CODEWITHSUNDEEP
rpandasdata.table
bogdanCsn
bogdanCsn

Reputation: 1325

Why is R's data.table so much faster than pandas?

I have a 12 million rows dataset, with 3 columns as unique identifiers and another 2 columns with values. I'm trying to do a rather simple task:
- group by the three identifiers. This yields about 2.6 million unique combinations
- Task 1: calculate the median for column Val1
- Task 2: calculate the mean for column Val1 given some condition on Val2

Here are my results, using pandas and data.table (both latest versions at the moment, on the same machine): 

+-----------------+-----------------+------------+
|                 |      pandas     | data.table |
+-----------------+-----------------+------------+
| TASK 1          | 150 seconds     | 4 seconds  |
| TASK 1 + TASK 2 |  doesn't finish | 5 seconds  |
+-----------------+-----------------+------------+

I think I may be doing something wrong with pandas - transforming Grp1 and Grp2 into categories didn't help a lot, nor did switching between .agg and .apply. Any ideas?

Below is the reproducible code.
Dataframe generation: 

import numpy as np
import pandas as pd
from collections import OrderedDict
import time

np.random.seed(123)
list1 = list(pd.util.testing.rands_array(10, 750))
list2 = list(pd.util.testing.rands_array(10, 700))
list3 = list(np.random.randint(100000,200000,5))

N = 12 * 10**6 # please make sure you have enough RAM
df = pd.DataFrame({'Grp1': np.random.choice(list1, N, replace = True),
                   'Grp2': np.random.choice(list2, N, replace = True),
                   'Grp3': np.random.choice(list3, N, replace = True),
                   'Val1': np.random.randint(0,100,N),
                   'Val2': np.random.randint(0,10,N)}) 


# this works and shows there are 2,625,000 unique combinations
df_test = df.groupby(['Grp1','Grp2','Grp3']).size()
print(df_test.shape[0]) # 2,625,000 rows

# export to feather so that same df goes into R
df.to_feather('file.feather')

Task 1 in Python: 

# TASK 1: 150 seconds (sorted / not sorted doesn't seem to matter)
df.sort_values(['Grp1','Grp2','Grp3'], inplace = True)
t0 = time.time()
df_agg1 = df.groupby(['Grp1','Grp2','Grp3']).agg({'Val1':[np.median]})
t1 = time.time()
print("Duration for complex: %s seconds ---" % (t1 - t0))

Task 1 + Task 2 in Python:

# TASK 1 + TASK 2: this kept running for 10 minutes to no avail
# (sorted / not sorted doesn't seem to matter)
def f(x):
    d = OrderedDict()
    d['Median_all'] = np.median(x['Val1'])
    d['Median_lt_5'] = np.median(x['Val1'][x['Val2'] < 5])
    return pd.Series(d)

t0 = time.time()
df_agg2 = df.groupby(['Grp1','Grp2','Grp3']).apply(f)
t1 = time.time()
print("Duration for complex: %s seconds ---" % (t1 - t0)) # didn't complete

Equivalent R code:

library(data.table)
library(feather)

DT = setDT(feater("file.feather"))
system.time({
DT_agg <- DT[,.(Median_all = median(Val1),
                Median_lt_5 = median(Val1[Val2 < 5])  ), by = c('Grp1','Grp2','Grp3')]
}) # 5 seconds

Upvotes: 21

Views: 4972

Answers (1)

Brad Miller
Brad Miller

Reputation: 97

I can't reproduce your R results, I fixed the typo where you misspelled feather, but I get the following:

Error in `[.data.table`(DT, , .(Median_all = median(Val1), Median_lt_5 = median(Val1[Val2 <  : 
column or expression 1 of 'by' or 'keyby' is type NULL. Do not quote column names. Usage: DT[,sum(colC),by=list(colA,month(colB))]

As to the python example, If you want to get the median for each group where Val2 is less than 5 then you should filter first, as in:

 df[df.Val2 < 5].groupby(['Grp1','Grp2','Grp3'])['Val2'].median()

This completes in under 8 seconds on my macbook pro.

Upvotes: 3

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