is interested in this answer, please also look at this issue. The author of the issue points out that mIoU can be computed in a different way (and that method is more accepted in literature). So, consider that before using the implementation for any formal publication.
Basically, the other method suggested by the issue-poster is to separately accumulate the intersections and unions over the entire dataset and divide them at the final step. The method in the below original answer computes intersection and union for a batch of images, then divides them to get IoU for the current batch, and then takes a mean of the IoUs over the entire dataset.
However, this below given original method is problematic because the final mean IoU would vary with the batch-size. On the other hand, the mIoU would not vary with the batch size for the method mentioned in the issue as the separate accumulation would ensure that batch size is irrelevant (though higher batch size can definitely help speed up the evaluation).
Original answer:
Given below is an implementation of mean IoU (Intersection over Union) in PyTorch.
def mIOU(label, pred, num_classes=19):
pred = F.softmax(pred, dim=1)
pred = torch.argmax(pred, dim=1).squeeze(1)
iou_list = list()
present_iou_list = list()
pred = pred.view(-1)
label = label.view(-1)
# Note: Following for loop goes from 0 to (num_classes-1)
# and ignore_index is num_classes, thus ignore_index is
# not considered in computation of IoU.
for sem_class in range(num_classes):
pred_inds = (pred == sem_class)
target_inds = (label == sem_class)
if target_inds.long().sum().item() == 0:
iou_now = float('nan')
else:
intersection_now = (pred_inds[target_inds]).long().sum().item()
union_now = pred_inds.long().sum().item() + target_inds.long().sum().item() - intersection_now
iou_now = float(intersection_now) / float(union_now)
present_iou_list.append(iou_now)
iou_list.append(iou_now)
return np.mean(present_iou_list)
Prediction of your model will be in one-hot form, so first take softmax (if your model doesn't already) followed by argmax to get the index with the highest probability at each pixel. Then, we calculate IoU for each class (and take the mean over it at the end).
We can reshape both the prediction and the label as 1-D vectors (I read that it makes the computation faster). For each class, we first identify the indices of that class using pred_inds = (pred == sem_class) and target_inds = (label == sem_class). The resulting pred_inds and target_inds will have 1 at pixels labelled as that particular class while 0 for any other class.
Then, there is a possibility that the target does not contain that particular class at all. This will make that class's IoU calculation invalid as it is not present in the target. So, you assign such classes a NaN IoU (so you can identify them later) and not involve them in the calculation of the mean.
If the particular class is present in the target, then pred_inds[target_inds] will give a vector of 1s and 0s where indices with 1 are those where prediction and target are equal and zero otherwise. Taking the sum of all elements of this will give us the intersection.
If we add all the elements of pred_inds and target_inds, we'll get the union + intersection of pixels of that particular class. So, we subtract the already calculated intersection to get the union. Then, we can divide the intersection and union to get the IoU of that particular class and add it to a list of valid IoUs.
At the end, you take the mean of the entire list to get the mIoU. If you want the Dice Coefficient, you can calculate it in a similar fashion.
I am implementing the Skipgram model, both in Pytorch and Tensorflow2. I am having doubts about the implementation of subsampling of frequent words. Verbatim from the paper, the probability of subsampling word wi is computed as
where t is a custom threshold (usually, a small value such as 0.0001) and f is the frequency of the word in the document. Although the authors implemented it in a different, but almost equivalent way, let's stick with this definition.
When computing the P(wi), we can end up with negative values. For example, assume we have 100 words, and one of them appears extremely more often than others (as it is the case for my dataset).
import numpy as np
import seaborn as sns
np.random.seed(12345)
# generate counts in [1, 20]
counts = np.random.randint(low=1, high=20, size=99)
# add an extremely bigger count
counts = np.insert(counts, 0, 100000)
# compute frequencies
f = counts/counts.sum()
# define threshold as in paper
t = 0.0001
# compute probabilities as in paper
probs = 1 - np.sqrt(t/f)
sns.distplot(probs);
Q: What is the correct way to implement subsampling using this "probability"?
