# how to print equation of line using scipy stats

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• Techknowledgy :

My favorite, given no other constraints is an arrow to the line, because then the reader has no doubt what the equation is actually referring to. To do this, use `annotate`:

```x0 = 20000
y0 = slope * x0 + intercept
pylab.annotate(line_eqn, xy = (x0, y0), xytext = (x0 - .4 * x0, y0 + .4 * y0),
arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=-0.5'))```

Suggestion : 2

Defines the alternative hypothesis. Default is ‘two-sided’. The following options are available:,Calculate a linear least-squares regression for two sets of measurements.,The p-value for a hypothesis test whose null hypothesis is that the slope is zero, using Wald Test with t-distribution of the test statistic. See alternative above for alternative hypotheses.,The Pearson correlation coefficient. The square of rvalue is equal to the coefficient of determination.

`slope, intercept, r, p, se = linregress(x, y)`
```result = linregress(x, y)
print(result.intercept, result.intercept_stderr)```
```>>>
import matplotlib.pyplot as plt >>>
from scipy
import stats
>>>
rng = np.random.default_rng()```
```>>> x = rng.random(10) >>>
y = 1.6 * x + rng.random(10)```
`>>> res = stats.linregress(x, y)`
```>>> print(f "R-squared: {res.rvalue**2:.6f}")
R - squared: 0.717533```

Suggestion : 3

My code performs a linear regression on 2 sets of data. It works fine but i do not know how i can print the equation of the line onto the graph itself with scipy or numpy.,Python – How to do a line break (line continuation) in Python,Where do you want the equation to go? To put it on the title, for example: plt.title('\$y=%3.7sx+%3.7s\$'%(slope, intercept)). To put it inside the plot use plot.text.,Python – How to print literal curly-brace characters in a string and also use .format on it

Here is my code:

```y = np.array([15, 1489, 859, 336, 277, 265, 229, 285, 391, 372, 5, 345])
x = np.array([196.16, 17762.47, 28542.19, 30170.5, 9384.06, 43210.29, 21819.2, 16978.2, 45767.54, 12328.78, 113.71, 19257.6])

print x
print y

slope, intercept, r_value, p_value, slope_std_error = stats.linregress(x, y)
print "slope = " + str(slope)
print "r_value = " + str(r_value)
print "r_squared = " + str(r_value ** 2)
print "p_value = " + str(p_value)
predict_y = intercept + slope * x
print predict_y
pred_error = y - predict_y
degrees_of_freedom = len(x) - 2
residual_std_error = np.sqrt(np.sum(pred_error ** 2) / degrees_of_freedom)

# Plotting
pylab.xlabel('cost')
pylab.ylabel('signups')
pylab.plot(x, y, 'o')
pylab.plot(x, predict_y, 'k-')
pylab.show()```

Suggestion : 4

Last Updated : 20 Jun, 2022

Output :

```Kurtosis
for normal distribution: -0.3073930877422071

Kurtosis
for normal distribution: 2.692606912257793

Kurtosis
for normal distribution: -0.3073930877422071```

Suggestion : 5

Calculating with python the slope and the intercept of a straight line from two points (x1,y1) and (x2,y2):,Using s simple regression with scipy:,Plot with matplotlib:,I have developed this web site from scratch with Django to share with everyone my notes. If you have any ideas or suggestions to improve the site, let me know ! (you can contact me using the form in the welcome page). Thanks!

Calculating with python the slope and the intercept of a straight line from two points (x1,y1) and (x2,y2):

```x1 = 2.0
y1 = 3.0

x2 = 6.0
y2 = 5.0

a = (y2 - y1) / (x2 - x1)
b = y1 - a * x1

print('slope: ', a)
print('intercept: ', b)```

Using a function

```def slope_intercept(x1, y1, x2, y2):
a = (y2 - y1) / (x2 - x1)
b = y1 - a * x1
return a, b

print(slope_intercept(x1, y1, x2, y2))```

Using s simple regression with scipy:

```from scipy.stats
import linregress

slope, intercept, r_value, p_value, std_err = linregress([x1, x2], [y1, y2])
print(slope, intercept)```