Enhancing Performance

Cython (Writing C extensions for pandas)

For many use cases writing pandas in pure python and numpy is sufficient. In some computationally heavy applications however, it can be possible to achieve sizeable speed-ups by offloading work to cython.

This tutorial assumes you have refactored as much as possible in python, for example trying to remove for loops and making use of numpy vectorization, it’s always worth optimising in python first.

This tutorial walks through a “typical” process of cythonizing a slow computation. We use an example from the cython documentation but in the context of pandas. Our final cythonized solution is around 100 times faster than the pure python.

Pure python

We have a DataFrame to which we want to apply a function row-wise.

In [1]: df = DataFrame({'a': randn(1000), 'b': randn(1000),'N': randint(100, 1000, (1000)), 'x': 'x'})

In [2]: df
Out[2]: 
       N         a         b  x
0    585  0.469112 -0.218470  x
1    841 -0.282863 -0.061645  x
2    251 -1.509059 -0.723780  x
3    972 -1.135632  0.551225  x
4    181  1.212112 -0.497767  x
5    458 -0.173215  0.837519  x
6    159  0.119209  1.103245  x
..   ...       ...       ... ..
993  190  0.131892  0.290162  x
994  931  0.342097  0.215341  x
995  374 -1.512743  0.874737  x
996  246  0.933753  1.120790  x
997  157 -0.308013  0.198768  x
998  977 -0.079915  1.757555  x
999  770 -1.010589 -1.115680  x

[1000 rows x 4 columns]

Here’s the function in pure python:

In [3]: def f(x):
   ...:     return x * (x - 1)
   ...: 

In [4]: def integrate_f(a, b, N):
   ...:     s = 0
   ...:     dx = (b - a) / N
   ...:     for i in range(N):
   ...:         s += f(a + i * dx)
   ...:     return s * dx
   ...: 

We achieve our result by using apply (row-wise):

In [7]: %timeit df.apply(lambda x: integrate_f(x['a'], x['b'], x['N']), axis=1)
10 loops, best of 3: 174 ms per loop

But clearly this isn’t fast enough for us. Let’s take a look and see where the time is spent during this operation (limited to the most time consuming four calls) using the prun ipython magic function:

In [5]: %prun -l 4 df.apply(lambda x: integrate_f(x['a'], x['b'], x['N']), axis=1)
         611748 function calls (609735 primitive calls) in 0.561 seconds

   Ordered by: internal time
   List reduced from 102 to 4 due to restriction <4>

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     1000    0.306    0.000    0.492    0.000 <ipython-input-4-91e33489f136>:1(integrate_f)
   552423    0.179    0.000    0.179    0.000 <ipython-input-3-bc41a25943f6>:1(f)
     1000    0.007    0.000    0.007    0.000 {range}
     3000    0.007    0.000    0.047    0.000 series.py:519(__getitem__)

By far the majority of time is spend inside either integrate_f or f, hence we’ll concentrate our efforts cythonizing these two functions.

Note

In python 2 replacing the range with its generator counterpart (xrange) would mean the range line would vanish. In python 3 range is already a generator.

Plain cython

First we’re going to need to import the cython magic function to ipython:

In [6]: %load_ext cythonmagic

Now, let’s simply copy our functions over to cython as is (the suffix is here to distinguish between function versions):

In [7]: %%cython
   ...: def f_plain(x):
   ...:     return x * (x - 1)
   ...: def integrate_f_plain(a, b, N):
   ...:     s = 0
   ...:     dx = (b - a) / N
   ...:     for i in range(N):
   ...:         s += f_plain(a + i * dx)
   ...:     return s * dx
   ...: 

Note

If you’re having trouble pasting the above into your ipython, you may need to be using bleeding edge ipython for paste to play well with cell magics.

In [4]: %timeit df.apply(lambda x: integrate_f_plain(x['a'], x['b'], x['N']), axis=1)
10 loops, best of 3: 85.5 ms per loop

Already this has shaved a third off, not too bad for a simple copy and paste.

