Python Pandas How to get rid of groupings with only 1 row? - python-3.x

Question

In my dataset, I am trying to get the margin between two values. The code below runs perfectly if the fourth race was not included. After grouping based on a column, it seems that sometimes, there will be only 1 value, therefore, no other value to get a margin out of. I want to ignore these groupings in that case. Here is my current code:
import pandas as pd
data = {'Name':['A', 'B', 'B', 'C', 'A', 'C', 'A'], 'RaceNumber':
[1, 1, 2, 2, 3, 3, 4], 'PlaceWon':['First', 'Second', 'First', 'Second', 'First', 'Second', 'First'], 'TimeRanInSec':[100, 98, 66, 60, 75, 70, 75]}
df = pd.DataFrame(data)
print(df)
def winning_margin(times):
times = list(times)
winner = min(times)
times.remove(winner)
return min(times) - winner
winning_margins = df[['RaceNumber', 'TimeRanInSec']] \
.groupby('RaceNumber').agg(winning_margin)
winning_margins.columns = ['margin']
winners = df.loc[df.PlaceWon == 'First', :]
winners = winners.join(winning_margins, on='RaceNumber')
avg_margins = winners[['Name', 'margin']].groupby('Name').mean()
avg_margins

Answer 1

How about returning a NaN if times does not have enough elements:
import numpy as np
def winning_margin(times):
if len(times) <= 1: # New code
return np.NaN # New code
times = list(times)
winner = min(times)
times.remove(winner)
return min(times) - winner
your code runs with this change and seem to produce sensible results. But you can furthermore remove NaNs later if you want eg in this line
winning_margins = df[['RaceNumber', 'TimeRanInSec']] \
.groupby('RaceNumber').agg(winning_margin).dropna() # note the addition of .dropna()

Answer 2

You could get the winner and margin in one step:
def get_margin(x):
if len(x) < 2:
return np.NaN
i = x['TimeRanInSec'].idxmin()
nl = x['TimeRanInSec'].nsmallest(2)
margin = nl.max()-nl.min()
return [x['Name'].loc[i], margin]
Then:
df.groupby('RaceNumber').apply(get_margin).dropna()
RaceNumber
1 [B, 2]
2 [C, 6]
3 [C, 5]
(the data has the 'First' indicator corresponding to the slower time in the data)