Create multiple columns in Pandas Dataframe from one function

Question

I\'m a python newbie, so I hope my two questions are clear and complete. I posted the actual code and a test data set in csv format below.

I\'ve been able to construct t

Answer 1

The trick to vectorize code is to not think in terms of rows, but instead think in terms of columns.

I almost have this working (I'll try to finish it later), but you want to do something along the lines of this:

from datetime import datetime
from math import sqrt, pi, log, exp, isnan
from numpy import inf, nan
from scipy.stats import norm
import pandas as pd
from pandas import Timestamp
from pandas.tseries.holiday import USFederalHolidayCalendar

# Initial parameters
rf = .0015                          # Get Fed Funds Rate https://research.stlouisfed.org/fred2/data/DFF.csv
tradingMinutesDay = 450             # 7.5 hours per day * 60 minutes per hour
tradingMinutesAnnum = 113400        # trading minutes per day * 252 trading days per year
cal = USFederalHolidayCalendar()    # Load US Federal holiday calendar
two_pi = 2 * pi                     # 2 * Pi (to reduce computations)
threshold = 1.0e-9                  # convergence threshold.

# Create sample data:
col_order = ['TimeStamp', 'OpraSymbol', 'RootSymbol', 'Expiry', 'Strike', 'OptType', 'RootPrice', 'Last', 'Bid', 'Ask', 'Volume', 'OpenInt', 'IV']
df = pd.DataFrame({'Ask': {0: 3.7000000000000002, 1: 2.4199999999999999, 2: 3.0, 3: 2.7999999999999998, 4: 2.4500000000000002, 5: 3.25, 6: 5.9500000000000002, 7: 6.2999999999999998},
                   'Bid': {0: 3.6000000000000001, 1: 2.3399999999999999, 2: 2.8599999999999999, 3: 2.7400000000000002, 4: 2.4399999999999999, 5: 3.1000000000000001, 6: 5.7000000000000002, 7: 6.0999999999999996},
                   'Expiry': {0: Timestamp('2015-10-16 16:00:00'), 1: Timestamp('2015-10-16 16:00:00'), 2: Timestamp('2015-10-16 16:00:00'), 3: Timestamp('2015-10-16 16:00:00'), 4: Timestamp('2015-10-16 16:00:00'), 5: Timestamp('2015-10-16 16:00:00'), 6: Timestamp('2015-11-20 16:00:00'), 7: Timestamp('2015-11-20 16:00:00')},
                   'IV': {0: 0.3497, 1: 0.3146, 2: 0.3288, 3: 0.3029, 4: 0.3187, 5: 0.2926, 6: 0.3635, 7: 0.3842},
                   'Last': {0: 3.46, 1: 2.34, 2: 3.0, 3: 2.81, 4: 2.35, 5: 3.20, 6: 5.90, 7: 6.15},
                   'OpenInt': {0: 1290.0, 1: 3087.0, 2: 28850.0, 3: 44427.0, 4: 2318.0, 5: 3773.0, 6: 17112.0, 7: 15704.0},
                   'OpraSymbol': {0: 'AAPL151016C00109000', 1: 'AAPL151016P00109000', 2: 'AAPL151016C00110000', 3: 'AAPL151016P00110000', 4: 'AAPL151016C00111000', 5: 'AAPL151016P00111000', 6: 'AAPL151120C00110000', 7: 'AAPL151120P00110000'},
                   'OptType': {0: 'C', 1: 'P', 2: 'C', 3: 'P', 4: 'C', 5: 'P', 6: 'C', 7: 'P'},
                   'RootPrice': {0: 109.95, 1: 109.95, 2: 109.95, 3: 109.95, 4: 109.95, 5: 109.95, 6: 109.95, 7: 109.95},
                   'RootSymbol': {0: 'AAPL', 1: 'AAPL', 2: 'AAPL', 3: 'AAPL', 4: 'AAPL', 5: 'AAPL', 6: 'AAPL', 7: 'AAPL'},
                   'Strike': {0: 109.0, 1: 109.0, 2: 110.0, 3: 110.0, 4: 111.0, 5: 111.0, 6: 110.0, 7: 110.0},
                   'TimeStamp': {0: Timestamp('2015-09-30 16:00:00'), 1: Timestamp('2015-09-30 16:00:00'), 2: Timestamp('2015-09-30 16:00:00'), 3: Timestamp('2015-09-30 16:00:00'), 4: Timestamp('2015-09-30 16:00:00'), 5: Timestamp('2015-09-30 16:00:00'), 6: Timestamp('2015-09-30 16:00:00'), 7: Timestamp('2015-09-30 16:00:00')},
                   'Volume': {0: 1565.0, 1: 3790.0, 2: 10217.0, 3: 12113.0, 4: 6674.0, 5: 2031.0, 6: 5330.0, 7: 3724.0}})
df = df[col_order]