As an additional info, I have seen that in keras the function keras.preprocessing.sequence.make_sampling_table takes a different approach:
def make_sampling_table(size, sampling_factor=1e-5):
"""Generates a word rank-based probabilistic sampling table.
Used for generating the `sampling_table` argument for `skipgrams`.
`sampling_table[i]` is the probability of sampling
the i-th most common word in a dataset
(more common words should be sampled less frequently, for balance).
The sampling probabilities are generated according
to the sampling distribution used in word2vec:
```
p(word) = (min(1, sqrt(word_frequency / sampling_factor) /
(word_frequency / sampling_factor)))
```
We assume that the word frequencies follow Zipf's law (s=1) to derive
a numerical approximation of frequency(rank):
`frequency(rank) ~ 1/(rank * (log(rank) + gamma) + 1/2 - 1/(12*rank))`
where `gamma` is the Euler-Mascheroni constant.
# Arguments
size: Int, number of possible words to sample.
sampling_factor: The sampling factor in the word2vec formula.
# Returns
A 1D Numpy array of length `size` where the ith entry
is the probability that a word of rank i should be sampled.
"""
gamma = 0.577
rank = np.arange(size)
rank[0] = 1
inv_fq = rank * (np.log(rank) + gamma) + 0.5 - 1. / (12. * rank)
f = sampling_factor * inv_fq
return np.minimum(1., f / np.sqrt(f))
I tend to trust deployed code more than paper write-ups, especially in a case like word2vec, where the original authors' word2vec.c code released by the paper's authors has been widely used & served as the template for other implementations. If we look at its subsampling mechanism...
if (sample > 0) {
real ran = (sqrt(vocab[word].cn / (sample * train_words)) + 1) * (sample * train_words) / vocab[word].cn;
next_random = next_random * (unsigned long long)25214903917 + 11;
if (ran < (next_random & 0xFFFF) / (real)65536) continue;
}
...we see that those words with tiny counts (.cn) that could give negative values in the original formula instead here give values greater-than 1.0, and thus can never be less than the long-random-masked-and-scaled to never be more than 1.0 ((next_random & 0xFFFF) / (real)65536). So, it seems the authors' intent was for all negative-values of the original formula to mean "never discard".
As per the keras make_sampling_table() comment & implementation, they're not consulting the actual word-frequencies at all. Instead, they're assuming a Zipf-like distribution based on word-rank order to synthesize a simulated word-frequency.
If their assumptions were to hold – the related words are from a natural-language corpus with a Zipf-like frequency-distribution – then I'd expect their sampling probabilities to be close to down-sampling probabilities that would have been calculated from true frequency information. And that's probably "close enough" for most purposes.
I'm not sure why they chose this approximation. Perhaps other aspects of their usual processes have not maintained true frequencies through to this step, and they're expecting to always be working with natural-language texts, where the assumed frequencies will be generally true.
(As luck would have it, and because people often want to impute frequencies to public sets of word-vectors which have dropped the true counts but are still sorted from most- to least-frequent, just a few days ago I wrote an answer about simulating a fake-but-plausible distribution using Zipf's law – similar to what this keras code is doing.)
But, if you're working with data that doesn't match their assumptions (as with your synthetic or described datasets), their sampling-probabilities will be quite different than what you would calculate yourself, with any form of the original formula that uses true word frequencies.
In particular, imagine a distribution with one token a million times, then a hundred tokens all appearing just 10 times each. Those hundred tokens' order in the "rank" list is arbitrary – truly, they're all tied in frequency. But the simulation-based approach, by fitting a Zipfian distribution on that ordering, will in fact be sampling each of them very differently. The one 10-occurrence word lucky enough to be in the 2nd rank position will be far more downsampled, as if it were far more frequent. And the 1st-rank "tall head" value, by having its true frequency *under-*approximated, will be less down-sampled than otherwise. Neither of those effects seem beneficial, or in the spirit of the frequent-word-downsampling option - which should only "thin out" very-frequent words, and in all cases leave words of the same frequency as each other in the original corpus roughly equivalently present to each other in the down-sampled corpus.
