# Enhancing performance

In this part of the tutorial, we will investigate how to speed up certain functions operating on pandas DataFrames using three different techniques: Cython, Numba and pandas.eval() (opens new window). We will see a speed improvement of ~200 when we use Cython and Numba on a test function operating row-wise on the DataFrame. Using pandas.eval() (opens new window) we will speed up a sum by an order of ~2.

# 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 sizable speed-ups by offloading work to cython (opens new window).

This tutorial assumes you have refactored as much as possible in Python, for example by 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 (opens new window) but in the context of pandas. Our final cythonized solution is around 100 times faster than the pure Python solution.

# Pure Python

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

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

In [2]: df
Out[2]: 
            a         b    N  x
0    0.469112 -0.218470  585  x
1   -0.282863 -0.061645  841  x
2   -1.509059 -0.723780  251  x
3   -1.135632  0.551225  972  x
4    1.212112 -0.497767  181  x
..        ...       ...  ... ..
995 -1.512743  0.874737  374  x
996  0.933753  1.120790  246  x
997 -0.308013  0.198768  157  x
998 -0.079915  1.757555  977  x
999 -1.010589 -1.115680  770  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 (opens new window):

In [5]: %prun -l 4 df.apply(lambda x: integrate_f(x['a'], x['b'], x['N']), axis=1)  # noqa E999
         672332 function calls (667306 primitive calls) in 0.285 seconds

   Ordered by: internal time
   List reduced from 221 to 4 due to restriction <4>

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     1000    0.144    0.000    0.217    0.000 <ipython-input-4-c2a74e076cf0>:1(integrate_f)
   552423    0.074    0.000    0.074    0.000 <ipython-input-3-c138bdd570e3>:1(f)
     3000    0.008    0.000    0.045    0.000 base.py:4695(get_value)
     6001    0.005    0.000    0.012    0.000 {pandas._libs.lib.values_from_object}

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 Cython

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)
         119905 function calls (114879 primitive calls) in 0.096 seconds

   Ordered by: internal time
   List reduced from 216 to 4 due to restriction <4>

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     3000    0.012    0.000    0.064    0.000 base.py:4695(get_value)
     6001    0.007    0.000    0.017    0.000 {pandas._libs.lib.values_from_object}
     3000    0.007    0.000    0.073    0.000 series.py:1061(__getitem__)
     3000    0.006    0.000    0.006    0.000 {method 'get_value' of 'pandas._libs.index.IndexEngine' objects}

# 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 minimize 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

You can not pass a Series directly as a ndarray typed parameter to a Cython function. Instead pass the actual ndarray using the Series.to_numpy() (opens new window). The reason is that the Cython definition is specific to an ndarray and not the passed Series.

So, do not do this:

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

But rather, use Series.to_numpy() (opens new window) to get the underlying ndarray:

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

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'].to_numpy(),
                                  df['b'].to_numpy(),
                                  df['N'].to_numpy())
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'].to_numpy(),
   ....:                              df['b'].to_numpy(),
   ....:                              df['N'].to_numpy())
   ....: 
  File "<ipython-input-11-613f5c6ec02d>", line 2
    df['b'].to_numpy(),
    ^
IndentationError: unexpected indent

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[int] col_N):
   ....:     cdef int 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'].to_numpy(),
                                       df['b'].to_numpy(),
                                       df['N'].to_numpy())
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. For more about boundscheck and wraparound, see the Cython docs on compiler directives (opens new window).

# 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 (opens new window).

Note

As of Numba version 0.20, pandas objects cannot be passed directly to Numba-compiled functions. Instead, one must pass the NumPy array underlying the pandas object to the Numba-compiled function as demonstrated below.

# Jit

We demonstrate how to use Numba to just-in-time compile our code. 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'].to_numpy(),
                                     df['b'].to_numpy(),
                                     df['N'].to_numpy())
    return pd.Series(result, index=df.index, name='result')

Note that we directly pass NumPy arrays to the Numba function. compute_numba is just a wrapper that provides a nicer interface by passing/returning pandas objects.

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

In this example, using Numba was faster than Cython.

# Vectorize

Numba can also be used to write vectorized functions that do not require the user to explicitly loop over the observations of a vector; a vectorized function will be applied to each row automatically. Consider the following toy example of doubling each observation:

import numba


def double_every_value_nonumba(x):
    return x * 2


@numba.vectorize
def double_every_value_withnumba(x):  # noqa E501
    return x * 2
# Custom function without numba
In [5]: %timeit df['col1_doubled'] = df.a.apply(double_every_value_nonumba)  # noqa E501
1000 loops, best of 3: 797 us per loop

# Standard implementation (faster than a custom function)
In [6]: %timeit df['col1_doubled'] = df.a * 2
1000 loops, best of 3: 233 us per loop

# Custom function with numba
In [7]: %timeit (df['col1_doubled'] = double_every_value_withnumba(df.a.to_numpy())
1000 loops, best of 3: 145 us per loop

# Caveats

Note

Numba will execute on any function, but can only accelerate certain classes of functions.

