I have two variables that I have plotted using matplotlib scatter function.
I would like to show the 68% confidence region by highlighting it in the plot. I know to show it in a histogram, but I don't know how to do it for a 2D plot like this (x vs y). In my case, the x is Mass and y is Ngal Mstar+2.
An example image of what I am looking for looks like this:
Here they have showed the 68% confidence region using dark blue and 95% confidence region using light blue.
Can it be achieved using one of thescipy.stats modules?
To plot a region between two curves, you could use pyplot.fill_between().
As for your confidence region, I was not sure what you wanted to achieve, so I exemplified with simultaneous confidence bands, by modifying the code from:
https://en.wikipedia.org/wiki/Confidence_and_prediction_bands#cite_note-2
import numpy as np
import matplotlib.pyplot as plt
import scipy.special as sp
## Sample size.
n = 50
## Predictor values.
XV = np.random.uniform(low=-4, high=4, size=n)
XV.sort()
## Design matrix.
X = np.ones((n,2))
X[:,1] = XV
## True coefficients.
beta = np.array([0, 1.], dtype=np.float64)
## True response values.
EY = np.dot(X, beta)
## Observed response values.
Y = EY + np.random.normal(size=n)*np.sqrt(20)
## Get the coefficient estimates.
u,s,vt = np.linalg.svd(X,0)
v = np.transpose(vt)
bhat = np.dot(v, np.dot(np.transpose(u), Y)/s)
## The fitted values.
Yhat = np.dot(X, bhat)
## The MSE and RMSE.
MSE = ((Y-EY)**2).sum()/(n-X.shape[1])
s = np.sqrt(MSE)
## These multipliers are used in constructing the intervals.
XtX = np.dot(np.transpose(X), X)
V = [np.dot(X[i,:], np.linalg.solve(XtX, X[i,:])) for i in range(n)]
V = np.array(V)
## The F quantile used in constructing the Scheffe interval.
QF = sp.fdtri(X.shape[1], n-X.shape[1], 0.95)
QF_2 = sp.fdtri(X.shape[1], n-X.shape[1], 0.68)
## The lower and upper bounds of the Scheffe band.
D = s*np.sqrt(X.shape[1]*QF*V)
LB,UB = Yhat-D,Yhat+D
D_2 = s*np.sqrt(X.shape[1]*QF_2*V)
LB_2,UB_2 = Yhat-D_2,Yhat+D_2
## Make the plot.
plt.clf()
plt.plot(XV, Y, 'o', ms=3, color='grey')
plt.hold(True)
a = plt.plot(XV, EY, '-', color='black', zorder = 4)
plt.fill_between(XV, LB_2, UB_2, where = UB_2 >= LB_2, facecolor='blue', alpha= 0.3, zorder = 0)
b = plt.plot(XV, LB_2, '-', color='blue', zorder=1)
plt.plot(XV, UB_2, '-', color='blue', zorder=1)
plt.fill_between(XV, LB, UB, where = UB >= LB, facecolor='blue', alpha= 0.3, zorder = 2)
b = plt.plot(XV, LB, '-', color='blue', zorder=3)
plt.plot(XV, UB, '-', color='blue', zorder=3)
d = plt.plot(XV, Yhat, '-', color='red',zorder=4)
plt.ylim([-8,8])
plt.xlim([-4,4])
plt.xlabel("X")
plt.ylabel("Y")
plt.show()
The output looks like this:
First of all thank you #snake_charmer for your answer, but I have found a simpler way of solving the issue using curve_fit from scipy.optimize
I fit my data sample using curve_fit which gives me my best fit parameters. What it also gives me is the estimated covariance of the parameters. The diagonals of the same provide the variance of the parameter estimate. To compute one standard deviation errors on the parameters we can use np.sqrt(np.diag(pcov)) where pcov is the covariance matrix.
def fitfunc(M,p1,p2):
N = p1+( (M)*p2 )
return N
The above is the fit function I use for the data.