Adding type

We get another huge improvement simply by providing type information:

In [8]: %%cython
   ...: cdef double f_typed(double x) except? -2:
   ...:     return x * (x - 1)
   ...: cpdef double integrate_f_typed(double a, double b, int N):
   ...:     cdef int i
   ...:     cdef double s, dx
   ...:     s = 0
   ...:     dx = (b - a) / N
   ...:     for i in range(N):
   ...:         s += f_typed(a + i * dx)
   ...:     return s * dx
   ...: 

In [4]: %timeit df.apply(lambda x: integrate_f_typed(x['a'], x['b'], x['N']), axis=1)
10 loops, best of 3: 20.3 ms per loop

Now, we’re talking! It’s now over ten times faster than the original python implementation, and we haven’t really modified the code. Let’s have another look at what’s eating up time:

In [9]: %prun -l 4 df.apply(lambda x: integrate_f_typed(x['a'], x['b'], x['N']), axis=1)
         58325 function calls (56312 primitive calls) in 0.078 seconds

   Ordered by: internal time
   List reduced from 100 to 4 due to restriction <4>

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     3000    0.008    0.000    0.052    0.000 series.py:519(__getitem__)
     3000    0.007    0.000    0.017    0.000 internals.py:3481(get_values)
     3000    0.007    0.000    0.039    0.000 index.py:1582(get_value)
     6000    0.007    0.000    0.026    0.000 {pandas.lib.values_from_object}

Using ndarray

It’s calling series... a lot! It’s creating a Series from each row, and get-ting from both the index and the series (three times for each row). Function calls are expensive in python, so maybe we could minimise these by cythonizing the apply part.

Note

We are now passing ndarrays into the cython function, fortunately cython plays very nicely with numpy.

In [10]: %%cython
   ....: cimport numpy as np
   ....: import numpy as np
   ....: cdef double f_typed(double x) except? -2:
   ....:     return x * (x - 1)
   ....: cpdef double integrate_f_typed(double a, double b, int N):
   ....:     cdef int i
   ....:     cdef double s, dx
   ....:     s = 0
   ....:     dx = (b - a) / N
   ....:     for i in range(N):
   ....:         s += f_typed(a + i * dx)
   ....:     return s * dx
   ....: cpdef np.ndarray[double] apply_integrate_f(np.ndarray col_a, np.ndarray col_b, np.ndarray col_N):
   ....:     assert (col_a.dtype == np.float and col_b.dtype == np.float and col_N.dtype == np.int)
   ....:     cdef Py_ssize_t i, n = len(col_N)
   ....:     assert (len(col_a) == len(col_b) == n)
   ....:     cdef np.ndarray[double] res = np.empty(n)
   ....:     for i in range(len(col_a)):
   ....:         res[i] = integrate_f_typed(col_a[i], col_b[i], col_N[i])
   ....:     return res
   ....: 

The implementation is simple, it creates an array of zeros and loops over the rows, applying our integrate_f_typed, and putting this in the zeros array.

Warning

In 0.13.0 since Series has internaly been refactored to no longer sub-class ndarray but instead subclass NDFrame, you can not pass a Series directly as a ndarray typed parameter to a cython function. Instead pass the actual ndarray using the .values attribute of the Series.

Prior to 0.13.0

apply_integrate_f(df['a'], df['b'], df['N'])

Use .values to get the underlying ndarray

apply_integrate_f(df['a'].values, df['b'].values, df['N'].values)

Note

Loops like this would be extremely slow in python, but in Cython looping over numpy arrays is fast.

In [4]: %timeit apply_integrate_f(df['a'].values, df['b'].values, df['N'].values)
1000 loops, best of 3: 1.25 ms per loop

We’ve gotten another big improvement. Let’s check again where the time is spent:

In [11]: %prun -l 4 apply_integrate_f(df['a'].values, df['b'].values, df['N'].values)
         190 function calls in 0.003 seconds

   Ordered by: internal time
   List reduced from 49 to 4 due to restriction <4>

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.002    0.002    0.002    0.002 {_cython_magic_073e22cb442403aaff864a05d833c10b.apply_integrate_f}
        3    0.000    0.000    0.000    0.000 internals.py:2872(iget)
        6    0.000    0.000    0.000    0.000 generic.py:2152(__setattr__)
        3    0.000    0.000    0.000    0.000 generic.py:1079(_get_item_cache)

As one might expect, the majority of the time is now spent in apply_integrate_f, so if we wanted to make anymore efficiencies we must continue to concentrate our efforts here.

More advanced techniques

There is still hope for improvement. Here’s an example of using some more advanced cython techniques:

In [12]: %%cython
   ....: cimport cython
   ....: cimport numpy as np
   ....: import numpy as np
   ....: cdef double f_typed(double x) except? -2:
   ....:     return x * (x - 1)
   ....: cpdef double integrate_f_typed(double a, double b, int N):
   ....:     cdef int i
   ....:     cdef double s, dx
   ....:     s = 0
   ....:     dx = (b - a) / N
   ....:     for i in range(N):
   ....:         s += f_typed(a + i * dx)
   ....:     return s * dx
   ....: @cython.boundscheck(False)
   ....: @cython.wraparound(False)
   ....: cpdef np.ndarray[double] apply_integrate_f_wrap(np.ndarray[double] col_a, np.ndarray[double] col_b, np.ndarray[Py_ssize_t] col_N):
   ....:     cdef Py_ssize_t i, n = len(col_N)
   ....:     assert len(col_a) == len(col_b) == n
   ....:     cdef np.ndarray[double] res = np.empty(n)
   ....:     for i in range(n):
   ....:         res[i] = integrate_f_typed(col_a[i], col_b[i], col_N[i])
   ....:     return res
   ....: 

In [4]: %timeit apply_integrate_f_wrap(df['a'].values, df['b'].values, df['N'].values)
1000 loops, best of 3: 987 us per loop

Even faster, with the caveat that a bug in our cython code (an off-by-one error, for example) might cause a segfault because memory access isn’t checked.

Using numba

A recent alternative to statically compiling cython code, is to use a dynamic jit-compiler, numba.

Numba gives you the power to speed up your applications with high performance functions written directly in Python. With a few annotations, array-oriented and math-heavy Python code can be just-in-time compiled to native machine instructions, similar in performance to C, C++ and Fortran, without having to switch languages or Python interpreters.

Numba works by generating optimized machine code using the LLVM compiler infrastructure at import time, runtime, or statically (using the included pycc tool). Numba supports compilation of Python to run on either CPU or GPU hardware, and is designed to integrate with the Python scientific software stack.

Note

You will need to install numba. This is easy with conda, by using: conda install numba, see installing using miniconda.

We simply take the plain python code from above and annotate with the @jit decorator.

import numba

@numba.jit
def f_plain(x):
   return x * (x - 1)

@numba.jit
def integrate_f_numba(a, b, N):
   s = 0
   dx = (b - a) / N
   for i in range(N):
       s += f_plain(a + i * dx)
   return s * dx

@numba.jit
def apply_integrate_f_numba(col_a, col_b, col_N):
   n = len(col_N)
   result = np.empty(n, dtype='float64')
   assert len(col_a) == len(col_b) == n
   for i in range(n):
      result[i] = integrate_f_numba(col_a[i], col_b[i], col_N[i])
   return result

def compute_numba(df):
   result = apply_integrate_f_numba(df['a'].values, df['b'].values, df['N'].values)
   return Series(result, index=df.index, name='result')

Similar to above, we directly pass numpy arrays directly to the numba function. Further we are wrapping the results to provide a nice interface by passing/returning pandas objects.

In [4]: %timeit compute_numba(df)
1000 loops, best of 3: 798 us per loop

Read more in the numba docs.

Expression Evaluation via eval() (Experimental)

New in version 0.13.

The top-level function pandas.eval() implements expression evaluation of Series and DataFrame objects.

Note

To benefit from using eval() you need to install numexpr. See the recommended dependencies section for more details.

The point of using eval() for expression evaluation rather than plain Python is two-fold: 1) large DataFrame objects are evaluated more efficiently and 2) large arithmetic and boolean expressions are evaluated all at once by the underlying engine (by default numexpr is used for evaluation).

Note

You should not use eval() for simple expressions or for expressions involving small DataFrames. In fact, eval() is many orders of magnitude slower for smaller expressions/objects than plain ol’ Python. A good rule of thumb is to only use eval() when you have a DataFrame with more than 10,000 rows.

eval() supports all arithmetic expressions supported by the engine in addition to some extensions available only in pandas.

Note

The larger the frame and the larger the expression the more speedup you will see from using eval().

Supported Syntax

These operations are supported by pandas.eval():

• Arithmetic operations except for the left shift (<<) and right shift (>>) operators, e.g., df + 2 * pi / s ** 4 % 42 - the_golden_ratio
• Comparison operations, including chained comparisons, e.g., 2 < df < df2
• Boolean operations, e.g., df < df2 and df3 < df4 or not df_bool
• list and tuple literals, e.g., [1, 2] or (1, 2)
• Attribute access, e.g., df.a
• Subscript expressions, e.g., df[0]
• Simple variable evaluation, e.g., pd.eval('df') (this is not very useful)

This Python syntax is not allowed:

• Expressions
  • Function calls
  • is/is not operations
  • if expressions
  • lambda expressions
  • list/set/dict comprehensions
  • Literal dict and set expressions
  • yield expressions
  • Generator expressions
  • Boolean expressions consisting of only scalar values
• Statements
  • Neither simple nor compound statements are allowed. This includes things like for, while, and if.

eval() Examples

pandas.eval() works well with expressions containing large arrays.

