How can i smooth data in Python? [closed]

Question

I'm using Python to detect some patterns on OHLC data. My problem is that the data I have is very noisy (I'm using Open data from the Open/High/Low/Close dataset), and it often leads me to incorrect or weak outcomes.

Is there any way to "smooth" this data, or to make it less noisy, to improve my results? What algorithms or libraries can I use for this task?

Here is a sample of my data, which is a normal array:

DataPoints = [6903.79, 6838.04, 6868.57, 6621.25, 7101.99, 7026.78, 7248.6, 7121.4, 6828.98, 6841.36, 7125.12, 7483.96, 7505.0, 7539.03, 7693.1, 7773.51, 7738.58, 8778.58, 8620.0, 8825.67, 8972.58, 8894.15, 8871.92, 9021.36, 9143.4, 9986.3, 9800.02, 9539.1, 8722.77, 8562.04, 8810.99, 9309.35, 9791.97, 9315.96, 9380.81, 9681.11, 9733.93, 9775.13, 9511.43, 9067.51, 9170.0, 9179.01, 8718.14, 8900.35, 8841.0, 9204.07, 9575.87, 9426.6, 9697.72, 9448.27, 10202.71, 9518.02, 9666.32, 9788.14, 9621.17, 9666.85, 9746.99, 9782.0, 9772.44, 9885.22, 9278.88, 9464.96, 9473.34, 9342.1, 9426.05, 9526.97, 9465.13, 9386.32, 9310.23, 9358.95, 9294.69, 9685.69, 9624.33, 9298.33, 9249.49, 9162.21, 9012.0, 9116.16, 9192.93, 9138.08, 9231.99, 9086.54, 9057.79, 9135.0, 9069.41, 9342.47, 9257.4, 9436.06, 9232.42, 9288.34, 9234.02, 9303.31, 9242.61, 9255.85, 9197.6, 9133.72, 9154.31, 9170.3, 9208.99, 9160.78, 9390.0, 9518.16, 9603.27, 9538.1, 9700.42, 9931.54, 11029.96, 10906.27, 11100.52, 11099.79, 11335.46, 11801.17, 11071.36, 11219.68, 11191.99, 11744.91, 11762.47, 11594.36, 11761.02, 11681.69, 11892.9, 11392.09, 11564.34, 11779.77, 11760.55, 11852.4, 11910.99, 12281.15, 11945.1, 11754.38]

plt.plot(DataPoints)

Answer 1

Smoothing is a pretty rich subject; there are several methods each with features and drawbacks. Here is one using scipy:

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from scipy.signal import savgol_filter    

# noisy data
x = [6903.79, 6838.04, 6868.57, 6621.25, 7101.99, 7026.78, 7248.6, 7121.4, 6828.98, 6841.36, 7125.12, 7483.96, 7505.0, 7539.03, 7693.1, 7773.51, 7738.58, 8778.58, 8620.0, 8825.67, 8972.58, 8894.15, 8871.92, 9021.36, 9143.4, 9986.3, 9800.02, 9539.1, 8722.77, 8562.04, 8810.99, 9309.35, 9791.97, 9315.96, 9380.81, 9681.11, 9733.93, 9775.13, 9511.43, 9067.51, 9170.0, 9179.01, 8718.14, 8900.35, 8841.0, 9204.07, 9575.87, 9426.6, 9697.72, 9448.27, 10202.71, 9518.02, 9666.32, 9788.14, 9621.17, 9666.85, 9746.99, 9782.0, 9772.44, 9885.22, 9278.88, 9464.96, 9473.34, 9342.1, 9426.05, 9526.97, 9465.13, 9386.32, 9310.23, 9358.95, 9294.69, 9685.69, 9624.33, 9298.33, 9249.49, 9162.21, 9012.0, 9116.16, 9192.93, 9138.08, 9231.99, 9086.54, 9057.79, 9135.0, 9069.41, 9342.47, 9257.4, 9436.06, 9232.42, 9288.34, 9234.02, 9303.31, 9242.61, 9255.85, 9197.6, 9133.72, 9154.31, 9170.3, 9208.99, 9160.78, 9390.0, 9518.16, 9603.27, 9538.1, 9700.42, 9931.54, 11029.96, 10906.27, 11100.52, 11099.79, 11335.46, 11801.17, 11071.36, 11219.68, 11191.99, 11744.91, 11762.47, 11594.36, 11761.02, 11681.69, 11892.9, 11392.09, 11564.34, 11779.77, 11760.55, 11852.4, 11910.99, 12281.15, 11945.1, 11754.38]

df = pd.DataFrame(dict(x=data))
x_filtered = df[["x"]].apply(savgol_filter,  window_length=31, polyorder=2)

plt.ion()
plt.plot(x)
plt.plot(x_filtered)
plt.show()

Answer 2

If you can plot it, use the moving average/rolling mean and identify the window size which suites your data.

from scipy.ndimage.filters import uniform_filter1d

data = [6903.79, 6838.04, 6868.57, 6621.25, 7101.99, 7026.78, 7248.6, 7121.4, 6828.98, 6841.36, 7125.12, 7483.96, 7505.0, 7539.03, 7693.1, 7773.51, 7738.58, 8778.58, 8620.0, 8825.67, 8972.58, 8894.15, 8871.92, 9021.36, 9143.4, 9986.3, 9800.02, 9539.1, 8722.77, 8562.04, 8810.99, 9309.35, 9791.97, 9315.96, 9380.81, 9681.11, 9733.93, 9775.13, 9511.43, 9067.51, 9170.0, 9179.01, 8718.14, 8900.35, 8841.0, 9204.07, 9575.87, 9426.6, 9697.72, 9448.27, 10202.71, 9518.02, 9666.32, 9788.14, 9621.17, 9666.85, 9746.99, 9782.0, 9772.44, 9885.22, 9278.88, 9464.96, 9473.34, 9342.1, 9426.05, 9526.97, 9465.13, 9386.32, 9310.23, 9358.95, 9294.69, 9685.69, 9624.33, 9298.33, 9249.49, 9162.21, 9012.0, 9116.16, 9192.93, 9138.08, 9231.99, 9086.54, 9057.79, 9135.0, 9069.41, 9342.47, 9257.4, 9436.06, 9232.42, 9288.34, 9234.02, 9303.31, 9242.61, 9255.85, 9197.6, 9133.72, 9154.31, 9170.3, 9208.99, 9160.78, 9390.0, 9518.16, 9603.27, 9538.1, 9700.42, 9931.54, 11029.96, 10906.27, 11100.52, 11099.79, 11335.46, 11801.17, 11071.36, 11219.68, 11191.99, 11744.91, 11762.47, 11594.36, 11761.02, 11681.69, 11892.9, 11392.09, 11564.34, 11779.77, 11760.55, 11852.4, 11910.99, 12281.15, 11945.1, 11754.38]

plt.figure(figsize=(15,15))
plt.plot(data)
for i in range(3, 100, 10):
  y = uniform_filter1d(data, size=i)
  plt.plot(y, '--', label=f"{i}")

plt.legend()

Output:

Answer 3

As mentioned in the comments, you can take the moving average, which sort of works like a convolutional layer. It averages the values from 0 to n and sets that as point 0. Then it averages values 1 to n+1, and sets that as point one. the larger the n, the less points you will have, yet the smoother it will be. You can get the moving average using the code below:

import numpy as np
def moving_avg(x, n):
    cumsum = np.cumsum(np.insert(x, 0, 0)) 
    return (cumsum[n:] - cumsum[:-n]) / float(n)

I found that code snippet here. With a n of 10, I got the following plot: