Hello! I am trying to code an algorythm that calculates long distance electrostatic interactions. I have a csv file where i have the average diatnce between two aminoacids, used to calculate the value of the free energy of these interactions. I introduce a sequence and, if there is a pair of charged aminoacids (not necessarly consecutive) i want it to get the value of r. It seems that it’s not doing this right now and i am getting really low values of G.

The csv code looks somethin like:

||Pos| i+1| i+2| i+3|…

|—|—|—|—|—|

DD| 5.5| 7.3| 5.5|

|DE| 5.8| 8.6| 6.4|

|DK| 7.3| 10.5| 6.2|

|DR| 7.2| 11.7| 5.8|

and so on.

This is my current code:

```
electroH_df = pd.read_csv("electro_helix.csv")
electroC_df = pd.read_csv("electro_coil.csv")
def calculate_energies(segment, electro_df, temperature_cels):
K = np.sqrt(8 * np.pi * (e**2) * N_A * I / (1000 * k_b * (temperature_cels + 273)))
length = len(segment)
energies = np.zeros(length) # Initialize energies array
for i in range(length):
total_energy = 0.0
for j in range(i + 1, length):
res1 = segment[i]
res2 = segment[j]
if res1 in charge_dict and res2 in charge_dict:
charge1 = charge_dict[res1]
charge2 = charge_dict[res2]
key = f"{res1}{res2}"
if key in electro_df['Pos'].values:
row = electro_df[electro_df['Pos'] == key].iloc[0]
position_difference = abs(j - i)
if position_difference < len(row):
r = row.iloc[position_difference]
r = r * 1e-10 # Convert to meters
if r > 0:
G_electro = ((e**2) * charge1 * charge2) / (3 * np.pi * epsilon_0 * epsilon_r * r) * np.exp(-K * r)
energies[i] += G_electro
energies[j] += G_electro # Add interaction to both residues
print(f"Calculated K: {K}")
print(f"Sample r values: {r}")
print(f"Sample G_electro values: {G_electro}")
return energies
def electrostatic_energy(segment, electroH_df, electroC_df, temperature_cels):
helix_energies = calculate_energies(segment, electroH_df, temperature_cels)
coil_energies = calculate_energies(segment, electroC_df, temperature_cels)
energy_difference = [h - c for h, c in zip(helix_energies, coil_energies)]
total_energy_difference = sum(energy_difference)
print(f"Helix energies: {helix_energies}")
print(f"Coil energies: {coil_energies}")
print(f"Energy difference: {energy_difference}")
print(f"Total energy difference: {total_energy_difference}")
return total_energy_difference
```

The main part of the code looks like:

```
# constant\
R = 1.987204258e-3
epsilon_0 = 8.854e-12 # Permittivity of free space in F/m
Q = 0.5*1.60217663e-19 # Assumed total charge of the helix macrodipole : "half a charge" from 1998 paper
e= 1.60217663e-19 #Coulombs
epsilon_r = 85.763 # Relative permittivity for water (this can be adjusted based on the environment)
N_A = 6.022e23 #Avogadro's number
I = 0.1 #DEfAULT (should be an input)
k_b = 1.380649e-23 # m^2 kg s^-2 K^-1
# define user input
def main():
print("Welcome to the helicity calculator!")
sequence = input("Enter your sequence: ").strip().upper()
while True:
try:
temperature_kelv = float(input("Enter your temperature in K: "))
break
except ValueError:
print("Invalid input. Please enter a numeric value for temperature.")
# convert kelvin to celsius
temperature_cels = temperature_kelv - 273.0
# extract segments after obtaining sequence
segments = extract_segments(sequence)
helix_length = calculate_alpha_helix_length(sequence)
# helicity
results = calculate_helicity(sequence, temperature_cels, helix_length)
# print results (aa with corresponding helicity)
print("\nHelical propensity per residue:")
for result in results['helical_propensity']:
aa = result[0]
propensity = result[1]
print(f"AA: {aa}, Hel: {propensity}") # AA for amino acid, Hel for helical propensity
print(f"\nAverage helical propensity: {results['percent_helix']}")
# Print electrostatic energies
print("\nElectrostatic energies per residue:")
for i, energy in enumerate(results['electrostatic_energies']):
print(f"Residue {sequence[i]} at position {i + 1}: Electrostatic energy = {energy}")
```

I believe I’m not accessing correctly to the csv file, but if someone sees another way to implement my problem i would really appreciate. Thank yoiu !