Numba is best at accelerating functions that apply numerical functions to NumPy arrays. When passed a function that only uses operations it knows how to accelerate, it will execute in nopython mode.

If Numba is passed a function that includes something it doesn’t know how to work with – a category that currently includes sets, lists, dictionaries, or string functions – it will revert to object mode. In object mode, Numba will execute but your code will not speed up significantly. If you would prefer that Numba throw an error if it cannot compile a function in a way that speeds up your code, pass Numba the argument nopython=True (e.g. @numba.jit(nopython=True)). For more on troubleshooting Numba modes, see the Numba troubleshooting page (opens new window).

Read more in the Numba docs (opens new window).

# Expression evaluation via eval()

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

Note

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

The point of using eval() (opens new window) for expression evaluation rather than plain Python is two-fold: 1) large DataFrame (opens new window) 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() (opens new window) for simple expressions or for expressions involving small DataFrames. In fact, eval() (opens new window) is many orders of magnitude slower for smaller expressions/objects than plain ol’ Python. A good rule of thumb is to only use eval() (opens new window) when you have a DataFrame with more than 10,000 rows.

eval() (opens new window) 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() (opens new window).

# Supported syntax

These operations are supported by pandas.eval() (opens new window):

  • 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)
  • Math functions: sin, cos, exp, log, expm1, log1p, sqrt, sinh, cosh, tanh, arcsin, arccos, arctan, arccosh, arcsinh, arctanh, abs, arctan2 and log10.

This Python syntax is not allowed:

  • Expressions

    • Function calls other than math functions.
    • 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

# eval() examples

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

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

In [13]: nrows, ncols = 20000, 100

In [14]: df1, df2, df3, df4 = [pd.DataFrame(np.random.randn(nrows, ncols)) for _ in range(4)]

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

In [15]: %timeit df1 + df2 + df3 + df4
21 ms +- 787 us per loop (mean +- std. dev. of 7 runs, 10 loops each)
In [16]: %timeit pd.eval('df1 + df2 + df3 + df4')
8.12 ms +- 249 us per loop (mean +- std. dev. of 7 runs, 100 loops each)

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

In [17]: %timeit (df1 > 0) & (df2 > 0) & (df3 > 0) & (df4 > 0)
272 ms +- 6.92 ms per loop (mean +- std. dev. of 7 runs, 1 loop each)
In [18]: %timeit pd.eval('(df1 > 0) & (df2 > 0) & (df3 > 0) & (df4 > 0)')
19.2 ms +- 1.87 ms per loop (mean +- std. dev. of 7 runs, 10 loops each)

eval() (opens new window) also works with unaligned pandas objects:

In [19]: s = pd.Series(np.random.randn(50))

In [20]: %timeit df1 + df2 + df3 + df4 + s
103 ms +- 12.7 ms per loop (mean +- std. dev. of 7 runs, 10 loops each)
In [21]: %timeit pd.eval('df1 + df2 + df3 + df4 + s')
10.2 ms +- 215 us per loop (mean +- std. dev. of 7 runs, 100 loops each)

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

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

In [22]: df = pd.DataFrame(np.random.randn(5, 2), columns=['a', 'b'])

In [23]: df.eval('a + b')
Out[23]: 
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() (opens new window) expression is also a valid DataFrame.eval() (opens new window) expression, with the added benefit that you don’t have to prefix the name of the DataFrame (opens new window) 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. The assignment target can be a new column name or an existing column name, and it must be a valid Python identifier.

New in version 0.18.0.

The inplace keyword determines whether this assignment will performed on the original DataFrame or return a copy with the new column.

Warning

For backwards compatibility, inplace defaults to True if not specified. This will change in a future version of pandas - if your code depends on an inplace assignment you should update to explicitly set inplace=True.

In [24]: df = pd.DataFrame(dict(a=range(5), b=range(5, 10)))

In [25]: df.eval('c = a + b', inplace=True)

In [26]: df.eval('d = a + b + c', inplace=True)

In [27]: df.eval('a = 1', inplace=True)

In [28]: df
Out[28]: 
   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

When inplace is set to False, a copy of the DataFrame with the new or modified columns is returned and the original frame is unchanged.

In [29]: df
Out[29]: 
   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

In [30]: df.eval('e = a - c', inplace=False)
Out[30]: 
   a  b   c   d   e
0  1  5   5  10  -4
1  1  6   7  14  -6
2  1  7   9  18  -8
3  1  8  11  22 -10
4  1  9  13  26 -12

In [31]: df
Out[31]: 
   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

New in version 0.18.0.

