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2019-nCoV prediction
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 #!/usr/bin/env python # Not meant to teach proper Python, or statistics, but it works import warnings import numpy import scipy.stats x = numpy.array([1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0]) y = numpy.array([4515.0, 5974.0, 7711.0, 9692.0, 11791.0, 14380.0, 17205.0, 20438.0, 24324.0, 28018.0]) z = numpy.polyfit(x,y,2) print("ax^2 + bx + c: " + str(z[0]) + " * x^2 + "+ str(z[1]) + " * x + " + str(z[2])) p = numpy.poly1d(z) # p(1) # p(2) predictions = numpy.array([p(1), p(2), p(3), p(4), p(5), p(6), p(7), p(8), p(9), p(10)]) slope, intercept, r_value, p_value, std_err = regres_result = scipy.stats.linregress(y, predictions) print(regres_result) # Nonsense # adjusted_r_squared = (1-(1-r_value) * ((10-1)/(10-3-1))) # print("Adjusted R squared: " + str(adjusted_r_squared)) # print("Remaining unexplained variance: " + str((1 - adjusted_r_squared) * 100) + "%") print("") print("Predictions:") print("Feb 7: " + str(p(11))) print("Feb 8: " + str(p(12))) print("Feb 9: " + str(p(13)))

### HenkPoley commented Feb 6, 2020 • edited

Current output of this script:

``````ax^2 + bx + c: 160.113636364 * x^2 + 837.174242424 * x + 3635.96666667
LinregressResult(slope=0.99977799747621676, intercept=3.1979019545997289, rvalue=0.99988899257678443, pvalue=6.6424499499822616e-16, stderr=0.0052672720484915241)
Adjusted R squared: 0.999888992577
Remaining unexplained variance: 0.0111007423216%

Predictions:
Feb 7: 32218.6333333
Feb 8: 36738.4212121
Feb 9: 41578.4363636
``````

### HenkPoley commented Feb 6, 2020 • edited

Made the script because of these tweets: https://twitter.com/evdefender/status/1225408294585393153

Just from playing with my script a bit, I don't think these statistical methods are appropriate for this limited dataset.

Easy to make a higher order polyfit with an R^2 that approaches 1, and super low p values, but get silly predictions.

But, ehh, I had limited stats edu.

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