How to find the best line Fit Python(banister-impulse model) - python

I have this formula that is used to predict athletic performance base on daily stress.
It is based on 5 constant unique to each person. I'm trying to find these based on daily stress and performance testing that has been done. I'm new to programming and I don't know where to start.
see the formula
Performance= Fitness(=daily stress+yesterday fitness put decay) - Fatigue(daily stress+yesterday fatigue put decay) +P0
This is a sample of the data: data
thank you
import pandas as pd
import numpy as np
import math
from scipy import optimize
data = pd.read_csv('data_mod1.csv')
TSS = data['stress'].fillna(0)
arr = np.array(TSS)
#data = data.dropna()
a = [arr[0]]
b = [arr[0]]
x = arr[1:]
def Banister(x, t1, t2,k1,k2, c):
for v in x:
a.append(a[-1]*np.exp(-1/t1) + v)
b.append(b[-1]*np.exp(-1/t2) + v)
data['fit'] = pd.Series(a)
data['fat'] = pd.Series(b)
data['perf'] = ((data['fit']*k1)-(data['fat']*k2))+c
return data['perf']
# In[ ]:
from scipy.optimize import curve_fit
fit = curve_fit(Banister, arr,data[data.index], p0=[20, 10,1 ,2, 50])

Related

How to add contrains in python when optimizing a portfolio?

I am new to python and does not have a lot of experience.
I am trying to add constrains in this code so that the weights of one stock cannot be equal to 0, but the weights of the same stock cannot be above 5% either. This constrain will be valid for all the stocks.
This is what I have so far, is there anyone that have any tips on how to add these constrains?
Thanks in advance!
import os
import pandas as pd
import numpy as np
from scipy.optimize import linprog
data = pd.read_excel("data.xlsm")
# change data['WGT_GLOBAL'] s.t. EUTax = 20
data['Weights screened'] = data['WGT_GLOBAL']*data['Positiv screening']
EUTax = (data['Weights screened']*data['EU tax']).sum()
# min = -(x*data['YTD Return']).sum()
# s.t. x >= 0, x <= 1, (x*data['Positiv screening']*data['EU tax']).sum() = 20
obj = -(data['YTD Return'].fillna(0).to_numpy())
bnd = [(0,1)]
lhs_eq = [data['Positiv screening']*data['EU tax'].to_numpy(),np.ones(len(data))]
rhs_eq = [[20],[1]]
opt = linprog(c=obj, A_eq=lhs_eq, b_eq=rhs_eq, bounds=bnd, method="revised simplex")
optimal_weights = opt.x

Is there a better way to solve this MINLP in pyscipopt?

I'm trying to solve the following MINLP, basically attempting to maximize the likelihood of a certain portfolio reaching a "ceiling" performance. My first attempt at the code is below.
EDIT: Math says maximize, should say minimize
from pyscipopt import Model, quicksum
import numpy as np
import pandas as pd
from random import uniform, normalvariate
model=Model()
t=20000
stocks_portfolio = {}
stocks_df = pd.DataFrame(np.zeros((150,4)),columns = {'ids','Mean','cost','stdev'})
noptions = len(stocks_df)
stocks_df['ids'] = [i for i in range(noptions)]
stocks_df['Mean'] = [uniform(500,2500) for i in range(noptions)]
stocks_df['cost'] = [stocks_df.loc[i,'Mean']*uniform(50,250) for i in range(noptions)]
stocks_df['stdev'] = [stocks_df.loc[i,'Mean']*uniform(0.2,0.5) for i in range(noptions)]
cov_mat = np.array([[normalvariate(0,0.3) for i in range(noptions)] for j in range(noptions)])
for i in range(len(stocks_df)):
stocks_portfolio[i] = model.addVar(vtype='B')
model.addCons(quicksum(stocks_portfolio[i] for i in range(noptions))==15)
model.addCons(quicksum(stocks_df.loc[i, 'cost']*stocks_portfolio[i] for i in range(noptions)) <= 600000)
stand_in = model.addVar(vtype='C')
model.addCons(stand_in>=(t-quicksum(stocks_df.loc[i,'Mean']*stocks_portfolio[i] for i in range(noptions)))/((quicksum(stocks_portfolio[i]*stocks_df.loc[i,'stdev']**2 for i in range(noptions))+quicksum(2*stocks_portfolio[i]*stocks_portfolio[j]*cov_mat[i,j] for i in range(noptions) for j in range(noptions)))**0.5))
model.setObjective(stand_in,'minimize')
model.optimize()
model.getCondition()
portfolios = []
for i in range(noptions):
if model.getVal(stocks_portfolio[i]) > 0.9:
portfolios.append(i)
The performance here has been slow and unwieldy, and I was wondering if I'm thinking about the question all wrong.

