问题:熊猫中的for循环真的不好吗?我什么时候应该在意?

for循环真正的“坏”?如果不是,在什么情况下它们会比使用更常规的“矢量化”方法更好?1个

我熟悉“矢量化”的概念,以及熊猫如何利用矢量化技术来加快计算速度。向量化功能在整个系列或DataFrame上广播操作,以实现比传统上迭代数据快得多的加速。

但是,我很惊讶地看到很多代码(包括来自Stack Overflow的答案)提供了解决问题的解决方案,这些问题涉及使用for循环和列表推导来遍历数据。文档和API指出循环是“不好的”循环,并且“绝不能”循环访问数组,序列或DataFrame。那么,为什么有时我会看到用户建议基于循环的解决方案?


1-虽然问题听起来似乎有些宽泛,但事实是,在某些非常特殊的情况下,for循环通常比传统上遍历数据更好。这篇文章的目的是为了后代。

Are for loops really “bad”? If not, in what situation(s) would they be better than using a more conventional “vectorized” approach?1

I am familiar with the concept of “vectorization”, and how pandas employs vectorized techniques to speed up computation. Vectorized functions broadcast operations over the entire series or DataFrame to achieve speedups much greater than conventionally iterating over the data.

However, I am quite surprised to see a lot of code (including from answers on Stack Overflow) offering solutions to problems that involve looping through data using for loops and list comprehensions. The documentation and API say that loops are “bad”, and that one should “never” iterate over arrays, series, or DataFrames. So, how come I sometimes see users suggesting loop-based solutions?


1 – While it is true that the question sounds somewhat broad, the truth is that there are very specific situations when for loops are usually better than conventionally iterating over data. This post aims to capture this for posterity.


回答 0

TLDR;不,for循环并非总会“坏”,至少并非总是如此。说某些矢量化操作比迭代慢,而不是说迭代快于某些矢量化操作,可能更准确。知道何时以及为什么是使代码获得最大性能的关键。简而言之,在以下情况下,值得考虑使用矢量化熊猫函数的替代方法:

  1. 当您的数据很小时(…取决于您的工作),
  2. 处理object/ mixed dtypes时
  3. 使用str/ regex访问器功能时

让我们分别检查这些情况。


小数据上的迭代v / s矢量化

熊猫在其API设计中遵循“配置惯例”方法。这意味着已经安装了相同的API,以适应广泛的数据和用例。

调用pandas函数时,该函数必须在内部处理以下各项(除其他事项外),以确保工作正常

  1. 索引/轴对齐
  2. 处理混合数据类型
  3. 处理丢失的数据

几乎每个函数都必须在不同程度上处理这些问题,这带来了开销。数字函数(例如Series.add)的开销较少,而字符串函数(例如Series.str.replace)的开销更为明显。

另一方面,循环比您想象的要快。更好的是列表理解(通过for循环创建列表)更快,因为它们是优化的列表创建迭代机制。

列表理解遵循模式

[f(x) for x in seq]

seq熊猫系列或DataFrame列在哪里。或者,当对多列进行操作时,

[f(x, y) for x, y in zip(seq1, seq2)]

seq1seq2列。

数值比较
考虑一个简单的布尔索引操作。列表推导方法已针对!=)和进行计时query。功能如下:

# Boolean indexing with Numeric value comparison.
df[df.A != df.B]                            # vectorized !=
df.query('A != B')                          # query (numexpr)
df[[x != y for x, y in zip(df.A, df.B)]]    # list comp

为简单起见,本文中使用了该perfplot包来运行所有的timeit测试。上述操作的时间安排如下:

在此处输入图片说明

query对于中等大小的N,列表理解要胜过,甚至对于较小的N而言,列表理解要胜过向量化不等于比较。不幸的是,列表理解是线性缩放的,因此对于较大的N而言,它不能提供很多性能提升。

注意
值得一提的是,列表理解的许多好处来自于不必担心索引对齐,但是这意味着,如果您的代码依赖于索引对齐,则此操作会中断。在某些情况下,可以将对基础NumPy数组的矢量化操作视为“两全其美”,从而实现了矢量化,而没有熊猫函数的所有不必要开销。这意味着您可以将上面的操作重写为

df[df.A.values != df.B.values]

