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I have a dataset similar to the following table:
The prediction target is going to be the 'score' column. I'm wondering how can I divide the testing set into different subgroups such as score between 1 to 3 or then check the accuracy on each subgroup.
Now what I have is as follows:
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3)
model = tree.DecisionTreeRegressor()
model.fit(X_train, y_train)
for i in (0,1,2,3,4):
y_new=y_test[(y_test>=i) & (y_test<=i+1)]
y_new_pred=model.predict(X_test)
print metrics.r2_score(y_new, y_new_pred)
However, my code did not work and this is the traceback that I get:
Found input variables with inconsistent numbers of samples: [14279,
55955]
I have tried the solution provided below, but it looks like that for the full score range (0-5) the r^2 is 0.67. but the subscore range for example (0-1,1-2,2-3,3-4,4-5) the r^2s are significantly lower than that of the full range. shouldn't some of the subscore r^2 be higher than 0.67 and some of them be lower than 0.67?
Could anyone kindly let me know where did I do wrong? Thanks a lot for all your help.
When you are computing the metrics, you have to filtered the predicted values (based on your subset condition).
Basically you are trying to compute
metrics.r2_score([1,3],[1,2,3,4,5])
which creates an error,
ValueError: Found input variables with inconsistent numbers of
samples: [2, 5]
Hence, my suggested solution would be
model.fit(X_train, y_train)
#compute the prediction only once.
y_pred = model.predict(X_test)
for i in (0,1,2,3,4):
#COMPUTE THE CONDITION FOR SUBSET HERE
subset = (y_test>=i) & (y_test<=i+1)
print metrics.r2_score(y_test [subset], y_pred[subset])
Dear pvlib users and devels.
I'm a researcher in computer science, not particularly expert in the simulation or modelling of solar panels. I'm interested in use pvlib since
we are trying to simulate the works of a small solar panel used for IoT
applications, in particular the panel spec are the following:
12.8% max efficiency, Vmp = 5.82V, size = 225 × 155 × 17 mm.
Before using pvlib, one of my collaborator wrote a code that compute the
irradiation directly from average monthly values calculated with PVWatt.
I was not really satisfied, so we are starting to use pvlib.
In the old code, we have the power and current of the panel calculated as:
W = Irradiation * PanelSize(m^2) * Efficiency
A = W / Vmp
The Irradiation, in Madrid, as been obtained with PVWatt, and this is
what my collaborator used:
DIrradiance = (2030.0,2960.0,4290.0,5110.0,5950.0,7090.0,7200.0,6340.0,4870.0,3130.0,2130.0,1700.0)
I'm trying to understand if pvlib compute values similar to the ones above, as averages over a day for each month. And the curve of production in day.
I wrote this to compare pvlib with our old model:
import math
import numpy as np
import datetime as dt
import matplotlib.pyplot as plt
import pandas as pd
import pvlib
from pvlib.location import Location
def irradiance(day,m):
DIrradiance =(2030.0,2960.0,4290.0,5110.0,5950.0,
7090.0,7200.0,6340.0,4870.0,3130.0,2130.0,1700.0)
madrid = Location(40.42, -3.70, 'Europe/Madrid', 600, 'Madrid')
times = pd.date_range(start=dt.datetime(2015,m,day,00,00),
end=dt.datetime(2015,m,day,23,59),
freq='60min')
spaout = pvlib.solarposition.spa_python(times, madrid.latitude, madrid.longitude)
spaout = spaout.assign(cosz=pd.Series(np.cos(np.deg2rad(spaout['zenith']))))
z = np.array(spaout['cosz'])
return z.clip(0)*(DIrradiance[m-1])
madrid = Location(40.42, -3.70, 'Europe/Madrid', 600, 'Madrid')
times = pd.date_range(start = dt.datetime(2015,8,15,00,00),
end = dt.datetime(2015,8,15,23,59),
freq='60min')
old = irradiance(15,8) # old model
new = madrid.get_clearsky(times) # pvlib irradiance
plt.plot(old,'r-') # compare them.
