Curve fitting for each column in Pandas + extrapolate values

Question

I have a data set with some 300 columns, each of them depth-dependent. The simplified version of the Pandas DataFrame would look something like this:

import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from scipy_optimize import curve_fit

df1 = pd.DataFrame({'depth': [1.65, 2.15, 2.65, 3.15, 3.65, 4.15, 4.65, 5.15, 5.65, 6.15, 6.65, 7.15, 7.65, 8.15, 8.65],
               '400.0': [13.909261, 7.758734, 3.513627, 2.095409, 1.628918, 0.782643, 0.278548, 0.160153, -0.155895, -0.152373, -0.147820, -0.023997, 0.010729, 0.006050, 0.002356],
               '401.0': [14.581624, 8.173803, 3.757856, 2.223524, 1.695623, 0.818065, 0.300235, 0.173674, -0.145402, -0.144456, -0.142969, -0.022471, 0.010802, 0.006181, 0.002641],
               '402.0': [15.253988, 8.588872, 4.002085, 2.351638, 1.762327, 0.853486, 0.321922, 0.187195, -0.134910, -0.136539, -0.138118, -0.020945, 0.010875, 0.006313, 0.002927],
               '403.0': [15.633908, 8.833914, 4.146499, 2.431543, 1.798185, 0.874350, 0.333470, 0.192128, -0.130119, -0.134795, -0.136049, -0.019307, 0.012037, 0.006674, 0.003002],
               '404.0': [15.991816, 9.066159, 4.283401, 2.507818, 1.831721, 0.894119, 0.344256, 0.196415, -0.125758, -0.133516  , -0.134189, -0.017659, -0.013281,0.007053, 0.003061],
               '405.0': [16.349725, 9.298403, 4.420303, 2.584094, 1.865257, 0.913887, 0.355041, 0.200702, -0.121396, -0.132237, -0.132330, -0.016012, 0.014525, 0.007433, 0.003120]
               })

What I need to do is to estimate the K in the equation below. Basically each column corresponds to a I(z) profile. The I(0) has to be calculated, for which I used the curve_fit, as a reference I'm using this helpful post: https://stackoverflow.com/a/15369787/7541421

x = df1.depth       # Column values as a function of depth
y = df1['400.0']

plt.plot(x, y, 'ro',label="Original Data")

def func(def func(x, I0, k):     # a = I0, b = k
    return I0 * np.exp(-k*x)    

popt, pcov = curve_fit(func, x, y)

print ("E0 = %s , k = %s" % (popt[0], popt[1]))

plt.plot(x, func(x, *popt), label="Fitted Curve")

Could this be done for each column separately and somehow saved as a new DataFrame?

Also, the new DataFrame needs to be propagated to the values towards z=0 for certain dz quotas. In this case I'm missing [0.15, 0.65, 1.15] in my depth column. So for every z I need to get per each column the I(z) from the function.

How can I automatize it since every data set has a different depth range in my case?

P.S. Alternatively, as it has been originally discussed in this post, a log-transfored linear regression fit can be applied, for which the solution is written in an answer below.

Answer 1

Some changes have been made after the conversation with the principal author of this answer and with his approval.

First of all, since we are dealing with log-transform quantities, it is necessary to find the range of values which correspond to non-negative values per column.

negative_idx_aux = df_drop_depth.apply(lambda x:(x<0).nonzero()[0][:1].tolist())   
negative_idx = [item for sublist in negative_idx_aux for item in sublist]

if len(negative_idx) > 0:
    max_idx = max_idx = np.min(negative_idx)
else:
    max_idx = None

Compared to the original, I only merge the loops to obtain both the slope and intercept.

iz_cols = df1.columns.difference(['depth'])
slp_int = {}
for c in iz_cols:
    slope, intercept, r_value, p_value, std_err = stats.linregress(df1['depth'][0:max_idx],np.log(df1[c][0:max_idx]))
    slp_int[c] = [intercept, slope]

slp_int = pd.DataFrame(, index = ['intercept', 'slope'])

Exponentiating intercept gives us the value of I at the surface:

slp_int.loc['intercept'] = np.exp(slp_int.loc['intercept'])

The last part of the post has been corrected due to a misunderstanding of the final concept. The dataframe is now recreated, with new values for the surface depths (above the depth range of df1, keeping the df1 for values below.

First a whole range between z = 0 and the maximum value of the depth column is recreated, with an assigned step plus keeping the value at z = 0:

depth = np.asarray(df1.depth)
depth_min = np.min(depth)    ;   
depth_min_arr = np.array([depth_min])
step = 0.5
missing_vals_aux = np.arange(depth_min - step, 0, -step)[::-1]
missing_vals = np.concatenate(([0.], missing_vals_aux), axis=0)
depth_tot = np.concatenate((missing_vals, depth), axis=0)

df_boundary = pd.DataFrame(columns = iz_cols) 
df_up = pd.DataFrame(columns = iz_cols) 

Create a dataframe with the range of the upward-propagated depth quotas:

for c in iz_cols: 
    df_up[c]       = missing_vals

Fill the data with the regression-obtained parameters:

upper_df = slp_int.loc['intercept']*np.exp(slp_int.loc['slope']*df_up)
upper_df['depth'] = missing_vals

Merge the df1 and the upper_df to obtain a whole profile:

lower_df = df1
lower_df['depth'] = depth

df_profile_tot = upper_df.append(lower_df, ignore_index=True)