如何将集群/类/组标签的数据帧转换成成对/非成对的数据帧？

Question

我试图找到聚类结果之间的一致性，但我很难有效地做到这一点。我想转换pandasDataFrame对象（或dict），它是i=node、j=iteration和[i,j]=cluster/group。我目前的方法是迭代所有的可能性，但我觉得有一个更有效的方法来做到这一点。对于大型数据集来说，这将花费很长时间

import string
import pandas as pd
import numpy as np
from collections import *

# Get alphabet as nodes
nodes = list(string.ascii_lowercase)

data = {0: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}, 1: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}, 2: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}, 3: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}, 4: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}, 5: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}, 6: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}, 7: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}, 8: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}, 9: {'a': 0, 'b': 0, 'c': 0, 'd': 1, 'e': 0, 'f': 0, 'g': 2, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 3, 'n': 0, 'o': 4, 'p': 0, 'q': 0, 'r': 3, 's': 0, 't': 0, 'u': 0, 'v': 4, 'w': 2, 'x': 2, 'y': 1, 'z': 0}} 
df_clusters = pd.DataFrame(data)

如何才能更有效地完成这一部分，而不是使用bruteforce迭代？有没有办法利用NumPy数组来实现这一点？

# Get pairs of nodes and determine if they are in the same cluster/community/group
d_pair_iteration = defaultdict(dict)
for iteration, communities in df_clusters.T.iterrows():
    # Iterate pairwise
    for i in range(len(nodes)):
        # Node A
        node_a = nodes[i]
        for j in range(i+1, len(nodes)):
            # Node B
            node_b = nodes[j]
            # Determine if they are in the same community
            d_pair_iteration[frozenset([node_a, node_b])][iteration] = int(communities[node_a] == communities[node_b])

# Create dataframe
df_pairs = pd.DataFrame(d_pair_iteration).T

Answer 1

使用numpy广播，我们可以将行a与整个数据帧进行比较，然后将b与整个数据帧进行比较，依此类推：

# `x` is a table of 26 rows and 10 columns
x = df_clusters.values

# `y` is an array of 26 tables, each having 1 row and 10 columns
y = x[:, None]

# Using numpy broadcasting, `z` contains the result of comparing each
# table in `y` against `x`. So the shape of `z` is 26 x 26 x 10
z = x == y

# Reshaping `z` by merging the first two dimensions
data = z.reshape((z.shape[0] * z.shape[1], z.shape[2])).astype('int')

# idx is the 2-permutation of values in `df_clusters.index`:
# (a,a), (a,b), ..., (a,z), (b,a), (b,b), ...
idx = pd.MultiIndex.from_product([df_clusters.index, df_clusters.index], names=['node1', 'node2'])
result = pd.DataFrame(data, index=idx, columns=df_clusters.columns)

# We don't want all permutations, only the unique combinations,
# so we have to slice the frame
from itertools import combinations
final_idx = list(combinations(df_clusters.index, 2))
result = result.loc[final_idx]

结果是一个325x10的dataframe，因为C（26，2）=325

             0  1  2  3  4  5  6  7  8  9
node1 node2                              
a     b      1  1  1  1  1  1  1  1  1  1
      c      1  1  1  1  1  1  1  1  1  1
      d      0  0  0  0  0  0  0  0  0  0
      e      1  1  1  1  1  1  1  1  1  1
      f      1  1  1  1  1  1  1  1  1  1