How to continue after gradient descent?

I am very new to Data Science and Python. After a few hours of Experimentation, I finally received values for my gradient descent (code below). I am having trouble to plotting bzw. How can I plot the regression line automatically after the algorithm?
import numpy as np;
import matplotlib.pyplot as plt;
import csv
import pandas as pd
def gradient_descent(x,y):
m_curr=b_curr=0
iterations = 5000
n=len(x)
learning_rate = 0.01
for i in range(iterations):
y_predicted = m_curr*x + b_curr
cost = (1/n)*sum([val**2 for val in (y-y_predicted)])
md = -(2/n)*sum(x*(y-y_predicted))
bd = -(2/n)*sum(y-y_predicted)
m_curr = m_curr - learning_rate*md
b_curr = b_curr - learning_rate*bd
print("m{}, b{}, cost {}, iteration {}".format(m_curr,b_curr,cost,i))
if __name__ == '__main__':
#Reading data -> Output: DataFrame in float64
data = pd.read_csv('ex1data1.txt', sep=',', header=None, names=['Feature', 'Label'])
data.plot(x='Feature', y='Label', kind = 'scatter')
#separating data frame to
feat_vec = pd.DataFrame(data['Feature'])
label_vec = pd.DataFrame(data['Label'])
#Finding the Best Fit Line for our given Dataset and convert the df to np.array
#because it's more convenient for matrix multiplication
x = np.array(feat_vec)
y = np.array(label_vec)
gradient_descent(x,y)

Calculating tvalue using numpy

As part of an exercise i needed to check whether a given sample's true mean is 1.75 or not by generating tvalue using numpy and compare with the output from scipy.
Code:
import numpy as np
import scipy.stats as stats
import matplotlib.pyplot as plt
np.random.seed(seed=42) # make example reproducible
n = 100
x = np.random.normal(loc=1.78, scale=.1, size=n) # the sample is here
tval, pval = stats.ttest_1samp(x, 1.75)
var_x = x.var(ddof=1)
std_x = np.sqrt(var_x)
tval1 = (x.mean() - 1.75)/(std_x*np.sqrt(n))
print("Scipy: ",tval,"\nNumpy: ",tval1)
The output from Scipy is 2.1598800019529265,
while output from numpy is 0.021598800019529265
I guess the logic i used is incorrect, Please suggest.
You made a mistake in the denominator. It should be
tval1 = (x.mean() - 1.75)/(std_x / np.sqrt(n)) # (std_x divided by root n)
That's why you will find there is a factor of 100 difference ((1/10)/10 = 1/100) between your Scipy and numpy output.
Here is the Wiki of Student's t-test
An example using another sample size:
np.random.seed(seed=42)
n = 369
x = np.random.normal(loc=1.78, scale=.1, size=n) # the sample is here
tval, pval = stats.ttest_1samp(x, 1.75)
var_x = x.var(ddof=1)
std_x = np.sqrt(var_x)
tval1 = (x.mean() - 1.75)/(std_x / np.sqrt(n))
print("Scipy: ",tval,"\nNumpy: ",tval1)
# Output:
# Scipy: 6.306500305262841
# Numpy: 6.306500305262841

