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# 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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval). 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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) 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](http://cython.org/).
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. Its 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](http://docs.cython.org/src/quickstart/cythonize.html)
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.
``` python
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]
```
Heres the function in pure Python:
``` 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):
``` python
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 isnt fast enough for us. Lets 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](http://ipython.org/ipython-doc/stable/api/generated/IPython.core.magics.execution.html#IPython.core.magics.execution.ExecutionMagics.prun):
``` python
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 well concentrate our efforts cythonizing these two functions.
::: tip 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 were going to need to import the Cython magic function to ipython:
``` python
In [6]: %load_ext Cython
```
Now, lets simply copy our functions over to Cython as is (the suffix
is here to distinguish between function versions):
``` python
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
...:
```
::: tip Note
If youre 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.
:::
``` python
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:
``` python
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
...:
```
``` python
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, were talking! Its now over ten times faster than the original python
implementation, and we havent *really* modified the code. Lets have another
look at whats eating up time:
``` python
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
Its calling series… a lot! Its 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.
::: tip Note
We are now passing ndarrays into the Cython function, fortunately Cython plays
very nicely with NumPy.
:::
``` python
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.
::: danger 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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.to_numpy.html#pandas.Series.to_numpy). The reason is that the Cython
definition is specific to an ndarray and not the passed ``Series``.
So, do not do this:
``` python
apply_integrate_f(df['a'], df['b'], df['N'])
```
But rather, use [``Series.to_numpy()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.to_numpy.html#pandas.Series.to_numpy) to get the underlying ``ndarray``:
``` python
apply_integrate_f(df['a'].to_numpy(),
df['b'].to_numpy(),
df['N'].to_numpy())
```
:::
::: tip Note
Loops like this would be *extremely* slow in Python, but in Cython looping
over NumPy arrays is *fast*.
:::
``` python
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
```
Weve gotten another big improvement. Lets check again where the time is spent:
``` python
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. Heres an example of using some more
advanced Cython techniques:
``` python
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
....:
```
``` python
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 isnt checked.
For more about ``boundscheck`` and ``wraparound``, see the Cython docs on
[compiler directives](http://cython.readthedocs.io/en/latest/src/reference/compilation.html?highlight=wraparound#compiler-directives).
## 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.
::: tip Note
You will need to install Numba. This is easy with ``conda``, by using: ``conda install numba``, see [installing using miniconda](https://pandas.pydata.org/pandas-docs/stable/install.html#install-miniconda).
:::
::: tip 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.
``` python
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.
``` python
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:
``` python
import numba
def double_every_value_nonumba(x):
return x * 2
@numba.vectorize
def double_every_value_withnumba(x): # noqa E501
return x * 2
```
``` python
# 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
::: tip 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 doesnt 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](http://numba.pydata.org/numba-doc/latest/user/troubleshoot.html#the-compiled-code-is-too-slow).
Read more in the [Numba docs](http://numba.pydata.org/).
## Expression evaluation via ``eval()``
The top-level function [``pandas.eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) implements expression evaluation of
[``Series``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.html#pandas.Series) and [``DataFrame``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.html#pandas.DataFrame) objects.
::: tip Note
To benefit from using [``eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) you need to
install ``numexpr``. See the [recommended dependencies section](https://pandas.pydata.org/pandas-docs/stable/install.html#install-recommended-dependencies) for more details.
:::
The point of using [``eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) for expression evaluation rather than
plain Python is two-fold: 1) large [``DataFrame``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.html#pandas.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).
::: tip Note
You should not use [``eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) for simple
expressions or for expressions involving small DataFrames. In fact,
[``eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) when you have a
``DataFrame`` with more than 10,000 rows.
:::
[``eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) supports all arithmetic expressions supported by the
engine in addition to some extensions available only in pandas.
::: tip Note
The larger the frame and the larger the expression the more speedup you will
see from using [``eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval).
:::
### Supported syntax
These operations are supported by [``pandas.eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#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)
- 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
- Neither [simple](https://docs.python.org/3/reference/simple_stmts.html)
nor [compound](https://docs.python.org/3/reference/compound_stmts.html)
statements are allowed. This includes things like ``for``, ``while``, and
``if``.
### ``eval()`` examples
[``pandas.eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) works well with expressions containing large arrays.
First lets create a few decent-sized arrays to play with:
``` python
In [13]: nrows, ncols = 20000, 100
In [14]: df1, df2, df3, df4 = [pd.DataFrame(np.random.randn(nrows, ncols)) for _ in range(4)]
```
Now lets compare adding them together using plain ol Python versus
[``eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval):
``` python
In [15]: %timeit df1 + df2 + df3 + df4
21 ms +- 787 us per loop (mean +- std. dev. of 7 runs, 10 loops each)
```
``` python
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 lets do the same thing but with comparisons:
``` python
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)
```
``` python
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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) also works with unaligned pandas objects:
``` python
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)
```
``` python
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)
```
::: tip Note
Operations such as
``` python
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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) function you can also
evaluate an expression in the “context” of a [``DataFrame``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.html#pandas.DataFrame).
``` python
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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) expression is also a valid
[``DataFrame.eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.eval.html#pandas.DataFrame.eval) expression, with the added benefit that you dont have to
prefix the name of the [``DataFrame``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.html#pandas.DataFrame) to the column(s) youre
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.
::: danger 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``.
:::
``` python
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.
``` python
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.
``` python
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
``` python
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.
``` python
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
```
::: danger 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,
``` python
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 dont prefix the local variable with ``@``, pandas will raise an
exception telling you the variable is undefined.
When using [``DataFrame.eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.eval.html#pandas.DataFrame.eval) and [``DataFrame.query()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.query.html#pandas.DataFrame.query), this allows you
to have a local variable and a [``DataFrame``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.html#pandas.DataFrame) column with the same
name in an expression.
``` python
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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) you cannot use the ``@`` prefix *at all*, because it
isnt defined in that context. ``pandas`` will let you know this if you try to
use ``@`` in a top-level call to [``pandas.eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval). For example,
``` python
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.
``` 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.
``` python
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``](https://docs.python.org/3/reference/expressions.html#and) as
well:
``` python
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
Theres also the option to make [``eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) operate identical to plain
ol Python.
::: tip 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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) with ``engine='python'`` and in fact may
incur a performance hit.
:::
You can see this by using [``pandas.eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) with the ``'python'`` engine. It
is a bit slower (not by much) than evaluating the same expression in Python
``` python
In [62]: %timeit df1 + df2 + df3 + df4
9.5 ms +- 241 us per loop (mean +- std. dev. of 7 runs, 100 loops each)
```
``` python
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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) is intended to speed up certain kinds of operations. In
particular, those operations involving complex expressions with large
[``DataFrame``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.html#pandas.DataFrame)/[``Series``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.html#pandas.Series) objects should see a
significant performance benefit. Here is a plot showing the running time of
[``pandas.eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#pandas.eval) as function of the size of the frame involved in the
computation. The two lines are two different engines.
![eval-perf](https://static.pypandas.cn/public/static/images/eval-perf.png)
::: tip Note
Operations with smallish objects (around 15k-20k rows) are faster using
plain Python:
![eval-perf-small](https://static.pypandas.cn/public/static/images/eval-perf-small.png)
:::
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
cant 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 expressionfor example
``` python
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()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.query.html#pandas.DataFrame.query)/[``pandas.eval()``](https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.eval.html#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.