它的性能优于熊猫和列表理解同等物:

NumPy矢量化不在本文讨论范围之内,但是如果性能很重要,则绝对值得考虑。

值计数
再举一个例子-这次,使用另一个比for循环快的香草python构造- 。通常的要求是计算值计数并将结果作为字典返回。这与做以及Counter

# Value Counts comparison.
ser.value_counts(sort=False).to_dict()           # value_counts
dict(zip(*np.unique(ser, return_counts=True)))   # np.unique
Counter(ser)                                     # Counter

在此处输入图片说明

结果更明显,Counter在较大的小N范围(〜3500)下胜过两种矢量化方法。

注意
更多琐事(由@ user2357112提供)。的Counter实现是使用C加速器实现的,因此尽管它仍必须使用python对象而不是底层的C数据类型,但它仍比for循环要快。Python的力量!

当然,这里的好处是性能取决于您的数据和用例。这些示例的目的是说服您不要将这些解决方案排除为合法选项。如果这些仍然不能满足您的需求,那么总会有cythonnumba。让我们将此测试添加到混合中。

from numba import njit, prange

@njit(parallel=True)
def get_mask(x, y):
    result = [False] * len(x)
    for i in prange(len(x)):
        result[i] = x[i] != y[i]

    return np.array(result)

df[get_mask(df.A.values, df.B.values)] # numba

在此处输入图片说明

Numba可以将循环python代码的JIT编译为功能非常强大的矢量化代码。了解如何使numba发挥作用需要学习。


混合/ objectdtype操作

基于字符串的比较再
来看第一部分的过滤示例,如果要比较的列是字符串怎么办?考虑上面相同的3个函数,但将输入DataFrame强制转换为字符串。

# Boolean indexing with string value comparison.
df[df.A != df.B]                            # vectorized !=
df.query('A != B')                          # query (numexpr)
df[[x != y for x, y in zip(df.A, df.B)]]    # list comp

在此处输入图片说明

那么,发生了什么变化?这里要注意的是,字符串操作本质上难以向量化。Pandas将字符串视为对象,并且对对象的所有操作都会回退到缓慢,循环的实现中。

现在,由于此循环实现被上述所有开销所包围,因此,即使这些解决方案按比例缩放,它们之间也存在恒定的幅度差异。

当涉及对可变/复杂对象的操作时,没有比较。列表理解胜过所有涉及字典和列表的操作。

通过键访问字典值
以下是从字典列中提取值的两个操作的时间安排:map和列表理解。该设置位于附录的“代码段”标题下。

# Dictionary value extraction.
ser.map(operator.itemgetter('value'))     # map
pd.Series([x.get('value') for x in ser])  # list comprehension

在此处输入图片说明


3个操作的位置列表索引计时,这些操作从列列表中提取第0个元素(处理异常)mapstr.get访问器方法和列表推导:

# List positional indexing. 
def get_0th(lst):
    try:
        return lst[0]
    # Handle empty lists and NaNs gracefully.
    except (IndexError, TypeError):
        return np.nan

ser.map(get_0th)                                          # map
ser.str[0]                                                # str accessor
pd.Series([x[0] if len(x) > 0 else np.nan for x in ser])  # list comp
pd.Series([get_0th(x) for x in ser])                      # list comp safe

注意
如果索引很重要,则需要执行以下操作:

pd.Series([...], index=ser.index)

重建系列时。

在此处输入图片说明

列表扁平
化最后一个例子是扁平化列表。这是另一个常见问题,它演示了纯python在这里有多么强大。

# Nested list flattening.
pd.DataFrame(ser.tolist()).stack().reset_index(drop=True)  # stack
pd.Series(list(chain.from_iterable(ser.tolist())))         # itertools.chain
pd.Series([y for x in ser for y in x])                     # nested list comp

