There are many questions (1, 2, 3) dealing with counting values in a single series.
However, there are fewer questions looking at the best way to count combinations of two or more series. Solutions are presented (1, 2), but when and why one should use each is not discussed.
Below is some benchmarking for three potential methods. I have two specific questions:
grouper
more efficient than count
? I expected count
to be the more efficient, as it is implemented in C. The superior performance of grouper
persists even if number of columns is increased from 2 to 4.value_counter
underperform grouper
by so much? Is this due to the cost of constructing a list, or series from list?I understand the outputs are different, and this should also inform choice. For example, filtering by count is more efficient with contiguous numpy
arrays versus a dictionary comprehension:
x, z = grouper(df), count(df)
%timeit x[x.values > 10] # 749µs
%timeit {k: v for k, v in z.items() if v > 10} # 9.37ms
However, the focus of my question is on performance of building comparable results in a series versus dictionary. My C knowledge is limited, yet I would appreciate any answer which can point to the logic underlying these methods.
Benchmarking code
import pandas as pd
import numpy as np
from collections import Counter
np.random.seed(0)
m, n = 1000, 100000
df = pd.DataFrame({'A': np.random.randint(0, m, n),
'B': np.random.randint(0, m, n)})
def grouper(df):
return df.groupby(['A', 'B'], sort=False).size()
def value_counter(df):
return pd.Series(list(zip(df.A, df.B))).value_counts(sort=False)
def count(df):
return Counter(zip(df.A.values, df.B.values))
x = value_counter(df).to_dict()
y = grouper(df).to_dict()
z = count(df)
assert (x == y) & (y == z), "Dictionary mismatch!"
for m, n in [(100, 10000), (1000, 10000), (100, 100000), (1000, 100000)]:
df = pd.DataFrame({'A': np.random.randint(0, m, n),
'B': np.random.randint(0, m, n)})
print(m, n)
%timeit grouper(df)
%timeit value_counter(df)
%timeit count(df)
Benchmarking results
Run on python 3.6.2, pandas 0.20.3, numpy 1.13.1
Machine specs: Windows 7 64-bit, Dual-Core 2.5 GHz, 4GB RAM.
Key: g = grouper
, v = value_counter
, c = count
.
m n g v c
100 10000 2.91 18.30 8.41
1000 10000 4.10 27.20 6.98[1]
100 100000 17.90 130.00 84.50
1000 100000 43.90 309.00 93.50
1 This is not a typo.
count() should be used when you want to find the frequency of valid values present in columns with respect to specified col . . value_counts() should be used to find the frequencies of a series.
Return a Series containing counts of unique values. The resulting object will be in descending order so that the first element is the most frequently-occurring element.
count returns a DataFrame when you call count on all column, while GroupBy. size returns a Series. The reason being that size is the same for all columns, so only a single result is returned. Meanwhile, the count is called for each column, as the results would depend on on how many NaNs each column has.
Use count() by Column Name Use pandas DataFrame. groupby() to group the rows by column and use count() method to get the count for each group by ignoring None and Nan values.
There's actually a bit of hidden overhead in zip(df.A.values, df.B.values)
. The key here comes down to numpy arrays being stored in memory in a fundamentally different way than Python objects.
A numpy array, such as np.arange(10)
, is essentially stored as a contiguous block of memory, and not as individual Python objects. Conversely, a Python list, such as list(range(10))
, is stored in memory as pointers to individual Python objects (i.e. integers 0-9). This difference is the basis for why numpy arrays are smaller in memory than the Python equivalent lists, and why you can perform faster computations on numpy arrays.
So, as Counter
is consuming the zip
, the associated tuples need to be created as Python objects. This means that Python needs to extract the tuple values from numpy data and create corresponding Python objects in memory. There is noticeable overhead to this, which is why you want to be very careful when combining pure Python functions with numpy data. A basic example of this pitfall that you might commonly see is using the built-in Python sum
on a numpy array: sum(np.arange(10**5))
is actually a bit slower than the pure Python sum(range(10**5))
, and both of which are of course significantly slower than np.sum(np.arange(10**5))
.
See this video for a more in depth discussion of this topic.
As an example specific to this question, observe the following timings comparing the performance of Counter
on zipped numpy arrays vs. the corresponding zipped Python lists.
In [2]: a = np.random.randint(10**4, size=10**6) ...: b = np.random.randint(10**4, size=10**6) ...: a_list = a.tolist() ...: b_list = b.tolist() In [3]: %timeit Counter(zip(a, b)) 455 ms ± 4.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) In [4]: %timeit Counter(zip(a_list, b_list)) 334 ms ± 4.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
The difference between these two timings gives you a reasonable estimate of the overhead discussed earlier.
This isn't quite the end of the story though. Constructing a groupby
object in pandas involves a some overhead too, at least as related to this problem, since there's some groupby
metadata that isn't strictly necessary just to get size
, whereas Counter
does the one singular thing you care about. Usually this overhead is far less than the overhead associated with Counter
, but from some quick experimentation I've found that you can actually get marginally better performance from Counter
when the majority of your groups just consist of single elements.
Consider the following timings (using @BallpointBen's sort=False
suggestion) that go along the spectrum of few large groups <--> many small groups:
def grouper(df): return df.groupby(['A', 'B'], sort=False).size() def count(df): return Counter(zip(df.A.values, df.B.values)) for m, n in [(10, 10**6), (10**3, 10**6), (10**7, 10**6)]: df = pd.DataFrame({'A': np.random.randint(0, m, n), 'B': np.random.randint(0, m, n)}) print(m, n) %timeit grouper(df) %timeit count(df)
Which gives me the following table:
m grouper counter 10 62.9 ms 315 ms 10**3 191 ms 535 ms 10**7 514 ms 459 ms
Of course, any gains from Counter
would be offset by converting back to a Series
, if that's what you want as your final object.
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