Principal Component Analysis (PCA) in Python

I have a (26424 x 144) array and I want to perform PCA over it using Python. However, there is no particular place on the web that explains about how to achieve this task (There are some sites which just do PCA according to their own - there is no generalized way of doing so that I can find). Anybody with any sort of help will do great.
I posted my answer even though another answer has already been accepted; the accepted answer relies on a deprecated function; additionally, this deprecated function is based on Singular Value Decomposition (SVD), which (although perfectly valid) is the much more memory- and processor-intensive of the two general techniques for calculating PCA. This is particularly relevant here because of the size of the data array in the OP. Using covariance-based PCA, the array used in the computation flow is just 144 x 144, rather than 26424 x 144 (the dimensions of the original data array).
Here's a simple working implementation of PCA using the linalg module from SciPy. Because this implementation first calculates the covariance matrix, and then performs all subsequent calculations on this array, it uses far less memory than SVD-based PCA.
(the linalg module in NumPy can also be used with no change in the code below aside from the import statement, which would be from numpy import linalg as LA.)
The two key steps in this PCA implementation are:
calculating the covariance matrix; and
taking the eivenvectors & eigenvalues of this cov matrix
In the function below, the parameter dims_rescaled_data refers to the desired number of dimensions in the rescaled data matrix; this parameter has a default value of just two dimensions, but the code below isn't limited to two but it could be any value less than the column number of the original data array.
def PCA(data, dims_rescaled_data=2):
"""
returns: data transformed in 2 dims/columns + regenerated original data
pass in: data as 2D NumPy array
"""
import numpy as NP
from scipy import linalg as LA
m, n = data.shape
# mean center the data
data -= data.mean(axis=0)
# calculate the covariance matrix
R = NP.cov(data, rowvar=False)
# calculate eigenvectors & eigenvalues of the covariance matrix
# use 'eigh' rather than 'eig' since R is symmetric,
# the performance gain is substantial
evals, evecs = LA.eigh(R)
# sort eigenvalue in decreasing order
idx = NP.argsort(evals)[::-1]
evecs = evecs[:,idx]
# sort eigenvectors according to same index
evals = evals[idx]
# select the first n eigenvectors (n is desired dimension
# of rescaled data array, or dims_rescaled_data)
evecs = evecs[:, :dims_rescaled_data]
# carry out the transformation on the data using eigenvectors
# and return the re-scaled data, eigenvalues, and eigenvectors
return NP.dot(evecs.T, data.T).T, evals, evecs
def test_PCA(data, dims_rescaled_data=2):
'''
test by attempting to recover original data array from
the eigenvectors of its covariance matrix & comparing that
'recovered' array with the original data
'''
_ , _ , eigenvectors = PCA(data, dim_rescaled_data=2)
data_recovered = NP.dot(eigenvectors, m).T
data_recovered += data_recovered.mean(axis=0)
assert NP.allclose(data, data_recovered)
def plot_pca(data):
from matplotlib import pyplot as MPL
clr1 = '#2026B2'
fig = MPL.figure()
ax1 = fig.add_subplot(111)
data_resc, data_orig = PCA(data)
ax1.plot(data_resc[:, 0], data_resc[:, 1], '.', mfc=clr1, mec=clr1)
MPL.show()
>>> # iris, probably the most widely used reference data set in ML
