在 Pandas DataFrame 中反转列顺序的大 O 复杂度是多少？

Question

假设我在 pandas 中有一个具有 m 行和 n 列的 DataFrame。还假设我想反转列的顺序，可以使用以下代码完成：

df_reversed = df[df.columns[::-1]]

这个操作的大 O 复杂度是多少？我假设这将取决于列数，但它也取决于行数吗？

Answer 1

我不知道 Pandas 是如何实现这一点的，但我确实根据经验对其进行了测试。我运行了以下代码（在 Jupyter 笔记本中）来测试操作的速度：

def get_dummy_df(n):
    return pd.DataFrame({'a': [1,2]*n, 'b': [4,5]*n, 'c': [7,8]*n})

df = get_dummy_df(100)
print df.shape
%timeit df_r = df[df.columns[::-1]]

df = get_dummy_df(1000)
print df.shape
%timeit df_r = df[df.columns[::-1]]

df = get_dummy_df(10000)
print df.shape
%timeit df_r = df[df.columns[::-1]]

df = get_dummy_df(100000)
print df.shape
%timeit df_r = df[df.columns[::-1]]

df = get_dummy_df(1000000)
print df.shape
%timeit df_r = df[df.columns[::-1]]

df = get_dummy_df(10000000)
print df.shape
%timeit df_r = df[df.columns[::-1]]

输出是：

(200, 3)
1000 loops, best of 3: 419 µs per loop
(2000, 3)
1000 loops, best of 3: 425 µs per loop
(20000, 3)
1000 loops, best of 3: 498 µs per loop
(200000, 3)
100 loops, best of 3: 2.66 ms per loop
(2000000, 3)
10 loops, best of 3: 25.2 ms per loop
(20000000, 3)
1 loop, best of 3: 207 ms per loop

如您所见，在前 3 种情况下，操作的开销是大部分时间（400-500μs）所花费的，但从第 4 种情况开始，它所花费的时间开始与数据，每次增加一个数量级。

所以，假设对n也一定有比例，看来我们处理的是O(m*n)

Answer 2

Big O 复杂度（截至 Pandas 0.24）为 m*n，其中 m 是列数，n 是行数。请注意，这是在使用 DataFrame.__getitem__ 方法（又名 []）和 Index（see relevant code, with other types that would trigger a copy）时。

这是一个有用的堆栈跟踪：

 <ipython-input-4-3162cae03863>(2)<module>()
      1 columns = df.columns[::-1]
----> 2 df_reversed = df[columns]

  pandas/core/frame.py(2682)__getitem__()
   2681             # either boolean or fancy integer index
-> 2682             return self._getitem_array(key)
   2683         elif isinstance(key, DataFrame):

  pandas/core/frame.py(2727)_getitem_array()
   2726             indexer = self.loc._convert_to_indexer(key, axis=1)
-> 2727             return self._take(indexer, axis=1)
   2728 

  pandas/core/generic.py(2789)_take()
   2788                                    axis=self._get_block_manager_axis(axis),
-> 2789                                    verify=True)
   2790         result = self._constructor(new_data).__finalize__(self)

  pandas/core/internals.py(4539)take()
   4538         return self.reindex_indexer(new_axis=new_labels, indexer=indexer,
-> 4539                                     axis=axis, allow_dups=True)
   4540 

  pandas/core/internals.py(4421)reindex_indexer()
   4420             new_blocks = self._slice_take_blocks_ax0(indexer,
-> 4421                                                      fill_tuple=(fill_value,))
   4422         else:

  pandas/core/internals.py(1254)take_nd()
   1253             new_values = algos.take_nd(values, indexer, axis=axis,
-> 1254                                        allow_fill=False)
   1255         else:

> pandas/core/algorithms.py(1658)take_nd()
   1657     import ipdb; ipdb.set_trace()
-> 1658     func = _get_take_nd_function(arr.ndim, arr.dtype, out.dtype, axis=axis,
   1659                                  mask_info=mask_info)
   1660     func(arr, indexer, out, fill_value)

pandas/core/algorithms 中 L1660 上的 func 调用最终调用了具有O(m * n) 复杂性的 cython 函数。这是将原始数据中的数据复制到out 的位置。 out 包含以相反顺序排列的原始数据的副本。

    inner_take_2d_axis0_template = """\
    cdef:
        Py_ssize_t i, j, k, n, idx
        %(c_type_out)s fv

    n = len(indexer)
    k = values.shape[1]

    fv = fill_value

    IF %(can_copy)s:
        cdef:
            %(c_type_out)s *v
            %(c_type_out)s *o