As a convenience, multiple assignments can be performed by using a multi-line string.

In [32]: df.eval("""
   ....: c = a + b
   ....: d = a + b + c
   ....: a = 1""", inplace=False)
   ....: 
Out[32]: 
   a  b   c   d
0  1  5   6  12
1  1  6   7  14
2  1  7   8  16
3  1  8   9  18
4  1  9  10  20

The equivalent in standard Python would be

In [33]: df = pd.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

New in version 0.18.0.

The query method gained the inplace keyword which determines whether the query modifies the original frame.

In [38]: df = pd.DataFrame(dict(a=range(5), b=range(5, 10)))

In [39]: df.query('a > 2')
Out[39]: 
   a  b
3  3  8
4  4  9

In [40]: df.query('a > 2', inplace=True)

In [41]: df
Out[41]: 
   a  b
3  3  8
4  4  9

Warning

Unlike with eval, the default value for inplace for query is False. This is consistent with prior versions of pandas.

# Local variables

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 [42]: df = pd.DataFrame(np.random.randn(5, 2), columns=list('ab'))

In [43]: newcol = np.random.randn(len(df))

In [44]: df.eval('b + @newcol')
Out[44]: 
0   -0.173926
1    2.493083
2   -0.881831
3   -0.691045
4    1.334703
dtype: float64

In [45]: df.query('b < @newcol')
Out[45]: 
          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() (opens new window) and DataFrame.query() (opens new window), this allows you to have a local variable and a DataFrame (opens new window) column with the same name in an expression.

In [46]: a = np.random.randn()

In [47]: df.query('@a < a')
Out[47]: 
          a         b
0  0.863987 -0.115998

In [48]: df.loc[a < df.a]  # same as the previous expression
Out[48]: 
          a         b
0  0.863987 -0.115998

With pandas.eval() (opens new window) 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() (opens new window). For example,

In [49]: a, b = 1, 2

In [50]: pd.eval('@a + b')
Traceback (most recent call last):

  File "/opt/conda/envs/pandas/lib/python3.7/site-packages/IPython/core/interactiveshell.py", line 3325, in run_code
    exec(code_obj, self.user_global_ns, self.user_ns)

  File "<ipython-input-50-af17947a194f>", line 1, in <module>
    pd.eval('@a + b')

  File "/pandas/pandas/core/computation/eval.py", line 311, in eval
    _check_for_locals(expr, level, parser)

  File "/pandas/pandas/core/computation/eval.py", line 166, in _check_for_locals
    raise SyntaxError(msg)

  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 [51]: pd.eval('a + b')
Out[51]: 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 [52]: expr = '(df1 > 0) & (df2 > 0) & (df3 > 0) & (df4 > 0)'

In [53]: x = pd.eval(expr, parser='python')

In [54]: expr_no_parens = 'df1 > 0 & df2 > 0 & df3 > 0 & df4 > 0'

In [55]: y = pd.eval(expr_no_parens, parser='pandas')

In [56]: np.all(x == y)
Out[56]: True

The same expression can be “anded” together with the word and (opens new window) as well:

In [57]: expr = '(df1 > 0) & (df2 > 0) & (df3 > 0) & (df4 > 0)'

In [58]: x = pd.eval(expr, parser='python')

In [59]: expr_with_ands = 'df1 > 0 and df2 > 0 and df3 > 0 and df4 > 0'

In [60]: y = pd.eval(expr_with_ands, parser='pandas')

In [61]: np.all(x == y)
Out[61]: 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() (opens new window) 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 achieve no performance benefits using eval() (opens new window) with engine='python' and in fact may incur a performance hit.

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

In [62]: %timeit df1 + df2 + df3 + df4
9.5 ms +- 241 us per loop (mean +- std. dev. of 7 runs, 100 loops each)
In [63]: %timeit pd.eval('df1 + df2 + df3 + df4', engine='python')
10.8 ms +- 898 us per loop (mean +- std. dev. of 7 runs, 100 loops each)

# pandas.eval() performance

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

eval-perf

Note

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

eval-perf-small

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 compatibility 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 expressions. So, if you have an expression–for example

In [64]: df = pd.DataFrame({'strings': np.repeat(list('cba'), 3),
   ....:                    'nums': np.repeat(range(3), 3)})
   ....: 

In [65]: df
Out[65]: 
  strings  nums
0       c     0
1       c     0
2       c     0
3       b     1
4       b     1
5       b     1
6       a     2
7       a     2
8       a     2

In [66]: df.query('strings == "a" and nums == 1')
Out[66]: 
Empty DataFrame
Columns: [strings, nums]
Index: []

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

In general, DataFrame.query() (opens new window)/pandas.eval() (opens new window) 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.