>>> df = "~/iris.csv"
>>> data = NP.loadtxt(df, delimiter=',')
>>> # remove class labels
>>> data = data[:,:-1]
>>> plot_pca(data)
The plot below is a visual representation of this PCA function on the iris data. As you can see, a 2D transformation cleanly separates class I from class II and class III (but not class II from class III, which in fact requires another dimension).
You can find a PCA function in the matplotlib module:
import numpy as np
from matplotlib.mlab import PCA
data = np.array(np.random.randint(10,size=(10,3)))
results = PCA(data)
results will store the various parameters of the PCA.
It is from the mlab part of matplotlib, which is the compatibility layer with the MATLAB syntax
EDIT:
on the blog nextgenetics I found a wonderful demonstration of how to perform and display a PCA with the matplotlib mlab module, have fun and check that blog!
Another Python PCA using numpy. The same idea as #doug but that one didn't run.
from numpy import array, dot, mean, std, empty, argsort
from numpy.linalg import eigh, solve
from numpy.random import randn
from matplotlib.pyplot import subplots, show
def cov(X):
"""
Covariance matrix
note: specifically for mean-centered data
note: numpy's `cov` uses N-1 as normalization
"""
return dot(X.T, X) / X.shape[0]
# N = data.shape[1]
# C = empty((N, N))
# for j in range(N):
# C[j, j] = mean(data[:, j] * data[:, j])
# for k in range(j + 1, N):
# C[j, k] = C[k, j] = mean(data[:, j] * data[:, k])
# return C
def pca(data, pc_count = None):
"""
Principal component analysis using eigenvalues
note: this mean-centers and auto-scales the data (in-place)
"""
data -= mean(data, 0)
data /= std(data, 0)
C = cov(data)
E, V = eigh(C)
key = argsort(E)[::-1][:pc_count]
E, V = E[key], V[:, key]
U = dot(data, V) # used to be dot(V.T, data.T).T
return U, E, V
""" test data """
data = array([randn(8) for k in range(150)])
data[:50, 2:4] += 5
data[50:, 2:5] += 5
""" visualize """
trans = pca(data, 3)[0]
fig, (ax1, ax2) = subplots(1, 2)
ax1.scatter(data[:50, 0], data[:50, 1], c = 'r')
ax1.scatter(data[50:, 0], data[50:, 1], c = 'b')
ax2.scatter(trans[:50, 0], trans[:50, 1], c = 'r')
ax2.scatter(trans[50:, 0], trans[50:, 1], c = 'b')
show()
Which yields the same thing as the much shorter
from sklearn.decomposition import PCA
def pca2(data, pc_count = None):
return PCA(n_components = 4).fit_transform(data)
As I understand it, using eigenvalues (first way) is better for high-dimensional data and fewer samples, whereas using Singular value decomposition is better if you have more samples than dimensions.
This is a job for numpy.
And here's a tutorial demonstrating how pincipal component analysis can be done using numpy's built-in modules like mean,cov,double,cumsum,dot,linalg,array,rank.
http://glowingpython.blogspot.sg/2011/07/principal-component-analysis-with-numpy.html
Notice that scipy also has a long explanation here
- https://github.com/scikit-learn/scikit-learn/blob/babe4a5d0637ca172d47e1dfdd2f6f3c3ecb28db/scikits/learn/utils/extmath.py#L105
with the scikit-learn library having more code examples -
https://github.com/scikit-learn/scikit-learn/blob/babe4a5d0637ca172d47e1dfdd2f6f3c3ecb28db/scikits/learn/utils/extmath.py#L105
Here are scikit-learn options. With both methods, StandardScaler was used because PCA is effected by scale
Method 1: Have scikit-learn choose the minimum number of principal components such that at least x% (90% in example below) of the variance is retained.
from sklearn.datasets import load_iris