plt.plot(old/6.0,'y-') # old seems 6 times more..I do not know why
plt.plot(new['ghi'].values,'b-')
plt.show()
The code above compute the old irradiance, using the zenit angle. and compute the ghi values using the clear_sky. I do not understand if the values in ghi must be multiplied by the cos of zenit too, or not. Anyway
they are smaller by a factor of 6. What I'd like to have at the end is the
power and current in output from the panel (DC) without any inverter, and
we are not really interested at modelling it exactly, but at least, to
have a reasonable curve. We are able to capture from the panel the ampere
produced, and we want to compare the values from the measurements putting
the panel on the roof top with the values calculated by pvlib.
Any help on this would be really appreachiated. Thanks
Sorry Will I do not care a lot about my previous model since I'd like to move all code to pvlib. I followed your suggestion and I'm using irradiance.total_irrad, the code now looks in this way:
madrid = Location(40.42, -3.70, 'Europe/Madrid', 600, 'Madrid')
times = pd.date_range(start=dt.datetime(2015,1,1,00,00),
end=dt.datetime(2015,1,1,23,59),
freq='60min')
ephem_data = pvlib.solarposition.spa_python(times, madrid.latitude,
madrid.longitude)
irrad_data = madrid.get_clearsky(times)
AM = atmosphere.relativeairmass(ephem_data['apparent_zenith'])
total = irradiance.total_irrad(40, 180,
ephem_data['apparent_zenith'], ephem_data['azimuth'],
dni=irrad_data['dni'], ghi=irrad_data['ghi'],
dhi=irrad_data['dhi'], airmass=AM,
surface_type='urban')
poa = total['poa_global'].values
Now, I know the irradiance on POA, and I want to compute the output in Ampere: It is just
(poa*PANEL_EFFICIENCY*AREA) / VOLT_OUTPUT ?
It's not clear to me how you arrived at your values for DIrradiance or what the units are, so I can't comment much the discrepancies between the values. I'm guessing that it's some kind of monthly data since there are 12 values. If so, you'd need to calculate ~hourly pvlib irradiance data and then integrate it to check for consistency.
If your module will be tilted, you'll need to convert your ~hourly irradiance GHI, DNI, DHI values to plane of array (POA) irradiance using a transposition model. The irradiance.total_irrad function is the easiest way to do that.
The next steps depend on the IV characteristics of your module, the rest of the circuit, and how accurate you need the model to be.
I am trying to calculate the perplexity for the data I have. The code I am using is:
import sys
sys.path.append("/usr/local/anaconda/lib/python2.7/site-packages/nltk")
from nltk.corpus import brown
from nltk.model import NgramModel
from nltk.probability import LidstoneProbDist, WittenBellProbDist
estimator = lambda fdist, bins: LidstoneProbDist(fdist, 0.2)
lm = NgramModel(3, brown.words(categories='news'), True, False, estimator)
print lm
But I am receiving the error,
File "/usr/local/anaconda/lib/python2.7/site-packages/nltk/model/ngram.py", line 107, in __init__
cfd[context][token] += 1
TypeError: 'int' object has no attribute '__getitem__'
I have already performed Latent Dirichlet Allocation for the data I have and I have generated the unigrams and their respective probabilities (they are normalized as the sum of total probabilities of the data is 1).
My unigrams and their probability looks like:
Negroponte 1.22948976891e-05
Andreas 7.11290670484e-07
Rheinberg 7.08255885794e-07
Joji 4.48481435106e-07
Helguson 1.89936727391e-07
CAPTION_spot 2.37395965468e-06
Mortimer 1.48540253778e-07
yellow 1.26582575863e-05
Sugar 1.49563800878e-06
four 0.000207196011781
This is just a fragment of the unigrams file I have. The same format is followed for about 1000s of lines. The total probabilities (second column) summed gives 1.
I am a budding programmer. This ngram.py belongs to the nltk package and I am confused as to how to rectify this. The sample code I have here is from the nltk documentation and I don't know what to do now. Please help on what I can do. Thanks in advance!