Now to fit the data using curve_fit
popt_1,pcov_1 = curve_fit(fitfunc,logx,logn,p0=(10.0,1.0),maxfev=2000)
p1_1 = popt_1[0]
p1_2 = popt_1[1]
sigma1 = [np.sqrt(pcov_1[0,0]),np.sqrt(pcov_1[1,1])] #THE 1 SIGMA CONFIDENCE INTERVALS
residuals1 = (logy) - fitfunc((logx),p1_1,p1_2)
xi_sq_1 = sum(residuals1**2) #THE CHI-SQUARE OF THE FIT
curve_y_1 = fitfunc((logx),p1_1,p1_2)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.scatter(logx,logy,c='r',label='$0.0<z<0.5$')
ax1.plot(logx,curve_y_1,'y')
ax1.plot(logx,fitfunc(logx,p1_1+sigma1[0],p1_2+sigma1[1]),'m',label='68% conf limits')
ax1.plot(logx,fitfunc(logx,p1_1-sigma1[0],p1_2-sigma1[1]),'m')
So just by using the square root the diagonal elements of the covariance matrix, I can obtain the 1 sigma confidence lines.
Related
I have written a code to fit the gaussian function in a dataset by scipy curve_fit. There are a few different datasets. One with 19 points and one with 21 points and both of them include different datasets in range of 0.5-0.7, 1.0-1.2 and 1.5-1.7.
Surprisingly, when I ran the code in 19 point datasets, all three of them executed successfully but in case of 21 point datasets, only 1.5-1.7 ranged data had the right fit. All others were given with horribly wrong fit.
Here is the code.
#function declaration
def gauss(x, amp, mu, sigma):
y = amp*np.exp(-(x-mu)**2/(2*sigma**2))
return y
#fitting
popt, pcov = curve_fit(f = gauss, xdata = x, ydata = y)
#print(popt)
amp = popt[0]
mu = popt[1]
sigma = popt[2]
print(amp,mu,sigma)
#krypton value
krypton_y = amp/((math.exp(1))**2)
#print(krypton_y)
krypton_x1 = mu + math.sqrt((-2*(sigma**2))*math.log(krypton_y/amp))
krypton_x2 = mu - math.sqrt((-2*(sigma**2))*math.log(krypton_y/amp))
print(krypton_x1-krypton_x2)
#print(gauss([krypton_x1, krypton_x2], popt[0], popt[1], popt[2]))
#horizontal line
horizontal_x = np.arange(min(x)-0.01, max(x)+0.02, 0.01)
horizontal_y = np.repeat(0, len(horizontal_x))
#build fit set
x_test = np.arange(min(x), max(x), 0.0000001)
y_test = gauss(x_test, popt[0], popt[1], popt[2])
y_krypton = []
for i in horizontal_x:
y_krypton.append(krypton_y)
#Vertical lines
vertical_y = np.arange(-20, amp+20, 0.01)
l = len(vertical_y)
vertical_mean = np.repeat(mu, l)
#fit data
fig = plt.figure()
fig = plt.scatter(x,y, label ='original data', color = 'red', marker = 'x')
fig = plt.plot(x_test, y_test, label = 'Gaussian fit curve')
fig = plt.plot(horizontal_x, y_krypton, color = '#830000', linewidth = 1)
fig = plt.plot(vertical_mean, vertical_y, color = '#0011ed')
fig = plt.xlabel('Distance in mm')
fig = plt.ylabel('Current in nA')
fig = plt.title('Intensity Profile for '+gas+' laser | Z = '+str(z)+'cm')
fig = plt.scatter(mu, amp, s = 25, color = '#0011ed')
fig = plt.scatter(krypton_x1, krypton_y, s = 25, color = '#830000')
fig = plt.scatter(krypton_x2, krypton_y, s = 25, color = '#830000')
plt.annotate('('+"{:.4f}".format(mu)+','+"{:.4f}".format(amp)+')', (mu, amp), xytext = (mu+0.002,amp+0.5))
plt.annotate('('+"{:.4f}".format(krypton_x1)+','+"{:.4f}".format(krypton_y)+')', (krypton_x1, krypton_y), xytext = (krypton_x1+0.002,krypton_y+0.5))
plt.annotate('('+"{:.4f}".format(krypton_x2)+','+"{:.4f}".format(krypton_y)+')', (krypton_x2, krypton_y), xytext = (krypton_x2+0.002,krypton_y+0.5))
plt.legend()
plt.margins(0)
plt.show()
I am also adding two images, the correct fit and the wrong fit.
In order to make clear the difficulty we will use an elementary regression method.
We see that the fitting involves ln(y) which is infinite at the points k<6 and k>16. Those points cannot be used for the numerical calculus. Also the point k=16 is not reliable because the small value of y=0.001 is not accurate enough (only one sigificative digit). So, we use only the points from k=6 to k=15 in the next calculus.
