摘要：问题描述：我知道如何在R中执行此操作。但是，pandas 中是否有任何函数可以将数据框转换为包含两个同时出现的方面计数的 nxn 共现矩阵。例如矩阵 df：import pandas as pd df = pd.DataFrame({'TFD' : ['AA', 'SL', 'BB', 'D0', 'Dk'...

问题描述：

我知道如何在R中执行此操作。但是，pandas 中是否有任何函数可以将数据框转换为包含两个同时出现的方面计数的 nxn 共现矩阵。

例如矩阵 df：

import pandas as pd

df = pd.DataFrame({'TFD' : ['AA', 'SL', 'BB', 'D0', 'Dk', 'FF'],
                    'Snack' : ['1', '0', '1', '1', '0', '0'],
                    'Trans' : ['1', '1', '1', '0', '0', '1'],
                    'Dop' : ['1', '0', '1', '0', '1', '1']}).set_index('TFD')

print df

>>> 
    Dop Snack Trans
TFD                
AA    1     1     1
SL    0     0     1
BB    1     1     1
D0    0     1     0
Dk    1     0     0
FF    1     0     1

[6 rows x 3 columns]

得出的结果为：

    Dop Snack Trans

Dop   0     2     3
Snack 2     0     2
Trans 3     2     0

由于矩阵在对角线上是镜像的，我猜想应该有一种方法可以优化代码。

解决方案 1：

这是一个简单的线性代数，你将矩阵与其转置相乘（您的示例包含字符串，请不要忘记将它们转换为整数）：

>>> df_asint = df.astype(int)
>>> coocc = df_asint.T.dot(df_asint)
>>> coocc
       Dop  Snack  Trans
Dop      4      2      3
Snack    2      3      2
Trans    3      2      4

如果如 R 答案中所述，您想要重置对角线，则可以使用 numpy 的fill_diagonal：

>>> import numpy as np
>>> np.fill_diagonal(coocc.values, 0)
>>> coocc
       Dop  Snack  Trans
Dop      0      2      3
Snack    2      0      2
Trans    3      2      0

解决方案 2：

NumPy 中的演示：

import numpy as np
np.random.seed(3) # for reproducibility

# Generate data: 5 labels, 10 examples, binary.
label_headers = 'Alice Bob Carol Dave Eve'.split(' ')
label_data = np.random.randint(0,2,(10,5)) # binary here but could be any integer.
print('labels:
{0}'.format(label_data))

# Compute cooccurrence matrix 
cooccurrence_matrix = np.dot(label_data.transpose(),label_data)
print('
cooccurrence_matrix:
{0}'.format(cooccurrence_matrix)) 

# Compute cooccurrence matrix in percentage
# FYI: http://stackoverflow.com/questions/19602187/numpy-divide-each-row-by-a-vector-element
#      http://stackoverflow.com/questions/26248654/numpy-return-0-with-divide-by-zero/32106804#32106804
cooccurrence_matrix_diagonal = np.diagonal(cooccurrence_matrix)
with np.errstate(divide='ignore', invalid='ignore'):
    cooccurrence_matrix_percentage = np.nan_to_num(np.true_divide(cooccurrence_matrix, cooccurrence_matrix_diagonal[:, None]))
print('
cooccurrence_matrix_percentage:
{0}'.format(cooccurrence_matrix_percentage))

输出：

labels:
[[0 0 1 1 0]
 [0 0 1 1 1]
 [0 1 1 1 0]
 [1 1 0 0 0]
 [0 1 1 0 0]
 [0 1 0 0 0]
 [0 1 0 1 1]
 [0 1 0 0 1]
 [1 0 0 1 0]
 [1 0 1 1 1]]

cooccurrence_matrix:
[[3 1 1 2 1]
 [1 6 2 2 2]
 [1 2 5 4 2]
 [2 2 4 6 3]
 [1 2 2 3 4]]

cooccurrence_matrix_percentage:
[[ 1.          0.33333333  0.33333333  0.66666667  0.33333333]
 [ 0.16666667  1.          0.33333333  0.33333333  0.33333333]
 [ 0.2         0.4         1.          0.8         0.4       ]
 [ 0.33333333  0.33333333  0.66666667  1.          0.5       ]
 [ 0.25        0.5         0.5         0.75        1.        ]]