from sklearn.decomposition import PCA
from sklearn.preprocessing import StandardScaler
iris = load_iris()
# mean-centers and auto-scales the data
standardizedData = StandardScaler().fit_transform(iris.data)
pca = PCA(.90)
principalComponents = pca.fit_transform(X = standardizedData)
# To get how many principal components was chosen
print(pca.n_components_)
Method 2: Choose the number of principal components (in this case, 2 was chosen)
from sklearn.datasets import load_iris
from sklearn.decomposition import PCA
from sklearn.preprocessing import StandardScaler
iris = load_iris()
standardizedData = StandardScaler().fit_transform(iris.data)
pca = PCA(n_components=2)
principalComponents = pca.fit_transform(X = standardizedData)
# to get how much variance was retained
print(pca.explained_variance_ratio_.sum())
Source: https://towardsdatascience.com/pca-using-python-scikit-learn-e653f8989e60
UPDATE: matplotlib.mlab.PCA is since release 2.2 (2018-03-06) indeed deprecated.
The library matplotlib.mlab.PCA (used in this answer) is not deprecated. So for all the folks arriving here via Google, I'll post a complete working example tested with Python 2.7.
Use the following code with care as it uses a now deprecated library!
from matplotlib.mlab import PCA
import numpy
data = numpy.array( [[3,2,5], [-2,1,6], [-1,0,4], [4,3,4], [10,-5,-6]] )
pca = PCA(data)
Now in `pca.Y' is the original data matrix in terms of the principal components basis vectors. More details about the PCA object can be found here.
>>> pca.Y
array([[ 0.67629162, -0.49384752, 0.14489202],
[ 1.26314784, 0.60164795, 0.02858026],
[ 0.64937611, 0.69057287, -0.06833576],
[ 0.60697227, -0.90088738, -0.11194732],
[-3.19578784, 0.10251408, 0.00681079]])
You can use matplotlib.pyplot to draw this data, just to convince yourself that the PCA yields "good" results. The names list is just used to annotate our five vectors.
import matplotlib.pyplot
names = [ "A", "B", "C", "D", "E" ]
matplotlib.pyplot.scatter(pca.Y[:,0], pca.Y[:,1])
for label, x, y in zip(names, pca.Y[:,0], pca.Y[:,1]):
matplotlib.pyplot.annotate( label, xy=(x, y), xytext=(-2, 2), textcoords='offset points', ha='right', va='bottom' )
matplotlib.pyplot.show()
Looking at our original vectors we'll see that data[0] ("A") and data[3] ("D") are rather similar as are data[1] ("B") and data[2] ("C"). This is reflected in the 2D plot of our PCA transformed data.
In addition to all the other answers, here is some code to plot the biplot using sklearn and matplotlib.
import numpy as np
import matplotlib.pyplot as plt
from sklearn import datasets
from sklearn.decomposition import PCA
import pandas as pd
from sklearn.preprocessing import StandardScaler
iris = datasets.load_iris()
X = iris.data
y = iris.target
#In general a good idea is to scale the data
scaler = StandardScaler()
scaler.fit(X)
X=scaler.transform(X)
pca = PCA()
x_new = pca.fit_transform(X)
def myplot(score,coeff,labels=None):
xs = score[:,0]
ys = score[:,1]
n = coeff.shape[0]
scalex = 1.0/(xs.max() - xs.min())
scaley = 1.0/(ys.max() - ys.min())
plt.scatter(xs * scalex,ys * scaley, c = y)
for i in range(n):
plt.arrow(0, 0, coeff[i,0], coeff[i,1],color = 'r',alpha = 0.5)
if labels is None:
plt.text(coeff[i,0]* 1.15, coeff[i,1] * 1.15, "Var"+str(i+1), color = 'g', ha = 'center', va = 'center')
else:
plt.text(coeff[i,0]* 1.15, coeff[i,1] * 1.15, labels[i], color = 'g', ha = 'center', va = 'center')
plt.xlim(-1,1)
plt.ylim(-1,1)
plt.xlabel("PC{}".format(1))