Perplexity is the inverse probability of the test set, normalized by the number of words. In the case of unigrams:
Now you say you have already constructed the unigram model, meaning, for each word you have the relevant probability. Then you only need to apply the formula. I assume you have a big dictionary unigram[word] that would provide the probability of each word in the corpus. You also need to have a test set. If your unigram model is not in the form of a dictionary, tell me what data structure you have used, so I could adapt it to my solution accordingly.
perplexity = 1
N = 0
for word in testset:
if word in unigram:
N += 1
perplexity = perplexity * (1/unigram[word])
perplexity = pow(perplexity, 1/float(N))
UPDATE:
As you asked for a complete working example, here's a very simple one.
Suppose this is our corpus:
corpus ="""
Monty Python (sometimes known as The Pythons) were a British surreal comedy group who created the sketch comedy show Monty Python's Flying Circus,
that first aired on the BBC on October 5, 1969. Forty-five episodes were made over four series. The Python phenomenon developed from the television series
into something larger in scope and impact, spawning touring stage shows, films, numerous albums, several books, and a stage musical.
The group's influence on comedy has been compared to The Beatles' influence on music."""
Here's how we construct the unigram model first:
import collections, nltk
# we first tokenize the text corpus
tokens = nltk.word_tokenize(corpus)
#here you construct the unigram language model
def unigram(tokens):
model = collections.defaultdict(lambda: 0.01)
for f in tokens:
try:
model[f] += 1
except KeyError:
model [f] = 1
continue
N = float(sum(model.values()))
for word in model:
model[word] = model[word]/N
return model
Our model here is smoothed. For words outside the scope of its knowledge, it assigns a low probability of 0.01. I already told you how to compute perplexity:
#computes perplexity of the unigram model on a testset
def perplexity(testset, model):
testset = testset.split()
perplexity = 1
N = 0
for word in testset:
N += 1
perplexity = perplexity * (1/model[word])
perplexity = pow(perplexity, 1/float(N))
return perplexity
Now we can test this on two different test sets:
testset1 = "Monty"
testset2 = "abracadabra gobbledygook rubbish"
model = unigram(tokens)
print perplexity(testset1, model)
print perplexity(testset2, model)
for which you get the following result:
>>>
49.09452736318415
99.99999999999997
Note that when dealing with perplexity, we try to reduce it. A language model that has less perplexity with regards to a certain test set is more desirable than one with a bigger perplexity. In the first test set, the word Monty was included in the unigram model, so the respective number for perplexity was also smaller.
Thanks for the code snippet! Shouldn't:
for word in model:
model[word] = model[word]/float(sum(model.values()))
be rather:
v = float(sum(model.values()))
for word in model:
model[word] = model[word]/v
Oh ... I see was already answered ...
As my title suggests, I'm trying to fit a Gaussian to some data and I'm just getting a straight line. I've been looking at these other discussion Gaussian fit for Python and Fitting a gaussian to a curve in Python which seem to suggest basically the same thing. I can make the code in those discussions work fine for the data they provide, but it won't do it for my data.
My code looks like this:
import pylab as plb
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy import asarray as ar,exp
y = y - y[0] # to make it go to zero on both sides
x = range(len(y))
max_y = max(y)
n = len(y)
mean = sum(x*y)/n
sigma = np.sqrt(sum(y*(x-mean)**2)/n)
# Someone on a previous post seemed to think this needed to have the sqrt.
# Tried it without as well, made no difference.
def gaus(x,a,x0,sigma):
return a*exp(-(x-x0)**2/(2*sigma**2))
popt,pcov = curve_fit(gaus,x,y,p0=[max_y,mean,sigma])