在此处输入图片说明

无论和嵌套列表理解是纯Python结构,并且规模比更好stack的解决方案。

这些时间点充分说明了熊猫没有为混合dtypes做好准备的事实,并且您可能应该避免使用它来这样做。数据应尽可能在单独的列中作为标量值(整数/浮点数/字符串)存在。

最后,这些解决方案的适用性在很大程度上取决于您的数据。因此,最好的办法是先决定对数据进行这些操作,然后再决定要做什么。请注意,我尚未apply对这些解决方案计时,因为它会使图形倾斜(是​​的,那太慢了)。


正则表达式操作和访问器.str方法

熊猫可以应用正则表达式的操作,如,以及其他的“矢量”字符串操作(例如str.split,str.find ,str.translate`,等等)的字符串列。这些功能比列表理解要慢,并且是比其他功能更方便的功能。

预编译正则表达式模式并使用遍历数据通常要快得多(另请参阅使用Python的re.compile是否值得?)。列表组合等效于str.contains如下所示:

p = re.compile(...)
ser2 = pd.Series([x for x in ser if p.search(x)])

要么,

ser2 = ser[[bool(p.search(x)) for x in ser]]

如果您需要处理NaN,则可以执行以下操作

ser[[bool(p.search(x)) if pd.notnull(x) else False for x in ser]]

相当于str.extract(无组)的列表组合看起来像:

df['col2'] = [p.search(x).group(0) for x in df['col']]

如果您需要处理不匹配和NaN,则可以使用自定义函数(速度更快!):

def matcher(x):
    m = p.search(str(x))
    if m:
        return m.group(0)
    return np.nan

df['col2'] = [matcher(x) for x in df['col']]

matcher功能是非常可扩展的。根据需要,它可以适合返回每个捕获组的列表。只需提取查询匹配对象的groupor groups属性即可。

对于str.extractall,请更改p.searchp.findall

字符串提取
考虑简单的过滤操作。这个想法是提取一个大写字母开头的4位数字。

# Extracting strings.
p = re.compile(r'(?<=[A-Z])(\d{4})')
def matcher(x):
    m = p.search(x)
    if m:
        return m.group(0)
    return np.nan

ser.str.extract(r'(?<=[A-Z])(\d{4})', expand=False)   #  str.extract
pd.Series([matcher(x) for x in ser])                  #  list comprehension

在此处输入图片说明

更多示例
完全公开-我是以下列出的这些帖子的作者(部分或全部)。


结论

如上面的示例所示,当处理少量的DataFrame,混合的数据类型和正则表达式时,迭代会发光。

您获得的提速取决于您的数据和问题,因此里程可能会有所不同。最好的办法是仔细运行测试,看看是否值得付出努力。

“向量化”功能的优点在于其简单性和可读性,因此,如果性能不是很关键,则绝对应该首选这些功能。

另一个注意事项是,某些字符串操作处理了一些使用NumPy的约束。以下是两个示例,其中仔细的NumPy向量化性能胜过python:

此外,有时.values相对于Series或DataFrame ,仅通过底层数组进行操作就可以为大多数常见情况提供足够健康的加速(请参见上面“ 数字比较”部分的“ 注意 ” )。因此,举例来说,即时性能会提高。使用可能并非在每种情况下都适用,但这是一个有用的技巧。df[df.A.values != df.B.values]df[df.A != df.B].values

如上所述,由您决定这些解决方案是否值得实施。


附录:代码段

import perfplot  
import operator 
import pandas as pd
import numpy as np
import re

from collections import Counter
from itertools import chain

# Boolean indexing with Numeric value comparison.
perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(1000, (n, 2)), columns=['A','B']),
    kernels=[
        lambda df: df[df.A != df.B],
        lambda df: df.query('A != B'),
        lambda df: df[[x != y for x, y in zip(df.A, df.B)]],
        lambda df: df[get_mask(df.A.values, df.B.values)]
    ],
    labels=['vectorized !=', 'query (numexpr)', 'list comp', 'numba'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N'
)