plt.ylabel("PC{}".format(2))
plt.grid()
#Call the function. Use only the 2 PCs.
myplot(x_new[:,0:2],np.transpose(pca.components_[0:2, :]))
plt.show()
I've made a little script for comparing the different PCAs appeared as an answer here:
import numpy as np
from scipy.linalg import svd
shape = (26424, 144)
repeat = 20
pca_components = 2
data = np.array(np.random.randint(255, size=shape)).astype('float64')
# data normalization
# data.dot(data.T)
# (U, s, Va) = svd(data, full_matrices=False)
# data = data / s[0]
from fbpca import diffsnorm
from timeit import default_timer as timer
from scipy.linalg import svd
start = timer()
for i in range(repeat):
(U, s, Va) = svd(data, full_matrices=False)
time = timer() - start
err = diffsnorm(data, U, s, Va)
print('svd time: %.3fms, error: %E' % (time*1000/repeat, err))
from matplotlib.mlab import PCA
start = timer()
_pca = PCA(data)
for i in range(repeat):
U = _pca.project(data)
time = timer() - start
err = diffsnorm(data, U, _pca.fracs, _pca.Wt)
print('matplotlib PCA time: %.3fms, error: %E' % (time*1000/repeat, err))
from fbpca import pca
start = timer()
for i in range(repeat):
(U, s, Va) = pca(data, pca_components, True)
time = timer() - start
err = diffsnorm(data, U, s, Va)
print('facebook pca time: %.3fms, error: %E' % (time*1000/repeat, err))
from sklearn.decomposition import PCA
start = timer()
_pca = PCA(n_components = pca_components)
_pca.fit(data)
for i in range(repeat):
U = _pca.transform(data)
time = timer() - start
err = diffsnorm(data, U, _pca.explained_variance_, _pca.components_)
print('sklearn PCA time: %.3fms, error: %E' % (time*1000/repeat, err))
start = timer()
for i in range(repeat):
(U, s, Va) = pca_mark(data, pca_components)
time = timer() - start
err = diffsnorm(data, U, s, Va.T)
print('pca by Mark time: %.3fms, error: %E' % (time*1000/repeat, err))
start = timer()
for i in range(repeat):
(U, s, Va) = pca_doug(data, pca_components)
time = timer() - start
err = diffsnorm(data, U, s[:pca_components], Va.T)
print('pca by doug time: %.3fms, error: %E' % (time*1000/repeat, err))
pca_mark is the pca in Mark's answer.
pca_doug is the pca in doug's answer.
Here is an example output (but the result depends very much on the data size and pca_components, so I'd recommend to run your own test with your own data. Also, facebook's pca is optimized for normalized data, so it will be faster and more accurate in that case):
svd time: 3212.228ms, error: 1.907320E-10
matplotlib PCA time: 879.210ms, error: 2.478853E+05
facebook pca time: 485.483ms, error: 1.260335E+04
sklearn PCA time: 169.832ms, error: 7.469847E+07
pca by Mark time: 293.758ms, error: 1.713129E+02
pca by doug time: 300.326ms, error: 1.707492E+02
EDIT:
The diffsnorm function from fbpca calculates the spectral-norm error of a Schur decomposition.
This will may be the simplest answer one can find for the PCA including easily understandable steps. Let say we want to retain 2 principal dimensions from the 144 which provides maximum information.
Firstly, convert your 2-D array to a dataframe:
import pandas as pd
# Here X is your array of size (26424 x 144)
data = pd.DataFrame(X)
Then, there are two methods one can go with:
Method 1: Manual calculation
Step 1: Apply column standardization on X
from sklearn import preprocessing
scalar = preprocessing.StandardScaler()
standardized_data = scalar.fit_transform(data)
Step 2: Find Co-variance matrix S of original matrix X
sample_data = standardized_data
covar_matrix = np.cov(sample_data)