First let’s create a few decent-sized arrays to play with:

In [13]: import pandas as pd

In [14]: from pandas import DataFrame, Series

In [15]: from numpy.random import randn

In [16]: import numpy as np

In [17]: nrows, ncols = 20000, 100

In [18]: df1, df2, df3, df4 = [DataFrame(randn(nrows, ncols)) for _ in range(4)]

Now let’s compare adding them together using plain ol’ Python versus eval():

In [19]: %timeit df1 + df2 + df3 + df4
10 loops, best of 3: 24.4 ms per loop

In [20]: %timeit pd.eval('df1 + df2 + df3 + df4')
100 loops, best of 3: 13.8 ms per loop

Now let’s do the same thing but with comparisons:

In [21]: %timeit (df1 > 0) & (df2 > 0) & (df3 > 0) & (df4 > 0)
10 loops, best of 3: 71.4 ms per loop

In [22]: %timeit pd.eval('(df1 > 0) & (df2 > 0) & (df3 > 0) & (df4 > 0)')
10 loops, best of 3: 24.5 ms per loop

eval() also works with unaligned pandas objects:

In [23]: s = Series(randn(50))

In [24]: %timeit df1 + df2 + df3 + df4 + s
10 loops, best of 3: 80.5 ms per loop

In [25]: %timeit pd.eval('df1 + df2 + df3 + df4 + s')
10 loops, best of 3: 57 ms per loop

Note

Operations such as

1 and 2  # would parse to 1 & 2, but should evaluate to 2
3 or 4  # would parse to 3 | 4, but should evaluate to 3
~1  # this is okay, but slower when using eval

should be performed in Python. An exception will be raised if you try to perform any boolean/bitwise operations with scalar operands that are not of type bool or np.bool_. Again, you should perform these kinds of operations in plain Python.

The DataFrame.eval method (Experimental)

New in version 0.13.

In addition to the top level pandas.eval() function you can also evaluate an expression in the “context” of a DataFrame.

In [26]: df = DataFrame(randn(5, 2), columns=['a', 'b'])

In [27]: df.eval('a + b')
Out[27]: 
0   -0.246747
1    0.867786
2   -1.626063
3   -1.134978
4   -1.027798
dtype: float64

Any expression that is a valid pandas.eval() expression is also a valid DataFrame.eval() expression, with the added benefit that you don’t have to prefix the name of the DataFrame to the column(s) you’re interested in evaluating.

In addition, you can perform assignment of columns within an expression. This allows for formulaic evaluation. Only a single assignment is permitted. The assignment target can be a new column name or an existing column name, and it must be a valid Python identifier.

In [28]: df = DataFrame(dict(a=range(5), b=range(5, 10)))

In [29]: df.eval('c = a + b')

In [30]: df.eval('d = a + b + c')

In [31]: df.eval('a = 1')

In [32]: df
Out[32]: 
   a  b   c   d
0  1  5   5  10
1  1  6   7  14
2  1  7   9  18
3  1  8  11  22
4  1  9  13  26

The equivalent in standard Python would be

In [33]: df = DataFrame(dict(a=range(5), b=range(5, 10)))

In [34]: df['c'] = df.a + df.b

In [35]: df['d'] = df.a + df.b + df.c

In [36]: df['a'] = 1

In [37]: df
Out[37]: 
   a  b   c   d
0  1  5   5  10
1  1  6   7  14
2  1  7   9  18
3  1  8  11  22
4  1  9  13  26

Local Variables

In pandas version 0.14 the local variable API has changed. In pandas 0.13.x, you could refer to local variables the same way you would in standard Python. For example,

df = DataFrame(randn(5, 2), columns=['a', 'b'])
newcol = randn(len(df))
df.eval('b + newcol')

UndefinedVariableError: name 'newcol' is not defined

As you can see from the exception generated, this syntax is no longer allowed. You must explicitly reference any local variable that you want to use in an expression by placing the @ character in front of the name. For example,

In [38]: df = DataFrame(randn(5, 2), columns=list('ab'))

In [39]: newcol = randn(len(df))

In [40]: df.eval('b + @newcol')
Out[40]: 
0   -0.173926
1    2.493083
2   -0.881831
3   -0.691045
4    1.334703
dtype: float64

In [41]: df.query('b < @newcol')
Out[41]: 
          a         b
0  0.863987 -0.115998
2 -2.621419 -1.297879

If you don’t prefix the local variable with @, pandas will raise an exception telling you the variable is undefined.

When using DataFrame.eval() and DataFrame.query(), this allows you to have a local variable and a DataFrame column with the same name in an expression.

