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How to speed up successive pd.apply with successive pd.DataFrame.loc calls?

def __link_price(row: pd.Series) -> Union[None, float]:
    if (row['fund'] == 'A') and (row['share_class'] == 'X'):
        return df_hist.loc[row['date'], 'AA']
    elif (row['fund'] == 'A') and (row['share_class'] == 'Y'):
        return df_hist.loc[row['date'], 'AB']
    elif (row['fund'] == 'B') and (row['share_class'] == 'X'):
        return df_hist.loc[row['date'], 'BA']
    elif (row['fund'] == 'B') and (row['share_class'] == 'Y'):
        return df_hist.loc[row['date'], 'BB']
    elif (row['fund'] == 'C') and (row['share_class'] == 'X'):
        return df_hist.loc[row['date'], 'CA']
    elif (row['fund'] == 'C') and (row['share_class'] == 'Y'):
        return df_hist.loc[row['date'], 'CB']
    else:
        return 0

df.loc[:, 'price'] = df.apply(__link_price_at_purchase, axis=1).values

df has 10,000+ lines, so this code is taking a long time. In addition for each row, I’m doing a df_hist.loc call to get the value.

I’m trying to speed up this section of code and then option I’ve found so far is using:

df.loc[:, 'price'] = df.apply(__link_price_at_purchase, axis=1, raw=True).values

But this forces me to use index based selection for row instead of value selection:

if (row[0] == 'A') and (row[1] == 'X')

which reduces the readability of the code.

I’m looking for an approach that both speeds up the code and still allows for readability of the code.

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Answer

In python, there is a certain cost for each attribute or item lookup and function call. And you don’t have a compiler that optimizes things for you.

Here are some general recommendations:

  1. Try creating a column that includes fund and share_class without using Python functions and then merge it with df_hist
# convert history from 'wide' format into 'long' format
hist = df_hist.set_index("date").stack()
prices = (
    # create key column for join
    df.assign(key=df["fund"] + df["share_class"].replace({"X": "A", "Y": "B"}))
    .set_index(["date", "key"])
    .join(hist)  # join by index
)
  1. If it’s not trivial to create a key column, minimize attribute lookups inside the apply function:
def __link_price(row):
    date, fund, share_class = row[["date", "fund", "share_class"]]
    
    if fund == 'A' and share_class == 'X':
        return df_hist.loc[date, 'AA']
    ...
  1. Optimize if conditions. For example, you need to check 6 conditions in case where (row['fund'] == 'C') and (row['share_class'] == 'Y'). You can reduce this number to … 1.
fund_and_share_class_to_key = {
    ("A", "X"): "AA",
    ("A", "Y"): "AB",
    ...
}

key = fund_and_share_class_to_key.get((fund, share_class))
return df_hist.loc[date, key] if key is not None else 0
  1. Pandas itself is pretty slow for non-vectorized and non-arithmetic operations. In your case it’s better to use standard python dicts for faster lookups.
# small benchmark
df = pd.DataFrame({"value": [4,5,6]})
d = df.to_dict(orient="index")
%timeit df.loc[1, "value"]  # 8.7ms
%timeit d[1]["value"]       # 50ns; ~170 times faster

# convert dataframe into the dict with format:
# {<date>: {"AA": <value>}}
history = df_hist.set_index("date").to_dict(orient="index")

def __link_price(row):
    ...
    price = history.get(date, {}).get(key, 0)
    return price
  1. It should be faster to pass history as an apply argument rather than search it in the non-local scope. It also makes the code cleaner.
def __link_price(row, history):
    ...
df.apply(__link_price, args=(history, ))

To summarize, a faster function would be something like this:

history = df_hist.set_index("date").to_dict(orient="index")
# we don't need to create a mapping on every __link_price call
fund_and_share_class_to_key = {
    ("A", "X"): "AA",
    ("A", "Y"): "AB",
    ...
}
def __link_price(row, history, fund_and_share_class_to_key):
    date, fund, share_class = row[["date", "fund", "share_class"]]
    key = fund_and_share_class_to_key.get((fund, share_class))
    return history.get(date, {}).get(key, 0)

df.apply(__link_price, args=(history, fund_and_share_class_to_key))
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