使用 matplotlib 的热图：

import numpy as np
np.random.seed(3) # for reproducibility

import matplotlib.pyplot as plt


def show_values(pc, fmt="%.2f", **kw):
    '''
    Heatmap with text in each cell with matplotlib's pyplot
    Source: http://stackoverflow.com/a/25074150/395857 
    By HYRY
    '''
    from itertools import izip
    pc.update_scalarmappable()
    ax = pc.get_axes()
    for p, color, value in izip(pc.get_paths(), pc.get_facecolors(), pc.get_array()):
        x, y = p.vertices[:-2, :].mean(0)
        if np.all(color[:3] > 0.5):
            color = (0.0, 0.0, 0.0)
        else:
            color = (1.0, 1.0, 1.0)
        ax.text(x, y, fmt % value, ha="center", va="center", color=color, **kw)

def cm2inch(*tupl):
    '''
    Specify figure size in centimeter in matplotlib
    Source: http://stackoverflow.com/a/22787457/395857
    By gns-ank
    '''
    inch = 2.54
    if type(tupl[0]) == tuple:
        return tuple(i/inch for i in tupl[0])
    else:
        return tuple(i/inch for i in tupl)

def heatmap(AUC, title, xlabel, ylabel, xticklabels, yticklabels):
    '''
    Inspired by:
    - http://stackoverflow.com/a/16124677/395857 
    - http://stackoverflow.com/a/25074150/395857
    '''

    # Plot it out
    fig, ax = plt.subplots()    
    c = ax.pcolor(AUC, edgecolors='k', linestyle= 'dashed', linewidths=0.2, cmap='RdBu', vmin=0.0, vmax=1.0)

    # put the major ticks at the middle of each cell
    ax.set_yticks(np.arange(AUC.shape[0]) + 0.5, minor=False)
    ax.set_xticks(np.arange(AUC.shape[1]) + 0.5, minor=False)

    # set tick labels
    #ax.set_xticklabels(np.arange(1,AUC.shape[1]+1), minor=False)
    ax.set_xticklabels(xticklabels, minor=False)
    ax.set_yticklabels(yticklabels, minor=False)

    # set title and x/y labels
    plt.title(title)
    plt.xlabel(xlabel)
    plt.ylabel(ylabel)      

    # Remove last blank column
    plt.xlim( (0, AUC.shape[1]) )

    # Turn off all the ticks
    ax = plt.gca()    
    for t in ax.xaxis.get_major_ticks():
        t.tick1On = False
        t.tick2On = False
    for t in ax.yaxis.get_major_ticks():
        t.tick1On = False
        t.tick2On = False

    # Add color bar
    plt.colorbar(c)

    # Add text in each cell 
    show_values(c)

    # Proper orientation (origin at the top left instead of bottom left)
    ax.invert_yaxis()
    ax.xaxis.tick_top()

    # resize 
    fig = plt.gcf()
    fig.set_size_inches(cm2inch(40, 20))



def main():

    # Generate data: 5 labels, 10 examples, binary.
    label_headers = 'Alice Bob Carol Dave Eve'.split(' ')
    label_data = np.random.randint(0,2,(10,5)) # binary here but could be any integer.
    print('labels:
{0}'.format(label_data))

    # Compute cooccurrence matrix 
    cooccurrence_matrix = np.dot(label_data.transpose(),label_data)
    print('
cooccurrence_matrix:
{0}'.format(cooccurrence_matrix)) 

    # Compute cooccurrence matrix in percentage
    # FYI: http://stackoverflow.com/questions/19602187/numpy-divide-each-row-by-a-vector-element
    #      http://stackoverflow.com/questions/26248654/numpy-return-0-with-divide-by-zero/32106804#32106804
    cooccurrence_matrix_diagonal = np.diagonal(cooccurrence_matrix)
    with np.errstate(divide='ignore', invalid='ignore'):
        cooccurrence_matrix_percentage = np.nan_to_num(np.true_divide(cooccurrence_matrix, cooccurrence_matrix_diagonal[:, None]))
    print('
cooccurrence_matrix_percentage:
{0}'.format(cooccurrence_matrix_percentage))