# It was suggested in one of the other posts I looked at to make the
# first element of p0 be the maximum value of y.
# I also tried it as 1, but that did not work either
plt.plot(x,y,'b:',label='data')
plt.plot(x,gaus(x,*popt),'r:',label='fit')
plt.legend()
plt.title('Fig. 3 - Fit for Time Constant')
plt.xlabel('Time (s)')
plt.ylabel('Voltage (V)')
plt.show()
The data I am trying to fit is as follows:
y = array([ 6.95301373e+12, 9.62971320e+12, 1.32501876e+13,
1.81150568e+13, 2.46111132e+13, 3.32321345e+13,
4.45978682e+13, 5.94819771e+13, 7.88394616e+13,
1.03837779e+14, 1.35888594e+14, 1.76677210e+14,
2.28196006e+14, 2.92781632e+14, 3.73133045e+14,
4.72340762e+14, 5.93892782e+14, 7.41632194e+14,
9.19750269e+14, 1.13278296e+15, 1.38551838e+15,
1.68291212e+15, 2.02996957e+15, 2.43161742e+15,
2.89259207e+15, 3.41725793e+15, 4.00937676e+15,
4.67187762e+15, 5.40667931e+15, 6.21440313e+15,
7.09421973e+15, 8.04366842e+15, 9.05855930e+15,
1.01328502e+16, 1.12585509e+16, 1.24257598e+16,
1.36226443e+16, 1.48356404e+16, 1.60496345e+16,
1.72482199e+16, 1.84140400e+16, 1.95291969e+16,
2.05757166e+16, 2.15360187e+16, 2.23933053e+16,
2.31320228e+16, 2.37385276e+16, 2.42009864e+16,
2.45114362e+16, 2.46427484e+16, 2.45114362e+16,
2.42009864e+16, 2.37385276e+16, 2.31320228e+16,
2.23933053e+16, 2.15360187e+16, 2.05757166e+16,
1.95291969e+16, 1.84140400e+16, 1.72482199e+16,
1.60496345e+16, 1.48356404e+16, 1.36226443e+16,
1.24257598e+16, 1.12585509e+16, 1.01328502e+16,
9.05855930e+15, 8.04366842e+15, 7.09421973e+15,
6.21440313e+15, 5.40667931e+15, 4.67187762e+15,
4.00937676e+15, 3.41725793e+15, 2.89259207e+15,
2.43161742e+15, 2.02996957e+15, 1.68291212e+15,
1.38551838e+15, 1.13278296e+15, 9.19750269e+14,
7.41632194e+14, 5.93892782e+14, 4.72340762e+14,
3.73133045e+14, 2.92781632e+14, 2.28196006e+14,
1.76677210e+14, 1.35888594e+14, 1.03837779e+14,
7.88394616e+13, 5.94819771e+13, 4.45978682e+13,
3.32321345e+13, 2.46111132e+13, 1.81150568e+13,
1.32501876e+13, 9.62971320e+12, 6.95301373e+12,
4.98705540e+12])
I would show you what it looks like, but apparently I don't have enough reputation points...
Anyone got any idea why it's not fitting properly?
Thanks for your help :)
The importance of the initial guess, p0 in curve_fit's default argument list, cannot be stressed enough.
Notice that the docstring mentions that
[p0] If None, then the initial values will all be 1
So if you do not supply it, it will use an initial guess of 1 for all parameters you're trying to optimize for.
The choice of p0 affects the speed at which the underlying algorithm changes the guess vector p0 (ref. the documentation of least_squares).
When you look at the data that you have, you'll notice that the maximum and the mean, mu_0, of the Gaussian-like dataset y, are
2.4e16 and 49 respectively. With the peak value so large, the algorithm, would need to make drastic changes to its initial guess to reach that large value.
When you supply a good initial guess to the curve fitting algorithm, convergence is more likely to occur.