# Value Counts comparison.
perfplot.show(
    setup=lambda n: pd.Series(np.random.choice(1000, n)),
    kernels=[
        lambda ser: ser.value_counts(sort=False).to_dict(),
        lambda ser: dict(zip(*np.unique(ser, return_counts=True))),
        lambda ser: Counter(ser),
    ],
    labels=['value_counts', 'np.unique', 'Counter'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=lambda x, y: dict(x) == dict(y)
)

# Boolean indexing with string value comparison.
perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(1000, (n, 2)), columns=['A','B'], dtype=str),
    kernels=[
        lambda df: df[df.A != df.B],
        lambda df: df.query('A != B'),
        lambda df: df[[x != y for x, y in zip(df.A, df.B)]],
    ],
    labels=['vectorized !=', 'query (numexpr)', 'list comp'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=None
)

# Dictionary value extraction.
ser1 = pd.Series([{'key': 'abc', 'value': 123}, {'key': 'xyz', 'value': 456}])
perfplot.show(
    setup=lambda n: pd.concat([ser1] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.map(operator.itemgetter('value')),
        lambda ser: pd.Series([x.get('value') for x in ser]),
    ],
    labels=['map', 'list comprehension'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=None
)

# List positional indexing. 
ser2 = pd.Series([['a', 'b', 'c'], [1, 2], []])        
perfplot.show(
    setup=lambda n: pd.concat([ser2] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.map(get_0th),
        lambda ser: ser.str[0],
        lambda ser: pd.Series([x[0] if len(x) > 0 else np.nan for x in ser]),
        lambda ser: pd.Series([get_0th(x) for x in ser]),
    ],
    labels=['map', 'str accessor', 'list comprehension', 'list comp safe'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=None
)

# Nested list flattening.
ser3 = pd.Series([['a', 'b', 'c'], ['d', 'e'], ['f', 'g']])
perfplot.show(
    setup=lambda n: pd.concat([ser2] * n, ignore_index=True),
    kernels=[
        lambda ser: pd.DataFrame(ser.tolist()).stack().reset_index(drop=True),
        lambda ser: pd.Series(list(chain.from_iterable(ser.tolist()))),
        lambda ser: pd.Series([y for x in ser for y in x]),
    ],
    labels=['stack', 'itertools.chain', 'nested list comp'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',    
    equality_check=None

)

# Extracting strings.
ser4 = pd.Series(['foo xyz', 'test A1234', 'D3345 xtz'])
perfplot.show(
    setup=lambda n: pd.concat([ser4] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.str.extract(r'(?<=[A-Z])(\d{4})', expand=False),
        lambda ser: pd.Series([matcher(x) for x in ser])
    ],
    labels=['str.extract', 'list comprehension'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=None
)

TLDR; No, for loops are not blanket “bad”, at least, not always. It is probably more accurate to say that some vectorized operations are slower than iterating, versus saying that iteration is faster than some vectorized operations. Knowing when and why is key to getting the most performance out of your code. In a nutshell, these are the situations where it is worth considering an alternative to vectorized pandas functions:

  1. When your data is small (…depending on what you’re doing),
  2. When dealing with object/mixed dtypes
  3. When using the str/regex accessor functions

Let’s examine these situations individually.


Iteration v/s Vectorization on Small Data

Pandas follows a “Convention Over Configuration” approach in its API design. This means that the same API has been fitted to cater to a broad range of data and use cases.

When a pandas function is called, the following things (among others) must internally be handled by the function, to ensure working

  1. Index/axis alignment
  2. Handling mixed datatypes
  3. Handling missing data

Almost every function will have to deal with these to varying extents, and this presents an overhead. The overhead is less for numeric functions (for example, Series.add), while it is more pronounced for string functions (for example, Series.str.replace).

loops, on the other hand, are faster then you think. What’s even better is list comprehensions (which create lists through for loops) are even faster as they are optimized iterative mechanisms for list creation.

List comprehensions follow the pattern

[f(x) for x in seq]

Where seq is a pandas series or DataFrame column. Or, when operating over multiple columns,

[f(x, y) for x, y in zip(seq1, seq2)]

Where seq1 and seq2 are columns.