    # Add count in labels
    label_header_with_count = [ '{0} ({1})'.format(label_header, cooccurrence_matrix_diagonal[label_number]) for label_number, label_header in enumerate(label_headers)]  
    print('
label_header_with_count: {0}'.format(label_header_with_count))

    # Plotting
    x_axis_size = cooccurrence_matrix_percentage.shape[0]
    y_axis_size = cooccurrence_matrix_percentage.shape[1]
    title = "Co-occurrence matrix
"
    xlabel= ''#"Labels"
    ylabel= ''#"Labels"
    xticklabels = label_header_with_count
    yticklabels = label_header_with_count
    heatmap(cooccurrence_matrix_percentage, title, xlabel, ylabel, xticklabels, yticklabels)
    plt.savefig('image_output.png', dpi=300, format='png', bbox_inches='tight') # use format='svg' or 'pdf' for vectorial pictures
    #plt.show()


if __name__ == "__main__":
    main()
    #cProfile.run('main()') # if you want to do some profiling

在此处输入图片描述

（PS：D3.js 中共现矩阵的简洁可视化。）

解决方案 3：

如果您有更大的语料库和词频矩阵，使用稀疏矩阵乘法可能会更有效。我使用algo与本页答案中相同的矩阵乘法技巧。

import scipy.sparse as sp
X = sp.csr_matrix(df.astype(int).values) # convert dataframe to sparse matrix
Xc = X.T * X # multiply sparse matrix # 
Xc.setdiag(0) # reset diagonal
print(Xc.todense()) # to print co-occurence matrix in dense format

Xc以下是稀疏 csr 格式的共生矩阵

解决方案 4：

为了进一步阐述这个问题，如果您想从句子构建共现矩阵，您可以这样做：

import numpy as np
import pandas as pd

def create_cooccurrence_matrix(sentences, window_size=2):
    """Create co occurrence matrix from given list of sentences.

    Returns:
    - vocabs: dictionary of word counts
    - co_occ_matrix_sparse: sparse co occurrence matrix

    Example:
    ===========
    sentences = ['I love nlp',    'I love to learn',
                 'nlp is future', 'nlp is cool']

    vocabs,co_occ = create_cooccurrence_matrix(sentences)

    df_co_occ  = pd.DataFrame(co_occ.todense(),
                              index=vocabs.keys(),
                              columns = vocabs.keys())

    df_co_occ = df_co_occ.sort_index()[sorted(vocabs.keys())]

    df_co_occ.style.applymap(lambda x: 'color: red' if x>0 else '')

    """
    import scipy
    import nltk

    vocabulary = {}
    data = []
    row = []
    col = []

    tokenizer = nltk.tokenize.word_tokenize

    for sentence in sentences:
        sentence = sentence.strip()
        tokens = [token for token in tokenizer(sentence) if token != u""]
        for pos, token in enumerate(tokens):
            i = vocabulary.setdefault(token, len(vocabulary))
            start = max(0, pos-window_size)
            end = min(len(tokens), pos+window_size+1)
            for pos2 in range(start, end):
                if pos2 == pos:
                    continue
                j = vocabulary.setdefault(tokens[pos2], len(vocabulary))
                data.append(1.)
                row.append(i)
                col.append(j)

    cooccurrence_matrix_sparse = scipy.sparse.coo_matrix((data, (row, col)))
    return vocabulary, cooccurrence_matrix_sparse

用法：

sentences = ['I love nlp',    'I love to learn',
             'nlp is future', 'nlp is cool']

vocabs,co_occ = create_cooccurrence_matrix(sentences)

df_co_occ  = pd.DataFrame(co_occ.todense(),
                          index=vocabs.keys(),
                          columns = vocabs.keys())

df_co_occ = df_co_occ.sort_index()[sorted(vocabs.keys())]

df_co_occ.style.applymap(lambda x: 'color: red' if x>0 else '')

# If not in jupyter notebook, print(df_co_occ)

输出