python - 计算按另一列值分组的列值在 Pandas 数据框中的共现

Question

题
我在 Python 3.7.7 上使用 Pandas。我想计算变量的分类值之间的互信息x按另一个变量的值分组 y .我的数据如下表所示:

+-----+-----+
|  x  |  y  |
+-----+-----+
| x_1 | y_1 |
| x_2 | y_1 |
| x_3 | y_1 |
| x_1 | y_2 |
| x_2 | y_2 |
| x_4 | y_3 |
| x_6 | y_3 |
| x_9 | y_3 |
| x_1 | y_4 |
| ... | ... |
+-----+-----+

我想要一个数据结构(pandas MultiIndex 系列/数据帧或 numpy 矩阵或任何合适的)来存储 的数量( x_i , x_j ) 对的同时出现给定 y_k值 .事实上，这会很好，例如，轻松计算 PMI :

+-----+-----+--------+-------+
| x_i | x_j |  cooc  |  pmi  |
+-----+-----+--------+-------+
| x_1 | x_2 |        |       |
| x_1 | x_3 |        |       |
| x_1 | x_4 |        |       |
| x_1 | x_5 |        |       |
| ... | ... |   ...  |  ...  |
+-----+-----+--------+-------+

有没有合适的内存有效方法？
边注 :我正在使用相当大的数据(40k 个不同的 x 值和 8k 个不同的 y 值，总共有 300k 个( x ， y )条目，因此内存友好和优化的方法会很棒(可能依赖在第三方库中为 Dask )
更新
非优化方案
我想出了一个使用 pd.crosstab 的解决方案.我在这里提供一个小例子:

import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(0,100,size=(100, 2)), columns=list('xy'))
"""
df:
  +-----+-----+
  |  x  |  y  |
  +-----+-----+
  | 4   | 99  |
  | 1   | 39  |
  | 39  | 56  |
  | ..  | ..  |
  | 59  | 20  |
  | 82  | 57  |
  +-----+-----+
 100 rows × 2 columns
"""
# Compute cross tabulation:
crosstab = pd.crosstab(df["x"], df["y"])
"""
crosstab:
  +------+-----+-----+-----+-----+
  |  y   |  0  |  2  |  3  | ... |
  |  x   +-----+-----+-----+-----+
  |  1   |  0  |  0  |  0  | ... |
  |  2   |  0  |  0  |  0  | ... |
  | ...  | ... | ... | ... | ... |
  +------+-----+-----+-----+-----+
 62 rows × 69 columns
"""
# Initialize a pandas MultiIndex Series storing PMI values
import itertools
x_pairs = list(itertools.combinations(crosstab.index, 2))
pmi = pd.Series(0, index = pd.MultiIndex.from_tuples(x_pairs))
"""
pmi:
  +-------------+-----+
  |    index    | val |
  +------+------|     |
  |  x_i |  x_j |     |
  +------+------+-----+
  |  1   |  2   |  0  |
  |      |  4   |  0  |
  |  ... |  ... | ... |
  |  95  |  98  |  0  |
  |      |  99  |  0  |
  |  96  |  98  |  0  |
  +------+------+-----+
 Length: 1891, dtype: int64
"""

然后，我用来填充系列的循环结构如下:

for x1, x2 in x_pairs:
    pmi.loc[x1, x2] = crosstab.loc[[x1, x2]].min().sum() / (crosstab.loc[x1].sum() * crosstab.loc[x2].sum())

这不是一个可选的解决方案，即使在小用例中也表现不佳。

Answer 1

优化方案
最后，我设法使用 scipy 稀疏矩阵进行中间计算，以内存友好的方式计算交叉出现:

import pandas as pd
import numpy as np
from scipy.sparse import csr_matrix

def df_compute_cooccurrences(df: pd.DataFrame, column1: str, column2: str) -> pd.DataFrame:
   
    # pd.factorize encode the object as an enumerated type or categorical variable, returning:
    # - `codes` (ndarray): an integer ndarray that’s an indexer into `uniques`.
    # - `uniques` (ndarray, Index, or Categorical): the unique valid values
    # see more at https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.factorize.html

    i, rows = pd.factorize(df[column1])
    # i    -> array([     0,      0,      0, ..., 449054,      0,      1])
    # rows -> Index(['column1_label1', 'column1_label2', ...])

    j, cols = pd.factorize(df[column2])
    # j    -> array([    0,     1,     2, ..., 28544,    -1,    -1])
    # cols -> Float64Index([column2_label1, column2_label2, ...])

    ij, tups = pd.factorize(list(zip(i, j)))
    # ij   -> array([      0,       1,       2, ..., 2878026, 2878027, 2878028])
    # tups -> array([(0, 0), (0, 1), (0, 2), ..., (449054, 28544), (0, -1), (1, -1)]

    # Then we can finally compute the crosstabulation matrix
    crosstab = csr_matrix((np.bincount(ij), tuple(zip(*tups))))
    # If we convert directly this into a Dataframe with
    # pd.DataFrame.sparse.from_spmatrix(crosstab, rows, cols)
    # we have the same result as using 
    # https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.crosstab.html
    # but we obtained it in a memory-friendly way (allowing big data processing)

    # In order to obtain the co-occurrences matrix for column 1, 
    # we have to multiply the crosstab matrix for its transposed 
    coocc = crosstab.dot(crosstab.transpose())
    
    # Then we can finally return the co-occurence matrix in in a DataFrame form
    return pd.DataFrame.sparse.from_spmatrix(coocc, rows, rows)

这里提供了一个小例子:

import pandas as pd
import numpy as np
from scipy.sparse import csr_matrix

def df_compute_cooccurrences(df: pd.DataFrame, column1: str, column2: str) -> pd.DataFrame:
    i, rows = pd.factorize(df[column1])
    j, cols = pd.factorize(df[column2])
    ij, tups = pd.factorize(list(zip(i, j)))
    crosstab = csr_matrix((np.bincount(ij), tuple(zip(*tups))))
    coocc = crosstab.dot(crosstab.transpose())
    return pd.DataFrame.sparse.from_spmatrix(coocc, rows, rows)

df = pd.DataFrame(zip([1,1,1,2,2,3,4],["a","a","a","a","a","b","b"]), columns=list('xy'))
"""
df:
  +-----+-----+
  ¦  x  ¦  y  ¦
  +-----+-----+
  |  1  |  a  |
  |  1  |  a  |
  |  1  |  a  |
  |  2  |  a  |
  |  2  |  a  |
  |  3  |  b  |
  |  4  |  b  |
  +-----+-----+
"""
cooc_df = df_compute_cooccurrences(df, "x", "y")
"""
cooc_df:
    +---+---+---+---+
    ¦ 1 | 2 | 3 | 4 |
+---+---+---+---+---+
¦ 1 ¦ 9 | 6 | 0 | 0 |
¦ 2 ¦ 6 | 4 | 0 | 0 |
¦ 3 ¦ 0 | 0 | 1 | 1 |
¦ 4 ¦ 0 | 0 | 1 | 1 |
+---+---+---+---+---+
"""
cooc_df2 = df_compute_cooccurrences(df, "y", "x")
"""
cooc_df2:
    +----+----+
    ¦  a ¦  b ¦
+---+----+----+
¦ a ¦ 13 |  0 |
¦ b ¦  0 |  2 |
+---+----+----+
"""