This shows that the non-significative points have to be eliminated. Of course more sophisticated methods implemented in nonlinear regression package with iterative calculus gives better fitting according to some particular criteria of fitting specified in the software.
I have a dataset for curvature and I need to find the tangent to the curve but unfortunately, this is a bit far from the curve. Kindly guide me the issue solution related to the problem. Thank you!
My code is as follows:
fig, ax1 = plt.subplots()
chData_m = efficient.get('Car.Road.y')
x_fit = chData_m.timestamps
y_fit = chData_m.samples
fittedParameters = np.polyfit(x_fit[:],y_fit[:],1)
f = plt.figure(figsize=(800/100.0, 600/100.0), dpi=100)
axes = f.add_subplot(111)
# first the raw data as a scatter plot
axes.plot(x_fit, y_fit, 'D')
# create data for the fitted equation plot
xModel = np.linspace(min(x_fit), max(x_fit))
yModel = np.polyval(fittedParameters, xModel)
# now the model as a line plot
axes.plot(xModel, yModel)
axes.set_xlabel('X Data') # X axis data label
axes.set_ylabel('Y Data') # Y axis data label
# polynomial derivative from numpy
deriv = np.polyder(fittedParameters)
# for plotting
minX = min(x_fit)
maxX = max(x_fit)
# value of derivative (slope) at a specific X value, so
# that a straight line tangent can be plotted at the point
# you might place this code in a loop to animate
pointVal = 10.0 # example X value
y_value_at_point = np.polyval(fittedParameters, pointVal)
slope_at_point = np.polyval(deriv, pointVal)
ylow = (minX - pointVal) * slope_at_point + y_value_at_point
yhigh = (maxX - pointVal) * slope_at_point + y_value_at_point
# now the tangent as a line plot
axes.plot([minX, maxX], [ylow, yhigh])
plt.show()
plt.close('all') # clean up after using pyplot
and the output is:
I am not sure how you wanted to use numpy polyfit/polyval to determine the tangent formula. I describe here a different approach. The advantage of this approach is that it does not have any assumptions about the nature of the function. The disadvantage is that it will not work for vertical tangents.
To be on the safe side, I have considered both cases, i.e., that the evaluated x-value is a data point in your series and that it is not. Some problems may arise because I see that you mention timestamps in your question without specifying their nature by providing a toy dataset - so, this version may or may not work with the datetime objects or timestamps of your original data:
import matplotlib.pyplot as plt
import numpy as np
#generate fake data with unique random x-values between 0 and 70
def func(x, a=0, b=100, c=1, n=3.5):
return a + (b/(1+(c/x)**n))
np.random.seed(123)
x = np.random.choice(range(700000), 100)/10000
x.sort()
y = func(x, 1, 2, 15, 2.4)
#data point to evaluate
xVal = 29
#plot original data
fig, ax = plt.subplots()
ax.plot(x, y, c="blue", label="data")
#calculate gradient
slope = np.gradient(y, x)
#determine slope and intercept at xVal
ind1 = (np.abs(x - xVal)).argmin()
#case 1 the value is a data point
if xVal == x[ind1]:
yVal, slopeVal = y[ind1], slope[ind1]
#case 2 the value lies between to data points
#in which case we approximate linearly from the two nearest data points
else:
if xVal < x[ind1]:
ind1, ind2 = ind1-1, ind1
else:
ind1, ind2 = ind1, ind1+1
yVal = y[ind1] + (y[ind2]-y[ind1]) * (xVal-x[ind1]) / (x[ind2]-x[ind1])
slopeVal = slope[ind1] + (slope[ind2]-slope[ind1]) * (xVal-x[ind1]) / (x[ind2]-x[ind1])
intercVal = yVal - slopeVal * xVal
ax.plot([x.min(), x.max()], [slopeVal*x.min()+intercVal, slopeVal*x.max()+intercVal], color="green",
label=f"tangent\nat point [{xVal:.1f}, {yVal:.1f}]\nwith slope {slopeVal:.2f}\nand intercept {intercVal:.2f}" )
ax.set_ylim(0.8 * y.min(), 1.2 * y.max())
ax.legend()
plt.show()
I have a heavily right-skewed histogram and would like to calculate the probabilities for a range of Lifetimevalues (Area under the curve, the PDF). For instance, the probability that the Lifetime value is in (0-0.01)
Dataframe consisting of LTV calculated by cumulative revenue/ cumulative installs:
df['LTV'] is
(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.208125,0.0558879,0.608348,0.212553,0.0865896,
0.728542,0,0.609512,0,0,0,0,0,0,0,0.0801339,0.140657,0.0194118,0,0,0.0634682,
0.339545,0.875902,0.8325,0.0260526,0.0711905,0.169894,0.202969,0.0761538,0,0.342055,
0.42781,0,0,0.192115,0,0,0,0,0,0,0,0,0,0,0,1.6473,0,0.232329,0,2.21329,0.748,0.0424286,
0.455439,0.210282,5.56453,0.427959,0,0.352059,0,0,0.567059,0,0,0,0.384462,1.29476,
0.0103125,0,0.0126923,1.03356,0,0,0.289785,0,0)
I have tried utilizing SKlearn's KernelDensity, however, after fitting it to the histogram it does not capture the over-represented 0s.
