输入：

Question

所以我试图做一些与这两个帖子非常相似的事情但是有一些差异。一，我不想要一个csv文件所以没有csv模块，我想用Python而不是R来完成它。

Convert Adjacency Matrix into Edgelist (csv file) for Cytoscape

输入：

AF001         AF002     AF003   AF004  AF005
AF001  1.000000  0.000000e+00  0.000000  0.0000      0
AF002  0.374449  1.000000e+00  0.000000  0.0000      0
AF003  0.000347  1.173926e-05  1.000000  0.0000      0
AF004  0.001030  1.494282e-07  0.174526  1.0000      0
AF005  0.001183  1.216664e-06  0.238497  0.7557      1

输出：

{('AF002', 'AF003'): 1.17392596672424e-05, ('AF004', 'AF005'): 0.75570008659397792, ('AF001', 'AF002'): 0.374449352805868, ('AF001', 'AF003'): 0.00034743953114899502, ('AF002', 'AF005'): 1.2166642639889999e-06, ('AF002', 'AF004'): 1.49428208843456e-07, ('AF003', 'AF004'): 0.17452569907144502, ('AF001', 'AF004'): 0.00103026903356954, ('AF003', 'AF005'): 0.238497202355299, ('AF001', 'AF005'): 0.0011830950375467401}

我有一个非冗余相关矩阵DF_sCorr，它是使用np.tril（由@jezrael礼貌提供的代码）从冗余矩阵处理的。

我想把它折叠成一个字典，其中密钥是样本的排序元组{即key=tuple(sorted([row_sample,col_sample])}及其价值。我在sif_format下面编写了一个示例函数，它生成一个类似于sif格式的字典（格式为sample_x interaction_value sample_y的3列表），但这需要很长时间。

我认为组织这种类型表的最佳方法是字典。我觉得有一种更有效的方法可以做到这一点。可能只处理布尔值？我使用的真实数据集是~7000x7000

我不确定numpy，pandas，scipy或networkx中是否有能够有效执行此类处理的功能。

import pandas as pd
import numpy as np

A_sCorr = np.array([[0.999999999999999, 0.0, 0.0, 0.0, 0.0], [0.374449352805868, 1.0, 0.0, 0.0, 0.0], [0.00034743953114899502, 1.17392596672424e-05, 1.0, 0.0, 0.0], [0.00103026903356954, 1.49428208843456e-07, 0.17452569907144502, 1.0, 0.0], [0.0011830950375467401, 1.2166642639889999e-06, 0.238497202355299, 0.75570008659397792, 1.0]])
sampleLabels = ['AF001', 'AF002', 'AF003', 'AF004', 'AF005']

DF_sCorr = pd.DataFrame(A_sCorr,columns=sampleLabels, index=sampleLabels)

#AF001         AF002     AF003   AF004  AF005
#AF001  1.000000  0.000000e+00  0.000000  0.0000      0
#AF002  0.374449  1.000000e+00  0.000000  0.0000      0
#AF003  0.000347  1.173926e-05  1.000000  0.0000      0
#AF004  0.001030  1.494282e-07  0.174526  1.0000      0
#AF005  0.001183  1.216664e-06  0.238497  0.7557      1

def sif_format(DF_var):
    D_interaction_corr = {}
    n,m = DF_var.shape 
    for i in range(n):
        row_sample = DF_var.index[i]
        for j in range(m):
            col_sample = DF_var.columns[j]
            if row_sample != col_sample:
                D_interaction_corr[tuple(sorted([row_sample,col_sample]))] = DF_var.iloc[i,j]
            if j==i:
                break
    return(D_interaction_corr)

D_interaction_corr = sif_format(DF_sCorr)  
{('AF002', 'AF003'): 1.17392596672424e-05, ('AF004', 'AF005'): 0.75570008659397792, ('AF001', 'AF002'): 0.374449352805868, ('AF001', 'AF003'): 0.00034743953114899502, ('AF002', 'AF005'): 1.2166642639889999e-06, ('AF002', 'AF004'): 1.49428208843456e-07, ('AF003', 'AF004'): 0.17452569907144502, ('AF001', 'AF004'): 0.00103026903356954, ('AF003', 'AF005'): 0.238497202355299, ('AF001', 'AF005'): 0.0011830950375467401}