Numeric Comparison
Consider a simple boolean indexing operation. The list comprehension method has been timed against (!=) and query. Here are the functions:

# Boolean indexing with Numeric value comparison.
df[df.A != df.B]                            # vectorized !=
df.query('A != B')                          # query (numexpr)
df[[x != y for x, y in zip(df.A, df.B)]]    # list comp

For simplicity, I have used the perfplot package to run all the timeit tests in this post. The timings for the operations above are below:

enter image description here

The list comprehension outperforms query for moderately sized N, and even outperforms the vectorized not equals comparison for tiny N. Unfortunately, the list comprehension scales linearly, so it does not offer much performance gain for larger N.

Note
It is worth mentioning that much of the benefit of list comprehension come from not having to worry about the index alignment, but this means that if your code is dependent on indexing alignment, this will break. In some cases, vectorised operations over the underlying NumPy arrays can be considered as bringing in the “best of both worlds”, allowing for vectorisation without all the unneeded overhead of the pandas functions. This means that you can rewrite the operation above as

df[df.A.values != df.B.values]

Which outperforms both the pandas and list comprehension equivalents:

NumPy vectorization is out of the scope of this post, but it is definitely worth considering, if performance matters.

Value Counts
Taking another example – this time, with another vanilla python construct that is faster than a for loop – . A common requirement is to compute the value counts and return the result as a dictionary. This is done with , , and Counter:

# Value Counts comparison.
ser.value_counts(sort=False).to_dict()           # value_counts
dict(zip(*np.unique(ser, return_counts=True)))   # np.unique
Counter(ser)                                     # Counter

enter image description here

The results are more pronounced, Counter wins out over both vectorized methods for a larger range of small N (~3500).

Note
More trivia (courtesy @user2357112). The Counter is implemented with a C accelerator, so while it still has to work with python objects instead of the underlying C datatypes, it is still faster than a for loop. Python power!

Of course, the take away from here is that the performance depends on your data and use case. The point of these examples is to convince you not to rule out these solutions as legitimate options. If these still don’t give you the performance you need, there is always cython and numba. Let’s add this test into the mix.

from numba import njit, prange

@njit(parallel=True)
def get_mask(x, y):
    result = [False] * len(x)
    for i in prange(len(x)):
        result[i] = x[i] != y[i]

    return np.array(result)

df[get_mask(df.A.values, df.B.values)] # numba

enter image description here

Numba offers JIT compilation of loopy python code to very powerful vectorized code. Understanding how to make numba work involves a learning curve.


Operations with Mixed/object dtypes

String-based Comparison
Revisiting the filtering example from the first section, what if the columns being compared are strings? Consider the same 3 functions above, but with the input DataFrame cast to string.

# Boolean indexing with string value comparison.
df[df.A != df.B]                            # vectorized !=
df.query('A != B')                          # query (numexpr)
df[[x != y for x, y in zip(df.A, df.B)]]    # list comp

enter image description here

So, what changed? The thing to note here is that string operations are inherently difficult to vectorize. Pandas treats strings as objects, and all operations on objects fall back to a slow, loopy implementation.

Now, because this loopy implementation is surrounded by all the overhead mentioned above, there is a constant magnitude difference between these solutions, even though they scale the same.

When it comes to operations on mutable/complex objects, there is no comparison. List comprehension outperforms all operations involving dicts and lists.

Accessing Dictionary Value(s) by Key
Here are timings for two operations that extract a value from a column of dictionaries: map and the list comprehension. The setup is in the Appendix, under the heading “Code Snippets”.

# Dictionary value extraction.
ser.map(operator.itemgetter('value'))     # map
pd.Series([x.get('value') for x in ser])  # list comprehension

enter image description here

Positional List Indexing
Timings for 3 operations that extract the 0th element from a list of columns (handling exceptions), map, str.get accessor method, and the list comprehension:

# List positional indexing. 
def get_0th(lst):
    try:
        return lst[0]
    # Handle empty lists and NaNs gracefully.
    except (IndexError, TypeError):
        return np.nan

ser.map(get_0th)                                          # map
ser.str[0]                                                # str accessor
pd.Series([x[0] if len(x) > 0 else np.nan for x in ser])  # list comp
pd.Series([get_0th(x) for x in ser])                      # list comp safe

Note
If the index matters, you would want to do:

pd.Series([...], index=ser.index)

When reconstructing the series.

enter image description here

List Flattening
A final example is flattening lists. This is another common problem, and demonstrates just how powerful pure python is here.