import gc
from sklearn.neighbors import KernelDensity
def plot_prob_density(df_lunch, field, x_start, x_end):
plt.figure(figsize = (10, 7))
unit = 0
x = np.linspace(df_lunch.min() - unit, df_lunch.max() + unit, 1000)[:, np.newaxis]
# Plot the data using a normalized histogram
plt.hist(df_lunch, bins=200, density=True, label='LTV', color='blue', alpha=0.2)
# Do kernel density estimation
kd_lunch = KernelDensity(kernel='gaussian', bandwidth=0.00187).fit(df_lunch) #0.00187
# Plot the estimated densty
kd_vals_lunch = np.exp(kd_lunch.score_samples(x))
plt.plot(x, kd_vals_lunch, color='orange')
plt.axvline(x=x_start,color='red',linestyle='dashed')
plt.axvline(x=x_end,color='red',linestyle='dashed')
# Show the plots
plt.xlabel(field, fontsize=15)
plt.ylabel('Probability Density', fontsize=15)
plt.legend(fontsize=15)
plt.show()
gc.collect()
return kd_lunch
kd_lunch = plot_prob_density(final_df['LTV'].values.reshape(-1,1), 'LTV', x_start=0, x_end=0.01)
Then finding the probabilities like this:
def get_probability(start_value, end_value, eval_points, kd):
# Number of evaluation points
N = eval_points
step = (end_value - start_value) / (N - 1) # Step size
x = np.linspace(start_value, end_value, N)[:, np.newaxis] # Generate values in the range
kd_vals = np.exp(kd.score_samples(x)) # Get PDF values for each x
probability = np.sum(kd_vals * step) # Approximate the integral of the PDF
return probability.round(4)
print('Probability of LTV 0-3 tips during LUNCH time: {}\n'
.format(get_probability(start_value = 0,
end_value = 0.01,
eval_points = 100,
kd = kd_lunch)))
However, this method does not yield the appropriate PDF values we were aiming for.
Any suggestions for alternative methods would be appreciated.
PLot:
I have used more or less similar script for my work, here is my script may be it will be helpful for you.
