kylecampbell

# METHOD 1 vs. METHOD 2

# test sample
l = [[1,2,3],[4,5,6], [7], [8,9]]*99

# METHOD 1: List Comprehension (slower)
flat_list = [item for sublist in l for item in sublist]
# tested in .ipynb with %timeit

# METHOD 2: itertools chain (~2x faster)

kylecampbell

void cubic_solver(double A, double B, double C, double D, double *roots) {
    double xN, yN;                // yN = f(xN)                        
    double del;
    double h;
    double theta;
    double alpha, beta, gamma;    // roots

    xN = -B/(3*A);
    yN = (A*xN*xN*xN) + (B*xN*xN) + (C*xN) + D;
    del = sqrt( (B*B - 3*A*C) / ((9*A*A)) );

kylecampbell

import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
import numpy as np
import pandas as pd
import seaborn as sns
#uncomment below if using Jupyter
#%config InlineBackend.figure_format = 'retina'

# get data

kylecampbell

import numpy as np
import pandas as pd
import tensorflow as tf
import matplotlib.pyplot as plt
import seaborn as sns
#uncomment below if using Jupyter
#%config InlineBackend.figure_format = 'retina'

# get data
df = pd.read_csv('../some/data/path')

kylecampbell

import tensorflow as tf
import numpy as np
import matplotlib.pyplot as plt

# setup dummy data
n = 1000
x = np.random.normal(0, 0.55, n)
y = 0.1*x + 0.3 + np.random.normal(0, 0.03, n)  #want to find a=0.1 and b=0.3
plt.plot(x, y)
plt.show()