# Nested list flattening.
pd.DataFrame(ser.tolist()).stack().reset_index(drop=True)  # stack
pd.Series(list(chain.from_iterable(ser.tolist())))         # itertools.chain
pd.Series([y for x in ser for y in x])                     # nested list comp

enter image description here

Both and the nested list comprehension are pure python constructs, and scale much better than the stack solution.

These timings are a strong indication of the fact that pandas is not equipped to work with mixed dtypes, and that you should probably refrain from using it to do so. Wherever possible, data should be present as scalar values (ints/floats/strings) in separate columns.

Lastly, the applicability of these solutions depend widely on your data. So, the best thing to do would be to test these operations on your data before deciding what to go with. Notice how I have not timed apply on these solutions, because it would skew the graph (yes, it’s that slow).


Regex Operations, and .str Accessor Methods

Pandas can apply regex operations such as , , and , as well as other “vectorized” string operations (such as str.split, str.find,str.translate`, and so on) on string columns. These functions are slower than list comprehensions, and are meant to be more convenience functions than anything else.

It is usually much faster to pre-compile a regex pattern and iterate over your data with (also see Is it worth using Python’s re.compile?). The list comp equivalent to str.contains looks something like this:

p = re.compile(...)
ser2 = pd.Series([x for x in ser if p.search(x)])

Or,

ser2 = ser[[bool(p.search(x)) for x in ser]]

If you need to handle NaNs, you can do something like

ser[[bool(p.search(x)) if pd.notnull(x) else False for x in ser]]

The list comp equivalent to str.extract (without groups) will look something like:

df['col2'] = [p.search(x).group(0) for x in df['col']]

If you need to handle no-matches and NaNs, you can use a custom function (still faster!):

def matcher(x):
    m = p.search(str(x))
    if m:
        return m.group(0)
    return np.nan

df['col2'] = [matcher(x) for x in df['col']]

The matcher function is very extensible. It can be fitted to return a list for each capture group, as needed. Just extract query the group or groups attribute of the matcher object.

For str.extractall, change p.search to p.findall.

String Extraction
Consider a simple filtering operation. The idea is to extract 4 digits if it is preceded by an upper case letter.

# Extracting strings.
p = re.compile(r'(?<=[A-Z])(\d{4})')
def matcher(x):
    m = p.search(x)
    if m:
        return m.group(0)
    return np.nan

ser.str.extract(r'(?<=[A-Z])(\d{4})', expand=False)   #  str.extract
pd.Series([matcher(x) for x in ser])                  #  list comprehension

enter image description here

More Examples
Full disclosure – I am the author (in part or whole) of these posts listed below.


Conclusion

As shown from the examples above, iteration shines when working with small rows of DataFrames, mixed datatypes, and regular expressions.

The speedup you get depends on your data and your problem, so your mileage may vary. The best thing to do is to carefully run tests and see if the payout is worth the effort.

The “vectorized” functions shine in their simplicity and readability, so if performance is not critical, you should definitely prefer those.

Another side note, certain string operations deal with constraints that favour the use of NumPy. Here are two examples where careful NumPy vectorization outperforms python:

Additionally, sometimes just operating on the underlying arrays via .values as opposed to on the Series or DataFrames can offer a healthy enough speedup for most usual scenarios (see the Note in the Numeric Comparison section above). So, for example df[df.A.values != df.B.values] would show instant performance boosts over df[df.A != df.B]. Using .values may not be appropriate in every situation, but it is a useful hack to know.

As mentioned above, it’s up to you to decide whether these solutions are worth the trouble of implementing.