import gc
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np
from sklearn.neighbors import KernelDensity
from scipy import stats
data1 = beta_95[0]
def plot_prob_density(data1, x_start, x_end):
plt.figure(figsize = (4, 3.5))
unit = 1.5
x = np.linspace(-20, 20, 1000)[:, np.newaxis]
# Plot the data using a normalized histogram
plt.hist(data1, bins=np.linspace(-20,20,40), density=True, color='r', alpha=0.4)
#plt.show
# Do kernel density estimation
kd_data1 = KernelDensity(kernel='gaussian', bandwidth=1.8).fit(data1)
# Plot the estimated densty
kd_vals_data1 = np.exp(kd_data1.score_samples(x))
plt.plot(x, kd_vals_data1, color='r', label='$N_a$', linewidth = 2)
plt.axvline(x=9.95,color='green',linestyle='dashed', linewidth = 2.0, label='$β_o$')
plt.axvline(x=1.9,color='black',linestyle='dashed', linewidth = 2.0, label='$β_b$')
plt.axvline(x=x_end,color='red',linestyle='dashed', linewidth = 2, label='$β_{95\%}$')
# Show the plots
plt.xlabel('Beta', fontsize=10)
plt.ylabel('Probability Density', fontsize=10)
plt.title('02 hours window', fontsize=12)
plt.xlim(-20, 20)
plt.ylim(0, 0.3)
plt.yticks([0, 0.1, 0.2, 0.3])
plt.legend(fontsize=12, loc='upper left', frameon=False)
plt.show()
gc.collect()
return kd_data1
def get_probability(start_value, end_value, eval_points, kd):
# Number of evaluation points
N = eval_points
step = (end_value - start_value) / (N - 1) # Step size
x = np.linspace(start_value, end_value, N)[:, np.newaxis] # Generate values in the range
kd_vals = np.exp(kd.score_samples(x)) # Get PDF values for each x
probability = np.sum(kd_vals * step) # Approximate the integral of the PDF
return probability.round(4)
data1 = np.array(data1).reshape(-1, 1)
kd_data1 = plot_prob_density(data1, x_start=3.0, x_end=13)
print('Beta-95%: {}\n'
.format(get_probability(start_value = -10,
end_value = 13,
eval_points = 1000,
kd = kd_data1)))
Would somebody be able to explain to me how to use the location parameter with the gamma.fit function in Scipy?
It seems to me that a location parameter (μ) changes the support of the distribution from x ≥ 0 to y = ( x - μ ) ≥ 0. If μ is positive then aren't we losing all the data which doesn't satisfy x - μ ≥ 0?
Thanks!
The fit function takes all of the data into consideration when finding a fit. Adding noise to your data will alter the fit parameters and can give a distribution that does not represent the data very well. So we have to be a bit clever when we are using fit.
Below is some code that generates data, y1, with loc=2 and scale=1 using numpy. It also adds noise to the data over the range 0 to 10 to create y2. Fitting y1 yield excellent results, but attempting to fit the noisy y2 is problematic. The noise we added smears out the distribution. However, we can also hold 1 or more parameters constant when fitting the data. In this case we pass floc=2 to the fit, which forces the location to be held at 2 when performing the fit, returning much better results.
from scipy.stats import gamma
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(0,10,.1)
y1 = np.random.gamma(shape=1, scale=1, size=1000) + 2 # sets loc = 2
y2 = np.hstack((y1, 10*np.random.rand(100))) # add noise from 0 to 10
# fit the distributions, get the PDF distribution using the parameters
shape1, loc1, scale1 = gamma.fit(y1)
g1 = gamma.pdf(x=x, a=shape1, loc=loc1, scale=scale1)
shape2, loc2, scale2 = gamma.fit(y2)
g2 = gamma.pdf(x=x, a=shape2, loc=loc2, scale=scale2)
# again fit the distribution, but force loc=2
shape3, loc3, scale3 = gamma.fit(y2, floc=2)
g3 = gamma.pdf(x=x, a=shape3, loc=loc3, scale=scale3)
And make some plots...
# plot the distributions and fits. to lazy to do iteration today
fig, axes = plt.subplots(1, 3, figsize=(13,4))
ax = axes[0]
ax.hist(y1, bins=40, normed=True);
ax.plot(x, g1, 'r-', linewidth=6, alpha=.6)
ax.annotate(s='shape = %.3f\nloc = %.3f\nscale = %.3f' %(shape1, loc1, scale1), xy=(6,.2))
ax.set_title('gamma fit')
ax = axes[1]
ax.hist(y2, bins=40, normed=True);
ax.plot(x, g2, 'r-', linewidth=6, alpha=.6)
ax.annotate(s='shape = %.3f\nloc = %.3f\nscale = %.3f' %(shape2, loc2, scale2), xy=(6,.2))
ax.set_title('gamma fit with noise')
ax = axes[2]
ax.hist(y2, bins=40, normed=True);
ax.plot(x, g3, 'r-', linewidth=6, alpha=.6)
ax.annotate(s='shape = %.3f\nloc = %.3f\nscale = %.3f' %(shape3, loc3, scale3), xy=(6,.2))
ax.set_title('gamma fit w/ noise, location forced')
I am following the StatsModels example here to plot quantile regression lines. With only slight modification for my data, the example works great, producing this plot (note that I have modified the code to only plot the 0.05, 0.25, 0.5, 0.75, and 0.95 quantiles) :
However, I would like to plot the OLS fit and corresponding quantiles for a 2nd order polynomial fit (instead of linear). For example, here is the 2nd-order OLS line for the same data:
How can I modify the code in the linked example to produce non-linear quantiles?
