Pandas中的一致表(Pearson每行对每行之间的相关性)

Question

使用下面的pandas数据框,取自dict的dict:

import numpy as np
import pandas as pd
from scipy.stats import pearsonr

NaN = np.nan


dd ={'A': {'A': '1', 'B': '2', 'C': '3'},
         'B': {'A': '4', 'B': '5', 'C': '6'},
         'C': {'A': '7', 'B': '8', 'C': '9'}}

df_link_link = pd.DataFrame.from_dict(dd, orient='index')

我想形成一个新的pandas DataFrame,每行的行之间有Pearson相关的结果,不包括相同行之间的Pearson相关性(将A与其自身相关联NaN.这是拼写为dicts的字典:

dict_congruent = {'A': {'A': NaN, 
                        'B': pearsonr([NaN,2,3],[4,5,6]), 
                        'C': pearsonr([NaN,2,3],[7,8,9])},
                  'B': {'A': pearsonr([4,NaN,6],[1,2,3]), 
                        'B': NaN, 
                        'C': pearsonr([4,NaN,6],[7,8,9])},
                  'C': {'A': pearsonr([7,8,NaN],[1,2,3]), 
                        'B': pearsonr([7,8,NaN],[4,5,6]), 
                        'C': NaN }}

这里NaN就是numpy.nan.有没有办法在pandas中执行此操作而无需迭代dicts的字典？我有~7600万对,所以非迭代方法会很好,如果存在的话.

Answer 1

规范但不可行的解决方案

df.corr().mask(np.equal.outer(df.index.values, df.columns.values))

默认方法corr是pearson.

TL; DR
转置你的数据使用它
用弓包裹起来

数据非常广泛

np.random.seed([3,1415])
m, n = 1000, 10000
df = pd.DataFrame(np.random.randn(m, n), columns=['s{}'.format(i) for i in range(n)])

神奇的功能

def corr(df, step=100, mask_diagonal=False):

    n = df.shape[0]

    def corr_closure(df):
        d = df.values
        sums = d.sum(0, keepdims=True)
        stds = d.std(0, keepdims=True)

        def corr_(k=0, l=10):
            d2 = d.T.dot(d[:, k:l])
            sums2 = sums.T.dot(sums[:, k:l])
            stds2 = stds.T.dot(stds[:, k:l])

            return pd.DataFrame((d2 - sums2 / n) / stds2 / n,
                                df.columns, df.columns[k:l])

        return corr_

    c = corr_closure(df)

    step = min(step, df.shape[1])

    tups = zip(range(0, n, step), range(step, n + step, step))

    corr_table = pd.concat([c(*t) for t in tups], axis=1)

    if mask_diagonal:
        np.fill_diagonal(corr_table.values, np.nan)

    return corr_table

示范

ct = corr(df, mask_diagonal=True)
ct.iloc[:10, :10]

魔术解决方案解释
逻辑:

• 使用闭包来预先计算列总和和标准偏差
• 返回一个函数,该函数获取要关联的列的位置

def corr_closure(df):
    d = df.values  # get underlying numpy array
    sums = d.sum(0, keepdims=True)  # pre calculate sums
    stds = d.std(0, keepdims=True)  # pre calculate standard deviations
    n = d.shape[0]  # grab number of rows

    def corr(k=0, l=10):
        d2 = d.T.dot(d[:, k:l])  # for this slice, run dot product
        sums2 = sums.T.dot(sums[:, k:l])  # dot pre computed sums with slice
        stds2 = stds.T.dot(stds[:, k:l])  # dot pre computed stds with slice

        # calculate correlations with the information I have
        return pd.DataFrame((d2 - sums2 / n) / stds2 / n,
                            df.columns, df.columns[k:l])

    return corr

计时
10列

100列

1000列

10000列
df.corr()没有在合理的时间内完成