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I have a Pandas series. I need to get sigma_i, which is the standard deviation of a series up to index i. Is there an existing function which efficiently calculates that?
I noticed that there are the cummax and cummin functions.
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The standard deviation is the square root of the average of the squared deviations from the mean, i.e., std = sqrt(mean(x)) , where x = abs(a - a. mean())**2 . The average squared deviation is typically calculated as x. sum() / N , where N = len(x) .
Coding a stdev() Function in Python sqrt() to take the square root of the variance. With this new implementation, we can use ddof=0 to calculate the standard deviation of a population, or we can use ddof=1 to estimate the standard deviation of a population using a sample of data.
stdev() method calculates the standard deviation from a sample of data. Standard deviation is a measure of how spread out the numbers are. A large standard deviation indicates that the data is spread out, - a small standard deviation indicates that the data is clustered closely around the mean.
Numpy is memory efficient. Pandas has a better performance when a number of rows is 500K or more. Numpy has a better performance when number of rows is 50K or less. Indexing of the pandas series is very slow as compared to numpy arrays.
See pandas.expanding_std.
For instance:
import numpy as np
import matplotlib.pyplot as plt
import pandas
%matplotlib inline
data = pandas.Series(np.random.normal(size=37))
full_std = np.std(data)
expand_std = pandas.expanding_std(data, min_periods=1)
fig, ax = plt.subplots()
expand_std.plot(ax=ax, color='k', linewidth=1.5, label='Expanded Std. Dev.')
ax.axhline(y=full_std, color='g', label='Full Value')
ax.legend()
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