Here is my relevant code modified from the linked example to produce the 1st plot:
d = {'temp': x, 'dens': y}
df = pd.DataFrame(data=d)
# Least Absolute Deviation
#
# The LAD model is a special case of quantile regression where q=0.5
mod = smf.quantreg('dens ~ temp', df)
res = mod.fit(q=.5)
print(res.summary())
# Prepare data for plotting
#
# For convenience, we place the quantile regression results in a Pandas DataFrame, and the OLS results in a dictionary.
quantiles = [.05, .25, .50, .75, .95]
def fit_model(q):
res = mod.fit(q=q)
return [q, res.params['Intercept'], res.params['temp']] + res.conf_int().ix['temp'].tolist()
models = [fit_model(x) for x in quantiles]
models = pd.DataFrame(models, columns=['q', 'a', 'b','lb','ub'])
ols = smf.ols('dens ~ temp', df).fit()
ols_ci = ols.conf_int().ix['temp'].tolist()
ols = dict(a = ols.params['Intercept'],
b = ols.params['temp'],
lb = ols_ci[0],
ub = ols_ci[1])
print(models)
print(ols)
x = np.arange(df.temp.min(), df.temp.max(), 50)
get_y = lambda a, b: a + b * x
for i in range(models.shape[0]):
y = get_y(models.a[i], models.b[i])
plt.plot(x, y, linestyle='dotted', color='grey')
y = get_y(ols['a'], ols['b'])
plt.plot(x, y, color='red', label='OLS')
plt.scatter(df.temp, df.dens, alpha=.2)
plt.xlim((-10, 40))
plt.ylim((0, 0.4))
plt.legend()
plt.xlabel('temp')
plt.ylabel('dens')
plt.show()
After a day of looking into this, came up with a solution, so posting my own answer. Much credit to Josef Perktold at StatsModels for assistance.
Here is the relevant code and plot:
d = {'temp': x, 'dens': y}
df = pd.DataFrame(data=d)
x1 = pd.DataFrame({'temp': np.linspace(df.temp.min(), df.temp.max(), 200)})
poly_2 = smf.ols(formula='dens ~ 1 + temp + I(temp ** 2.0)', data=df).fit()
plt.plot(x, y, 'o', alpha=0.2)
plt.plot(x1.temp, poly_2.predict(x1), 'r-',
label='2nd order poly fit, $R^2$=%.2f' % poly_2.rsquared,
alpha=0.9)
plt.xlim((-10, 50))
plt.ylim((0, 0.25))
plt.xlabel('mean air temp')
plt.ylabel('density')
plt.legend(loc="upper left")
# with quantile regression
# Least Absolute Deviation
# The LAD model is a special case of quantile regression where q=0.5
mod = smf.quantreg('dens ~ temp + I(temp ** 2.0)', df)
res = mod.fit(q=.5)
print(res.summary())
# Quantile regression for 5 quantiles
quantiles = [.05, .25, .50, .75, .95]
# get all result instances in a list
res_all = [mod.fit(q=q) for q in quantiles]
res_ols = smf.ols('dens ~ temp + I(temp ** 2.0)', df).fit()
plt.figure()
# create x for prediction
x_p = np.linspace(df.temp.min(), df.temp.max(), 50)
df_p = pd.DataFrame({'temp': x_p})
for qm, res in zip(quantiles, res_all):
# get prediction for the model and plot
# here we use a dict which works the same way as the df in ols
plt.plot(x_p, res.predict({'temp': x_p}), linestyle='--', lw=1,
color='k', label='q=%.2F' % qm, zorder=2)
y_ols_predicted = res_ols.predict(df_p)
plt.plot(x_p, y_ols_predicted, color='red', zorder=1)
#plt.scatter(df.temp, df.dens, alpha=.2)
plt.plot(df.temp, df.dens, 'o', alpha=.2, zorder=0)
plt.xlim((-10, 50))
plt.ylim((0, 0.25))
#plt.legend(loc="upper center")
plt.xlabel('mean air temp')
plt.ylabel('density')
plt.title('')
plt.show()
red line: 2nd order polynomial fit
black dashed lines: 5th, 25th, 50th, 75th, 95th percentiles