So for your case, I would go with the original formula (probability-of-discarding-that-requires-special-handling-of-negative-values), or the word2vec.c practical/inverted implementation (probability-of-keeping-that-saturates-at-1.0), rather than the keras-style approximation.
(As a totally-separate note that nonetheless may be relevant for your dataset/purposes, if you're using negative-sampling: there's another parameter controlling the relative sampling of negative examples, often fixed at 0.75 in early implementations, that one paper has suggested can usefully vary for non-natural-language token distributions & recommendation-related end-uses. This parameter is named ns_exponent in the Python gensim implementation, but simply a fixed power value internal to a sampling-table pre-calculation in the original word2vec.c code.)
I'm running a physics simulation related to visible light, and the resulting wave function has a very, very high frequency -- cyclic frequency is on the order of 1.0e15, and the spatial frequency k is on the order of 1.0e7. Thankfully, I only use the spatial frequency, but when I calculate it for later usage (using either math or numpy), I get something that resembles a beat wave, unless I use N ~= k sample points, because I have to calculate it over a much greater range (on the order of 1.0e-3 - 1.0e-1). It produces a beat wave so consistently I spent a few hours to make sure I'm not actually calculating one. I'll also have to use fft() on the resulting wave and I'm afraid it won't work properly with a misrepresented wave.
I've tried using various amounts of sample points, but unless it's extraordinarily high (takes a good minute or two to calculate), only the prominence of beating changes. Just in case I'm misusing numpy, I tried the same thing with appending wave.value calculated by math.sin to a float array, but it had the same result.
import numpy as np
import matplotlib.pyplot as plt
mmScale = 1.0e-3
nmScale = 1.0e-9
c = 3.0e8
N = 1000
class Wave:
def __init__(self, amplitude, wavelength):
self.wavelength = wavelength*nmScale
self.amplitude = amplitude
self.omega = 2*pi*c/self.wavelength
self.k = 2*pi/self.wavelength
def value(self, time, travel):
return self.amplitude*np.sin(self.omega*time - self.k*travel)
x = np.linspace(50, 250, N)*mmScale
wave = Wave(1, 400)
y = wave.value(0.1, x)
plt.plot(x,y)
plt.show()
The code above produces a graph of the function, and you can put in different values for N to see how it gives different waveforms.
Your sampling spatial frequency is:
1/Ts = 1 / ((250-50)*mmScale) / N) = 5000 [samples/meter]
Your wave's spatial frequency is:
1/Tw = 1 / wavelength = 1 / (400e-9) = 2500000 [wavelengths/meter]
You fail to satisfy Nyquist criterion by a factor of (2*2500000 ) / 5000 = 1000.
Thus you must expect serious aliasing effects. See https://en.wikipedia.org/wiki/Aliasing.
Not much can be done to battle it. But there are some tricks that may help you depending on application. One is to represent a wave as a complex envelop around carier frequency, which is 400e-9. Please provide more detail on what you do with the wave.
I would like to compare the output of an algorithm with different preprocessed data: NMF and PCA.
In order to get somehow a comparable result, instead of choosing just the same number of components for each PCA and NMF, I would like to pick the amount that explains e.g 95% of retained variance.
I was wondering if its possible to identify the variance retained in each component of NMF.
For instance using PCA this would be given by:
retainedVariance(i) = eigenvalue(i) / sum(eigenvalue)
Any ideas?
TL;DR
You should loop over different n_components and estimate explained_variance_score of the decoded X at each iteration. This will show you how many components do you need to explain 95% of variance.
Now I will explain why.
Relationship between PCA and NMF
NMF and PCA, as many other unsupervised learning algorithms, are aimed to do two things:
encode input X into a compressed representation H;
decode H back to X', which should be as close to X as possible.
They do it in a somehow similar way:
Decoding is similar in PCA and NMF: they output X' = dot(H, W), where W is a learned matrix parameter.
Encoding is different. In PCA, it is also linear: H = dot(X, V), where V is also a learned parameter. In NMF, H = argmin(loss(X, H, W)) (with respect to H only), where loss is mean squared error between X and dot(H, W), plus some additional penalties. Minimization is performed by coordinate descent, and result may be nonlinear in X.