        #GH3130
        if (values.strides[1] == out.strides[1] and
            values.strides[1] == sizeof(%(c_type_out)s) and
            sizeof(%(c_type_out)s) * n >= 256):

            for i from 0 <= i < n:
                idx = indexer[i]
                if idx == -1:
                    for j from 0 <= j < k:
                        out[i, j] = fv
                else:
                    v = &values[idx, 0]
                    o = &out[i, 0]
                    memmove(o, v, <size_t>(sizeof(%(c_type_out)s) * k))
            return

    for i from 0 <= i < n:
        idx = indexer[i]
        if idx == -1:
            for j from 0 <= j < k:
                out[i, j] = fv
        else:
            for j from 0 <= j < k:
                out[i, j] = %(preval)svalues[idx, j]%(postval)s
"""

请注意，在上面的模板函数中，有一个使用memmove的路径（这是本例中采用的路径，因为我们正在从int64映射到int64，并且输出的维度与我们只是在切换索引）。请注意，memmove is still O(n) 与它必须复制的字节数成正比，尽管可能比直接写入索引要快。

Answer 3

我使用big_O拟合库here进行了实证测试

注意：所有测试均在自变量扫描 6 个数量级（即

• rows 从 10 到 10^6 与 3 的常量 column 大小，
• columns 从 10 到 10^6 与常量 row 大小为 10

结果显示columns逆向运算.columns[::-1]在DataFrame中的复杂度为

1. 立方：O(n^3) 其中 n 是 rows 的数量
2. 立方：O(n^3) 其中 n 是 columns 的数量

先决条件：您需要使用终端命令pip install big_o安装big_o()

代码

import big_o
import pandas as pd
import numpy as np

SWEAP_LOG10 = 6
COLUMNS = 3
ROWS = 10

def build_df(rows, columns):
    # To isolated the creation of the DataFrame from the inversion operation.
    narray = np.zeros(rows*columns).reshape(rows, columns)
    df = pd.DataFrame(narray)
    return df

def flip_columns(df):
    return df[df.columns[::-1]]

def get_row_df(n, m=COLUMNS):
    return build_df(1*10**n, m)

def get_column_df(n, m=ROWS):
    return build_df(m, 1*10**n)


# infer the big_o on columns[::-1] operation vs. rows
best, others = big_o.big_o(flip_columns, get_row_df, min_n=1, max_n=SWEAP_LOG10,n_measures=SWEAP_LOG10, n_repeats=10)

# print results
print('Measuring .columns[::-1] complexity against rapid increase in # rows')
print('-'*80 + '\nBig O() fits: {}\n'.format(best) + '-'*80)

for class_, residual in others.items():
    print('{:<60s}  (res: {:.2G})'.format(str(class_), residual))

print('-'*80)

# infer the big_o on columns[::-1] operation vs. columns
best, others = big_o.big_o(flip_columns, get_column_df, min_n=1, max_n=SWEAP_LOG10,n_measures=SWEAP_LOG10, n_repeats=10)

# print results
print()
print('Measuring .columns[::-1] complexity against rapid increase in # columns')
print('-'*80 + '\nBig O() fits: {}\n'.format(best) + '-'*80)

for class_, residual in others.items():
    print('{:<60s}  (res: {:.2G})'.format(str(class_), residual))
    
print('-'*80)

结果

Measuring .columns[::-1] complexity against rapid increase in # rows
--------------------------------------------------------------------------------
Big O() fits: Cubic: time = -0.017 + 0.00067*n^3
--------------------------------------------------------------------------------
Constant: time = 0.032                                        (res: 0.021)
Linear: time = -0.051 + 0.024*n                               (res: 0.011)
Quadratic: time = -0.026 + 0.0038*n^2                         (res: 0.0077)
Cubic: time = -0.017 + 0.00067*n^3                            (res: 0.0052)
Polynomial: time = -6.3 * x^1.5                               (res: 6)
Logarithmic: time = -0.026 + 0.053*log(n)                     (res: 0.015)
Linearithmic: time = -0.024 + 0.012*n*log(n)                  (res: 0.0094)
Exponential: time = -7 * 0.66^n                               (res: 3.6)
--------------------------------------------------------------------------------


Measuring .columns[::-1] complexity against rapid increase in # columns
--------------------------------------------------------------------------------
Big O() fits: Cubic: time = -0.28 + 0.009*n^3
--------------------------------------------------------------------------------
Constant: time = 0.38                                         (res: 3.9)
Linear: time = -0.73 + 0.32*n                                 (res: 2.1)
Quadratic: time = -0.4 + 0.052*n^2                            (res: 1.5)
Cubic: time = -0.28 + 0.009*n^3                               (res: 1.1)
Polynomial: time = -6 * x^2.2                                 (res: 16)
Logarithmic: time = -0.39 + 0.71*log(n)                       (res: 2.8)
Linearithmic: time = -0.38 + 0.16*n*log(n)                    (res: 1.8)
Exponential: time = -7 * 1^n                                  (res: 9.7)
--------------------------------------------------------------------------------