Using your data, you can supply a good initial guess for the peak_value, the mean and sigma, by writing them like this:
y = np.array([...]) # starting from the original dataset
x = np.arange(len(y))
peak_value = y.max()
mean = x[y.argmax()] # observation of the data shows that the peak is close to the center of the interval of the x-data
sigma = mean - np.where(y > peak_value * np.exp(-.5))[0][0] # when x is sigma in the gaussian model, the function evaluates to a*exp(-.5)
popt,pcov = curve_fit(gaus, x, y, p0=[peak_value, mean, sigma])
print(popt) # prints: [ 2.44402560e+16 4.90000000e+01 1.20588976e+01]
Note that in your code, for the mean you take sum(x*y)/n , which is strange, because this would modulate the gaussian by a polynome of degree 1 (it multiplies a gaussian with a monotonously increasing line of constant slope) before taking the mean. That will offset the mean value of y (in this case to the right). A similar remark can be made for your calculation of sigma.
Final remark: the histogram of y will not resemble a Gaussian, as y is already a Gaussian. The histogram will merely bin (count) values into different categories (answering the question "how many datapoints in y reach a value between [a, b]?").
Word2vec is a open source tool to calculate the words distance provided by Google. It can be used by inputting a word and output the ranked word lists according to the similarity. E.g.
Input:
france
Output:
Word Cosine distance
spain 0.678515
belgium 0.665923
netherlands 0.652428
italy 0.633130
switzerland 0.622323
luxembourg 0.610033
portugal 0.577154
russia 0.571507
germany 0.563291
catalonia 0.534176
However, what I need to do is to calculate the similarity distance by giving 2 words. If I give the 'france' and 'spain', how can I get the score 0.678515 without reading the whole words list by giving just 'france'.
gensim has a Python implementation of Word2Vec which provides an in-built utility for finding similarity between two words given as input by the user. You can refer to the following:
Intro: http://radimrehurek.com/gensim/models/word2vec.html
Tutorial: http://radimrehurek.com/2014/02/word2vec-tutorial/
UPDATED: Gensim 4.0.0 and above
The syntax in Python for finding similarity between two words goes like this:
>> from gensim.models import Word2Vec
>> model = Word2Vec.load(path/to/your/model)
>> model.wv.similarity('france', 'spain')
As you know word2vec can represent a word as a mathematical vector. So once you train the model, you can obtain the vectors of the words spain and france and compute the cosine distance (dot product).
An easy way to do this is to use this Python wrapper of word2vec. You can obtain the vector using this:
>>> model['computer'] # raw numpy vector of a word
array([-0.00449447, -0.00310097, 0.02421786, ...], dtype=float32)
To compute the distances between two words, you can do the following:
>>> import numpy
>>> cosine_similarity = numpy.dot(model['spain'], model['france'])/(numpy.linalg.norm(model['spain'])* numpy.linalg.norm(model['france']))
I just stumbled on this while looking for how to do this by modifying the original distance.c version, not by using another library like gensim.
I didn't find an answer so I did some research, and am sharing it here for others who also want to know how to do it in the original implementation.
After looking through the C source, you will find that 'bi' is an array of indexes. If you provide two words, the index for word1 will be in bi[0] and the index of word2 will be in bi[1].
The model 'M' is an array of vectors. Each word is represented as a vector with dimension 'size'.
Using these two indexes and the model of vectors, look them up and calculate the cosine distance (which is the same as the dot product) like this:
dist = 0;
for (a = 0; a < size; a++) {
dist += M[a + bi[0] * size] * M[a + bi[1] * size];
}
after this completes, the value 'dist' is the cosine similarity between the two words.
I have developed a code to help with calculating cosine similarity for 2 sentences / SKUs using gensim. The code can be found here
https://github.com/aviralmathur/Word2Vec
The code is using data for Kaggle competition on Crowdflower
It has been developed using Code for Kaggle Tutorial on Word2Vec available here
https://www.kaggle.com/c/word2vec-nlp-tutorial
I hope this helps
If you look at the source code of the Gensim's native method to calculate word similarities, you will find that it calculates word similarities using the following method:
import numpy as np
from gensim import matutils # utility fnc for pickling, common scipy operations etc
def similarity_cosine(vec1, vec2):
cosine_similarity = np.dot(matutils.unitvec(vec1), matutils.unitvec(vec2))
return cosine_similarity
similarity_cosine(model.wv['space'], model.wv['france'])