Step 3: Find eigen values and eigen vectors of S (here 2D, so 2 of each)
from scipy.linalg import eigh
# eigh() function will provide eigen-values and eigen-vectors for a given matrix.
# eigvals=(low value, high value) takes eigen value numbers in ascending order
values, vectors = eigh(covar_matrix, eigvals=(142,143))
# Converting the eigen vectors into (2,d) shape for easyness of further computations
vectors = vectors.T
Step 4: Transform the data
# Projecting the original data sample on the plane formed by two principal eigen vectors by vector-vector multiplication.
new_coordinates = np.matmul(vectors, sample_data.T)
print(new_coordinates.T)
This new_coordinates.T will be of size (26424 x 2) with 2 principal components.
Method 2: Using Scikit-Learn
Step 1: Apply column standardization on X
from sklearn import preprocessing
scalar = preprocessing.StandardScaler()
standardized_data = scalar.fit_transform(data)
Step 2: Initializing the pca
from sklearn import decomposition
# n_components = numbers of dimenstions you want to retain
pca = decomposition.PCA(n_components=2)
Step 3: Using pca to fit the data
# This line takes care of calculating co-variance matrix, eigen values, eigen vectors and multiplying top 2 eigen vectors with data-matrix X.
pca_data = pca.fit_transform(sample_data)
This pca_data will be of size (26424 x 2) with 2 principal components.
For the sake def plot_pca(data): will work, it is necessary to replace the lines
data_resc, data_orig = PCA(data)
ax1.plot(data_resc[:, 0], data_resc[:, 1], '.', mfc=clr1, mec=clr1)
with lines
newData, data_resc, data_orig = PCA(data)
ax1.plot(newData[:, 0], newData[:, 1], '.', mfc=clr1, mec=clr1)
this sample code loads the Japanese yield curve, and creates PCA components.
It then estimates a given date's move using the PCA and compares it against the actual move.
%matplotlib inline
import numpy as np
import scipy as sc
from scipy import stats
from IPython.display import display, HTML
import pandas as pd
import matplotlib
import matplotlib.pyplot as plt
import datetime
from datetime import timedelta
import quandl as ql
start = "2016-10-04"
end = "2019-10-04"
ql_data = ql.get("MOFJ/INTEREST_RATE_JAPAN", start_date = start, end_date = end).sort_index(ascending= False)
eigVal_, eigVec_ = np.linalg.eig(((ql_data[:300]).diff(-1)*100).cov()) # take latest 300 data-rows and normalize to bp
print('number of PCA are', len(eigVal_))
loc_ = 10
plt.plot(eigVec_[:,0], label = 'PCA1')
plt.plot(eigVec_[:,1], label = 'PCA2')
plt.plot(eigVec_[:,2], label = 'PCA3')
plt.xticks(range(len(eigVec_[:,0])), ql_data.columns)
plt.legend()
plt.show()
x = ql_data.diff(-1).iloc[loc_].values * 100 # set the differences
x_ = x[:,np.newaxis]
a1, _, _, _ = np.linalg.lstsq(eigVec_[:,0][:, np.newaxis], x_) # linear regression without intercept
a2, _, _, _ = np.linalg.lstsq(eigVec_[:,1][:, np.newaxis], x_)
a3, _, _, _ = np.linalg.lstsq(eigVec_[:,2][:, np.newaxis], x_)
pca_mv = m1 * eigVec_[:,0] + m2 * eigVec_[:,1] + m3 * eigVec_[:,2] + c1 + c2 + c3
pca_MV = a1[0][0] * eigVec_[:,0] + a2[0][0] * eigVec_[:,1] + a3[0][0] * eigVec_[:,2]
pca_mV = b1 * eigVec_[:,0] + b2 * eigVec_[:,1] + b3 * eigVec_[:,2]
display(pd.DataFrame([eigVec_[:,0], eigVec_[:,1], eigVec_[:,2], x, pca_MV]))
print('PCA1 regression is', a1, a2, a3)
plt.plot(pca_MV)
plt.title('this is with regression and no intercept')
plt.plot(ql_data.diff(-1).iloc[loc_].values * 100, )
plt.title('this is with actual moves')
plt.show()

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