Appendix: Code Snippets

import perfplot  
import operator 
import pandas as pd
import numpy as np
import re

from collections import Counter
from itertools import chain

# Boolean indexing with Numeric value comparison.
perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(1000, (n, 2)), columns=['A','B']),
    kernels=[
        lambda df: df[df.A != df.B],
        lambda df: df.query('A != B'),
        lambda df: df[[x != y for x, y in zip(df.A, df.B)]],
        lambda df: df[get_mask(df.A.values, df.B.values)]
    ],
    labels=['vectorized !=', 'query (numexpr)', 'list comp', 'numba'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N'
)

# Value Counts comparison.
perfplot.show(
    setup=lambda n: pd.Series(np.random.choice(1000, n)),
    kernels=[
        lambda ser: ser.value_counts(sort=False).to_dict(),
        lambda ser: dict(zip(*np.unique(ser, return_counts=True))),
        lambda ser: Counter(ser),
    ],
    labels=['value_counts', 'np.unique', 'Counter'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=lambda x, y: dict(x) == dict(y)
)

# Boolean indexing with string value comparison.
perfplot.show(
    setup=lambda n: pd.DataFrame(np.random.choice(1000, (n, 2)), columns=['A','B'], dtype=str),
    kernels=[
        lambda df: df[df.A != df.B],
        lambda df: df.query('A != B'),
        lambda df: df[[x != y for x, y in zip(df.A, df.B)]],
    ],
    labels=['vectorized !=', 'query (numexpr)', 'list comp'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=None
)

# Dictionary value extraction.
ser1 = pd.Series([{'key': 'abc', 'value': 123}, {'key': 'xyz', 'value': 456}])
perfplot.show(
    setup=lambda n: pd.concat([ser1] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.map(operator.itemgetter('value')),
        lambda ser: pd.Series([x.get('value') for x in ser]),
    ],
    labels=['map', 'list comprehension'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=None
)

# List positional indexing. 
ser2 = pd.Series([['a', 'b', 'c'], [1, 2], []])        
perfplot.show(
    setup=lambda n: pd.concat([ser2] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.map(get_0th),
        lambda ser: ser.str[0],
        lambda ser: pd.Series([x[0] if len(x) > 0 else np.nan for x in ser]),
        lambda ser: pd.Series([get_0th(x) for x in ser]),
    ],
    labels=['map', 'str accessor', 'list comprehension', 'list comp safe'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=None
)

# Nested list flattening.
ser3 = pd.Series([['a', 'b', 'c'], ['d', 'e'], ['f', 'g']])
perfplot.show(
    setup=lambda n: pd.concat([ser2] * n, ignore_index=True),
    kernels=[
        lambda ser: pd.DataFrame(ser.tolist()).stack().reset_index(drop=True),
        lambda ser: pd.Series(list(chain.from_iterable(ser.tolist()))),
        lambda ser: pd.Series([y for x in ser for y in x]),
    ],
    labels=['stack', 'itertools.chain', 'nested list comp'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',    
    equality_check=None

)

# Extracting strings.
ser4 = pd.Series(['foo xyz', 'test A1234', 'D3345 xtz'])
perfplot.show(
    setup=lambda n: pd.concat([ser4] * n, ignore_index=True),
    kernels=[
        lambda ser: ser.str.extract(r'(?<=[A-Z])(\d{4})', expand=False),
        lambda ser: pd.Series([matcher(x) for x in ser])
    ],
    labels=['str.extract', 'list comprehension'],
    n_range=[2**k for k in range(0, 15)],
    xlabel='N',
    equality_check=None
)

回答 1

简而言之

  • for循环+ iterrows非常慢。大约1k行的开销并不重要,但超过10k行的开销却很明显。
  • for loop + itertuplesiterrowsor 快得多apply
  • 向量化通常比 itertuples

基准测试 在此处输入图片说明

In short

  • for loop + iterrows is extremely slow. Overhead isn’t significant on ~1k rows, but noticeable on 10k+ rows.
  • for loop + itertuples is much faster than iterrows or apply.
  • vectorization is usually much faster than itertuples

Benchmark enter image description here


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