# Vectorize columns
df['mark'] = (df.Bid + df.Ask) / 2
df['cp'] = df.OptType.map({'C': 1, 'P': -1})
df['Log_S_K'] = (df.RootPrice / df.Strike).apply(log)
df['divs'] = 0  # TODO: Get dividend value.
df['vega'] = 0.
df['converged'] = False

# Vectorized datetime calculations
date_pairs = set(zip(df.TimeStamp, df.Expiry))
total_days = {(t1, t2): len(pd.bdate_range(t1, t2)) 
                        for t1, t2 in date_pairs}
hols = {(t1, t2): len(cal.holidays(t1, t2).to_pydatetime()) 
                  for t1, t2 in date_pairs}
del date_pairs

df['total_days'] = [total_days.get((t1, t2))
                    for t1, t2 in zip(df.TimeStamp, df.Expiry)]
df['hols'] = [hols.get((t1, t2))
              for t1, t2 in zip(df.TimeStamp, df.Expiry)]
df['days_to_exp'] = df.total_days - df.hols - 1
df.loc[df.days_to_exp < 0, 'days_to_exp'] = 0  # Min zero.
df.drop(['total_days', 'hols'], axis='columns', inplace=True)
df['years_to_expiry'] = (df.days_to_exp * tradingMinutesDay / tradingMinutesAnnum)

# Initial implied vol 'guess'
df['implied_vol'] = (two_pi / df.years_to_expiry) ** 0.5 * df.mark / df.RootPrice  

for i in xrange(100):  # range(100) in Python 3.x
    # Create mask of options where the vol has not converged.
    mask = [not c for c in df.converged.values]
    if df.converged.all():
        break

    # Aliases.
    data = df.loc[mask, :]
    cp = data.cp
    mark = data.mark
    S = data.RootPrice
    K = data.Strike
    d = data.divs
    T = data.years_to_expiry
    log_S_K = data.Log_S_K
    iv = data.implied_vol

    # Calcs.
    d1 = (log_S_K + T * (rf - d + .5 * iv ** 2)) / (iv * T ** 0.5)
    d2 = d1 - iv * T ** 0.5
    df.loc[mask, 'vega'] = vega = S * d1.apply(norm.pdf) * T ** 0.5
    model = cp * (S * (cp * d1).apply(norm.cdf)
                  - K * (-rf * T).apply(exp) * (cp * d2).apply(norm.cdf))
    iv_delta = (model - mark) / vega
    df.loc[mask, 'implied_vol'] = iv - iv_delta

    # Clean-up and check for convergence.
    df.loc[df.implied_vol < 0, 'implied_vol'] = 0
    idx = model[(model - mark).abs() < threshold].index
    df.ix[idx, 'converged'] = True
    df.loc[:, 'implied_vol'].fillna(0, inplace=True)
    df.loc[:, 'implied_vol'].replace([inf, -inf], nan, inplace=True)
    df.loc[:, 'vega'].fillna(0, inplace=True)
    df.loc[:, 'vega'].replace([inf, -inf], nan, inplace=True)

Answer 2

If I understand you right, what you should be doing is returning a Series from your function. Something like:

return pandas.Series({"IV": iv, "Vega": vega})

If you want to put the result into new columns of the same input DataFrame, then just do:

df[["IV", "Vega"]] = df.apply(newtonRap, axis=1)

Answer 3

As far as the performance with numba is concerned, numba doesn't know anything about pandas dataframes and cannot compile operations on them down to fast machine code. Your best bet is to profile what part of your method is slow (using line_profiler for example), and then offload that part to another method that you construct the inputs using the .values attributes of the dataframe columns, which gives you access to the underlying numpy array. Otherwise numba is just going to operate mostly in "object mode" (see the numba glossary) and won't improve performance drastically