Training is also different. PCA learns sequentially: the first component minimizes MSE without constraints, each next kth component minimizes residual MSE subject to being orthogonal with the previous components. NMF minimizes the same loss(X, H, W) as when encoding, but now with respect to both H and W.
How to measure performance of dimensionality reduction
If you want to measure performance of an encoding/decoding algorithm, you can follow the usual steps:
Train your encoder+decoder on X_train
To measure in-sample performance, compare X_train'=decode(encode(X_train)) with X_train using your preferred metric (e.g. MAE, RMSE, or explained variance)
To measure out-of-sample performance (generalizing ability) of your algorithm, do step 2 with the unseen X_test.
Let's try it with PCA and NMF!
from sklearn import decomposition, datasets, model_selection, preprocessing, metrics
# use the well-known Iris dataset
X, _ = datasets.load_iris(return_X_y=True)
# split the dataset, to measure overfitting
X_train, X_test = model_selection.train_test_split(X, test_size=0.5, random_state=1)
# I scale the data in order to give equal importance to all its dimensions
# NMF does not allow negative input, so I don't center the data
scaler = preprocessing.StandardScaler(with_mean=False).fit(X_train)
X_train_sc = scaler.transform(X_train)
X_test_sc = scaler.transform(X_test)
# train the both decomposers
pca = decomposition.PCA(n_components=2).fit(X_train_sc)
nmf = decomposition.NMF(n_components=2).fit(X_train_sc)
print(sum(pca.explained_variance_ratio_))
It will print you explained variance ratio of 0.9536930834362043 - the default metric of PCA, estimated using its eigenvalues. We can measure it in a more direct way - by applying a metric to actual and "predicted" values:
def get_score(model, data, scorer=metrics.explained_variance_score):
""" Estimate performance of the model on the data """
prediction = model.inverse_transform(model.transform(data))
return scorer(data, prediction)
print('train set performance')
print(get_score(pca, X_train_sc))
print(get_score(nmf, X_train_sc))
print('test set performance')
print(get_score(pca, X_test_sc))
print(get_score(nmf, X_test_sc))
which gives
train set performance
0.9536930834362043 # same as before!
0.937291711378812
test set performance
0.9597828443047842
0.9590555069007827
You can see that on the training set PCA performs better than NMF, but on the test set their performance is almost identical. This happens, because NMF applies lots of regularization:
H and W (the learned parameter) must be non-negative
H should be as small as possible (L1 and L2 penalties)
W should be as small as possible (L1 and L2 penalties)
These regularizations make NMF fit worse than possible to the training data, but they might improve its generalizing ability, which happened in our case.
How to choose the number of components
In PCA, it is simple, because its components h_1, h_2, ... h_k are learned sequentially. If you add the new component h_(k+1), the first k will not change. Thus, you can estimate performance of each component, and these estimates will not depent on the number of components. This makes it possible for PCA to output the explained_variance_ratio_ array after only a single fit to data.
NMF is more complex, because all its components are trained at the same time, and each one depends on all the rest. Thus, if you add the k+1th component, the first k components will change, and you cannot match each particular component with its explained variance (or any other metric).
But what you can to is to fit a new instance of NMF for each number of components, and compare the total explained variance:
ks = [1,2,3,4]
perfs_train = []
perfs_test = []
for k in ks:
nmf = decomposition.NMF(n_components=k).fit(X_train_sc)
perfs_train.append(get_score(nmf, X_train_sc))
perfs_test.append(get_score(nmf, X_test_sc))
print(perfs_train)
print(perfs_test)
which would give
[0.3236945680665101, 0.937291711378812, 0.995459457205891, 0.9974027602663655]
[0.26186701106012833, 0.9590555069007827, 0.9941424954209546, 0.9968456603914185]
Thus, three components (judging by the train set performance) or two components (by the test set) are required to explain at least 95% of variance. Please notice that this case is unusual and caused by a small size of training and test data: usually performance degrades a little bit on the test set, but in my case it actually improved a little.