How to create a dataframe column based on a conditional calculation

Question

I am trying to create a new column that will perform either 1 function or another depending on the value in the row in the df['Call/Put'] column. I am having difficulties performing a calculation based on certain row values as well as it determining which function to perform. Below is the last function I tried but it appears to not execute the formula properly. I have tried several ways to no avail but this is the last one I tried.

I am trying to create a new column called 'Black Scholes' and perform either bs_call if df['Call/Put']=='Call' in that row or perform bs_put if df['Call/Put']=='Put' in that row.

for index, value in df.iterrows():
        df['Black Scholes'][index]=np.where((df['Call/Put']=='Call')|(df['Call/Put']==' Put'),bs_call(df['Close'][index],df['Strike Price'][index],df['Days to Expiry'][index],rf,df['Volatility'][index]),bs_put(df['Close'][index],df['Strike Price'][index],df['Days to Expiry'][index],rf,df['Volatility'][index]))

Below are the functions I use for the calculation along with a dataframe that contains 3 rows

def bs_call(S,K,T,r,sigma):
    T=T/365
    d1=(log(S/K)+(r+sigma**2/2)*T)/(sigma*sqrt(T))
    d2= d1-sigma*sqrt(T)
    ans = S*norm.cdf(d1)-K*exp(-r*T)*norm.cdf(d2)
    return ans

def bs_put(S,K,T,r,sigma):
    T=T/365
    d1=(log(S/K)+(r+sigma**2/2)*T)/(sigma*sqrt(T))
    d2= d1-sigma*sqrt(T)
    ans = S*norm.cdf(d1)-K*exp(-r*T)*norm.cdf(d2)
    return K*exp(-r*T)-S+ans

df = [{'Close': 27.3,
  'Company': 'Barrick Gold Corporation (ABX)',
  'Ticker': 'ABX',
  'Yahoo Ticker': 'ABX.TO',
  'Expiry Date': Timestamp('2020-03-01 00:00:00'),
  'Strike Price': 19.5,
  'Call/Put': 'Put',
  'Days to Expiry': 2,
  'Volume': 1,
  'Bid Price': 0.0,
  'Ask Price': 0.11,
  'Open Interest': 24,
  'Implied Volatility': 2.4757,
  'Spread %': 100.0,
  'Volatility': 0.41140252083455864},
 {'Close': 27.3,
  'Company': 'Barrick Gold Corporation (ABX)',
  'Ticker': 'ABX',
  'Yahoo Ticker': 'ABX.TO',
  'Expiry Date': Timestamp('2020-03-01 00:00:00'),
  'Strike Price': 23.0,
  'Call/Put': 'Call',
  'Days to Expiry': 2,
  'Volume': 5,
  'Bid Price': 4.1,
  'Ask Price': 5.9,
  'Open Interest': 5,
  'Implied Volatility': 3.0017,
  'Spread %': 30.508474576271194,
  'Volatility': 0.41140252083455864},
 {'Close': 27.3,
  'Company': 'Barrick Gold Corporation (ABX)',
  'Ticker': 'ABX',
  'Yahoo Ticker': 'ABX.TO',
  'Expiry Date': Timestamp('2020-03-01 00:00:00'),
  'Strike Price': 24.0,
  'Call/Put': 'Put',
  'Days to Expiry': 2,
  'Volume': 5,
  'Bid Price': 0.06,
  'Ask Price': 0.17,
  'Open Interest': 5,
  'Implied Volatility': 1.3371,
  'Spread %': 64.70588235294117,
  'Volatility': 0.41140252083455864}]

Answer 1

I think you could go with apply :

df["Black Scholes"] = df.apply(lambda r : bs_call(r) if r["Call/Put"] == "Call" else bs_put(r),axis=1)

That would work fine if you only have two possible value for the column "Call/Put" but if you're planning on having more, you should define a function that does this :

def foo(row) :
   if row["Call/put"] == "value_1" : 
        return func_1(r)
   elif ...
df["Black Scholes"] = df.apply(foo,axis=1)