CODEWITHSUNDEEP
pythonperformancepandasdataframeindexing
Syntax_Error
Syntax_Error

Reputation: 6220

pandas dataframe: loc vs query performance

I have 2 dataframes in python that I would like to query for data.

Both queries return a single record. The simulation was done by running the same operation 10K times.

Running python 2.7 and pandas 0.16.0

Any recommendations to improve the query speed?

Upvotes: 42

Views: 38877

Answers (2)

Rafael Braga
Rafael Braga

Reputation: 195

I've added the .loc function and used the same code @jezrael used to test the performance in python 3.9.5 and pandas 1.2.5.

import numpy as np
import pandas as pd
import numexpr
import perfplot
np.random.seed(125)

def ne(x):
    return x[numexpr.evaluate('(x > 5)')]

def bex(x):
    return x[x.A.values > 5]

def be(x):
    return x[x.A > 5]

def lc(x):
    return x.loc[x.A > 5]

def lcx(x):
    return x.loc[x.A.values > 5]

def q(x):
    return x.query('A > 5')

def ev(x):
    return x[x.eval('A > 5')]

def make_df(n):
    df = pd.DataFrame(np.random.randint(10, size=n), columns=['A'])
    return df


perfplot.show(
    setup=make_df,
    kernels=[ne, lc, lcx, be, bex, q, ev],
    n_range=[2**k for k in range(2, 25)],
    logx=True,
    logy=True,
    equality_check=False,  
    xlabel='len(df)')

The using the .loc function does not afect the result of the be funcion and until 10^6 lines, using te column .values seens to be fastest option. The query method was the slowest.

top5

here just the loc and be, to see the overlap. loc vs. be

Upvotes: 10

jezrael
jezrael

Reputation: 862511

For improve performance is possible use numexpr:

import numexpr

np.random.seed(125)
N = 40000000
df = pd.DataFrame({'A':np.random.randint(10, size=N)})

def ne(df):
    x = df.A.values
    return df[numexpr.evaluate('(x > 5)')]
print (ne(df))

In [138]: %timeit (ne(df))
1 loop, best of 3: 494 ms per loop

In [139]: %timeit df[df.A > 5]
1 loop, best of 3: 536 ms per loop

In [140]: %timeit df.query('A > 5')
1 loop, best of 3: 781 ms per loop

In [141]: %timeit df[df.eval('A > 5')]
1 loop, best of 3: 770 ms per loop

import numexpr
np.random.seed(125)

def ne(x):
    x = x.A.values
    return x[numexpr.evaluate('(x > 5)')]

def be(x):
    return x[x.A > 5]

def q(x):
    return x.query('A > 5')

def ev(x):
    return x[x.eval('A > 5')]


def make_df(n):
    df = pd.DataFrame(np.random.randint(10, size=n), columns=['A'])
    return df


perfplot.show(
    setup=make_df,
    kernels=[ne, be, q, ev],
    n_range=[2**k for k in range(2, 25)],
    logx=True,
    logy=True,
    equality_check=False,  
    xlabel='len(df)')

graph

Edit:

Graph with modified ne and improved be:

def ne(x):
    return x[numexpr.evaluate('(x > 5)')]

def bex(x):
    return x[x.A.values > 5]

enter image description here

Upvotes: 42

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