What is the most efficient way to loop through dataframes with pandas?

Question

I want to perform my own complex operations on financial data in dataframes in a sequential manner.

For example I am using the following MSFT CSV file taken from Yahoo Finance:

Date,Open,High,Low,Close,Volume,Adj Close
2011-10-19,27.37,27.47,27.01,27.13,42880000,27.13
2011-10-18,26.94,27.40,26.80,27.31,52487900,27.31
2011-10-17,27.11,27.42,26.85,26.98,39433400,26.98
2011-10-14,27.31,27.50,27.02,27.27,50947700,27.27

....

I then do the following:

#!/usr/bin/env python
from pandas import *

df = read_csv('table.csv')

for i, row in enumerate(df.values):
    date = df.index[i]
    open, high, low, close, adjclose = row
    #now perform analysis on open/close based on date, etc..

Is that the most efficient way? Given the focus on speed in pandas, I would assume there must be some special function to iterate through the values in a manner that one also retrieves the index (possibly through a generator to be memory efficient)? df.iteritems unfortunately only iterates column by column.

Solution 1

The newest versions of pandas now include a built-in function for iterating over rows.

for index, row in df.iterrows():

    # do some logic here

Or, if you want it faster use itertuples()

But, unutbu's suggestion to use numpy functions to avoid iterating over rows will produce the fastest code.

Solution 2

Pandas is based on NumPy arrays. The key to speed with NumPy arrays is to perform your operations on the whole array at once, never row-by-row or item-by-item.

For example, if close is a 1-d array, and you want the day-over-day percent change,

pct_change = close[1:]/close[:-1]

This computes the entire array of percent changes as one statement, instead of

pct_change = []
for row in close:
    pct_change.append(...)

So try to avoid the Python loop for i, row in enumerate(...) entirely, and think about how to perform your calculations with operations on the entire array (or dataframe) as a whole, rather than row-by-row.

Solution 3

Like what has been mentioned before, pandas object is most efficient when process the whole array at once. However for those who really need to loop through a pandas DataFrame to perform something, like me, I found at least three ways to do it. I have done a short test to see which one of the three is the least time consuming.

t = pd.DataFrame({'a': range(0, 10000), 'b': range(10000, 20000)})
B = []
C = []
A = time.time()
for i,r in t.iterrows():
    C.append((r['a'], r['b']))
B.append(time.time()-A)

C = []
A = time.time()
for ir in t.itertuples():
    C.append((ir[1], ir[2]))    
B.append(time.time()-A)

C = []
A = time.time()
for r in zip(t['a'], t['b']):
    C.append((r[0], r[1]))
B.append(time.time()-A)

print B

Result:

[0.5639059543609619, 0.017839908599853516, 0.005645036697387695]

This is probably not the best way to measure the time consumption but it's quick for me.

Here are some pros and cons IMHO:

• .iterrows(): return index and row items in separate variables, but significantly slower
• .itertuples(): faster than .iterrows(), but return index together with row items, ir[0] is the index
• zip: quickest, but no access to index of the row

EDIT 2020/11/10

For what it is worth, here is an updated benchmark with some other alternatives (perf with MacBookPro 2,4 GHz Intel Core i9 8 cores 32 Go 2667 MHz DDR4)

import sys
import tqdm
import time
import pandas as pd

B = []
t = pd.DataFrame({'a': range(0, 10000), 'b': range(10000, 20000)})
for _ in tqdm.tqdm(range(10)):
    C = []
    A = time.time()
    for i,r in t.iterrows():
        C.append((r['a'], r['b']))
    B.append({"method": "iterrows", "time": time.time()-A})

    C = []
    A = time.time()
    for ir in t.itertuples():
        C.append((ir[1], ir[2]))
    B.append({"method": "itertuples", "time": time.time()-A})

    C = []
    A = time.time()
    for r in zip(t['a'], t['b']):
        C.append((r[0], r[1]))
    B.append({"method": "zip", "time": time.time()-A})

    C = []
    A = time.time()
    for r in zip(*t.to_dict("list").values()):
        C.append((r[0], r[1]))
    B.append({"method": "zip + to_dict('list')", "time": time.time()-A})

    C = []
    A = time.time()
    for r in t.to_dict("records"):
        C.append((r["a"], r["b"]))
    B.append({"method": "to_dict('records')", "time": time.time()-A})

    A = time.time()
    t.agg(tuple, axis=1).tolist()
    B.append({"method": "agg", "time": time.time()-A})

    A = time.time()
    t.apply(tuple, axis=1).tolist()
    B.append({"method": "apply", "time": time.time()-A})

print(f'Python {sys.version} on {sys.platform}')
print(f"Pandas version {pd.__version__}")
print(
    pd.DataFrame(B).groupby("method").agg(["mean", "std"]).xs("time", axis=1).sort_values("mean")
)

## Output

Python 3.7.9 (default, Oct 13 2020, 10:58:24) 
[Clang 12.0.0 (clang-1200.0.32.2)] on darwin
Pandas version 1.1.4
                           mean       std
method                                   
zip + to_dict('list')  0.002353  0.000168
zip                    0.003381  0.000250
itertuples             0.007659  0.000728
to_dict('records')     0.025838  0.001458
agg                    0.066391  0.007044
apply                  0.067753  0.006997
iterrows               0.647215  0.019600

Solution 4

You can loop through the rows by transposing and then calling iteritems:

for date, row in df.T.iteritems():
   # do some logic here

I am not certain about efficiency in that case. To get the best possible performance in an iterative algorithm, you might want to explore writing it in Cython, so you could do something like:

def my_algo(ndarray[object] dates, ndarray[float64_t] open,
            ndarray[float64_t] low, ndarray[float64_t] high,
            ndarray[float64_t] close, ndarray[float64_t] volume):
    cdef:
        Py_ssize_t i, n
        float64_t foo
    n = len(dates)

    for i from 0 <= i < n:
        foo = close[i] - open[i] # will be extremely fast

I would recommend writing the algorithm in pure Python first, make sure it works and see how fast it is-- if it's not fast enough, convert things to Cython like this with minimal work to get something that's about as fast as hand-coded C/C++.

Solution 5

You have three options:

By index (simplest):

>>> for index in df.index:
...     print ("df[" + str(index) + "]['B']=" + str(df['B'][index]))

With iterrows (most used):

>>> for index, row in df.iterrows():
...     print ("df[" + str(index) + "]['B']=" + str(row['B']))

With itertuples (fastest):

>>> for row in df.itertuples():
...     print ("df[" + str(row.Index) + "]['B']=" + str(row.B))

Three options display something like:

df[0]['B']=125
df[1]['B']=415
df[2]['B']=23
df[3]['B']=456
df[4]['B']=189
df[5]['B']=456
df[6]['B']=12

Source: alphons.io

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