In [42]: a = randn()

In [43]: df.query('@a < a')
Out[43]: 
          a         b
0  0.863987 -0.115998

In [44]: df.loc[a < df.a]  # same as the previous expression
Out[44]: 
          a         b
0  0.863987 -0.115998

With pandas.eval() you cannot use the @ prefix at all, because it isn’t defined in that context. pandas will let you know this if you try to use @ in a top-level call to pandas.eval(). For example,

In [45]: a, b = 1, 2

In [46]: pd.eval('@a + b')
  File "<string>", line unknown
SyntaxError: The '@' prefix is not allowed in top-level eval calls, 
please refer to your variables by name without the '@' prefix

In this case, you should simply refer to the variables like you would in standard Python.

In [47]: pd.eval('a + b')
Out[47]: 3

pandas.eval() Parsers

There are two different parsers and two different engines you can use as the backend.

The default 'pandas' parser allows a more intuitive syntax for expressing query-like operations (comparisons, conjunctions and disjunctions). In particular, the precedence of the & and | operators is made equal to the precedence of the corresponding boolean operations and and or.

For example, the above conjunction can be written without parentheses. Alternatively, you can use the 'python' parser to enforce strict Python semantics.

In [48]: expr = '(df1 > 0) & (df2 > 0) & (df3 > 0) & (df4 > 0)'

In [49]: x = pd.eval(expr, parser='python')

In [50]: expr_no_parens = 'df1 > 0 & df2 > 0 & df3 > 0 & df4 > 0'

In [51]: y = pd.eval(expr_no_parens, parser='pandas')

In [52]: np.all(x == y)
Out[52]: True

The same expression can be “anded” together with the word and as well:

In [53]: expr = '(df1 > 0) & (df2 > 0) & (df3 > 0) & (df4 > 0)'

In [54]: x = pd.eval(expr, parser='python')

In [55]: expr_with_ands = 'df1 > 0 and df2 > 0 and df3 > 0 and df4 > 0'

In [56]: y = pd.eval(expr_with_ands, parser='pandas')

In [57]: np.all(x == y)
Out[57]: True

The and and or operators here have the same precedence that they would in vanilla Python.

pandas.eval() Backends

There’s also the option to make eval() operate identical to plain ol’ Python.

Note

Using the 'python' engine is generally not useful, except for testing other evaluation engines against it. You will acheive no performance benefits using eval() with engine='python' and in fact may incur a performance hit.

You can see this by using pandas.eval() with the 'python' engine. It is a bit slower (not by much) than evaluating the same expression in Python

In [58]: %timeit df1 + df2 + df3 + df4
10 loops, best of 3: 24.3 ms per loop

In [59]: %timeit pd.eval('df1 + df2 + df3 + df4', engine='python')
10 loops, best of 3: 26.1 ms per loop

pandas.eval() Performance

eval() is intended to speed up certain kinds of operations. In particular, those operations involving complex expressions with large DataFrame/Series objects should see a significant performance benefit. Here is a plot showing the running time of pandas.eval() as function of the size of the frame involved in the computation. The two lines are two different engines.

Note

Operations with smallish objects (around 15k-20k rows) are faster using plain Python:

This plot was created using a DataFrame with 3 columns each containing floating point values generated using numpy.random.randn().

Technical Minutia Regarding Expression Evaluation

Expressions that would result in an object dtype or involve datetime operations (because of NaT) must be evaluated in Python space. The main reason for this behavior is to maintain backwards compatbility with versions of numpy < 1.7. In those versions of numpy a call to ndarray.astype(str) will truncate any strings that are more than 60 characters in length. Second, we can’t pass object arrays to numexpr thus string comparisons must be evaluated in Python space.

The upshot is that this only applies to object-dtype’d expressions. So, if you have an expression–for example

In [60]: df = DataFrame({'strings': np.repeat(list('cba'), 3),
   ....:                 'nums': np.repeat(range(3), 3)})
   ....: 

In [61]: df
Out[61]: 
   nums strings
0     0       c
1     0       c
2     0       c
3     1       b
4     1       b
5     1       b
6     2       a
7     2       a
8     2       a

In [62]: df.query('strings == "a" and nums == 1')
Out[62]: 
Empty DataFrame
Columns: [nums, strings]
Index: []

the numeric part of the comparison (nums == 1) will be evaluated by numexpr.

In general, DataFrame.query()/pandas.eval() will evaluate the subexpressions that can be evaluated by numexpr and those that must be evaluated in Python space transparently to the user. This is done by inferring the result type of an expression from its arguments and operators.