# 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()`

. 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()`

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.

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 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:

```
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()`

. 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()`

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.

## 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.

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.

Read more in the Numba docs.

`eval()`

Expression evaluation via 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) - 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()`

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()`

:

```
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()`

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.

`DataFrame.eval`

method

The In addition to the top level `pandas.eval()`

function you can also evaluate an expression in the “context” of a `DataFrame`

.

```
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()`

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*. 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()`

and `DataFrame.query()`

, this allows you to have a local variable and a `DataFrame`

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()`

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 [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`

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()`

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()`

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 [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()`

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 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()`

/`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.