DataFrame.to_dict（）不会为此

工作

DF_sCorr.to_dict()
{'AF002': {'AF002': 1.0, 'AF003': 1.17392596672424e-05, 'AF001': 0.0, 'AF004': 1.49428208843456e-07, 'AF005': 1.2166642639889999e-06}, 'AF003': {'AF002': 0.0, 'AF003': 1.0, 'AF001': 0.0, 'AF004': 0.17452569907144502, 'AF005': 0.238497202355299}, 'AF001': {'AF002': 0.374449352805868, 'AF003': 0.00034743953114899502, 'AF001': 0.999999999999999, 'AF004': 0.00103026903356954, 'AF005': 0.0011830950375467401}, 'AF004': {'AF002': 0.0, 'AF003': 0.0, 'AF001': 0.0, 'AF004': 1.0, 'AF005': 0.75570008659397792}, 'AF005': {'AF002': 0.0, 'AF003': 0.0, 'AF001': 0.0, 'AF004': 0.0, 'AF005': 1.0}}

Answer 1

在熊猫中，这是unstack;如果您在结果上致电to_dict，您将获得所需内容：

In [43]: df.unstack()
Out[43]:
AF001  AF001    1.000000e+00
       AF002    3.744490e-01
       AF003    3.470000e-04
       AF004    1.030000e-03
       AF005    1.183000e-03
AF002  AF001    0.000000e+00
       AF002    1.000000e+00
       AF003    1.173926e-05
       AF004    1.494282e-07
       AF005    1.216664e-06
AF003  AF001    0.000000e+00
       AF002    0.000000e+00
       AF003    1.000000e+00
       AF004    1.745260e-01
       AF005    2.384970e-01
AF004  AF001    0.000000e+00
       AF002    0.000000e+00
       AF003    0.000000e+00
       AF004    1.000000e+00
       AF005    7.557000e-01
AF005  AF001    0.000000e+00
       AF002    0.000000e+00
       AF003    0.000000e+00
       AF004    0.000000e+00
       AF005    1.000000e+00
dtype: float64

In [45]: df.unstack().to_dict()
Out[45]:
{('AF001', 'AF001'): 1.0,
 ('AF001', 'AF002'): 0.37444899999999998,
 ('AF001', 'AF003'): 0.00034700000000000003,
 ('AF001', 'AF004'): 0.0010300000000000001,
 ('AF001', 'AF005'): 0.001183,
 ('AF002', 'AF001'): 0.0,
 ('AF002', 'AF002'): 1.0,
 ('AF002', 'AF003'): 1.1739260000000002e-05,
 ('AF002', 'AF004'): 1.4942820000000002e-07,
 ('AF002', 'AF005'): 1.216664e-06,
 ('AF003', 'AF001'): 0.0,
 ('AF003', 'AF002'): 0.0,
 ('AF003', 'AF003'): 1.0,
 ('AF003', 'AF004'): 0.17452599999999999,
 ('AF003', 'AF005'): 0.23849699999999999,
 ('AF004', 'AF001'): 0.0,
 ('AF004', 'AF002'): 0.0,
 ('AF004', 'AF003'): 0.0,
 ('AF004', 'AF004'): 1.0,
 ('AF004', 'AF005'): 0.75570000000000004,
 ('AF005', 'AF001'): 0.0,
 ('AF005', 'AF002'): 0.0,
 ('AF005', 'AF003'): 0.0,
 ('AF005', 'AF004'): 0.0,
 ('AF005', 'AF005'): 1.0}

如果您想对数据进行其他操作，例如删除诊断或空值，则需要在未堆叠的数据框上调用reset_index，然后使用您需要的任何转换，

u = df.unstack().reset_index()
u[(u['level_0']!=u['level_1']) & (u[0] != 0)]

   level_0 level_1             0
1    AF001   AF002  3.744490e-01
2    AF001   AF003  3.470000e-04
3    AF001   AF004  1.030000e-03
4    AF001   AF005  1.183000e-03
7    AF002   AF003  1.173926e-05
8    AF002   AF004  1.494282e-07
9    AF002   AF005  1.216664e-06
13   AF003   AF004  1.745260e-01
14   AF003   AF005  2.384970e-01
19   AF004   AF005  7.557000e-01

将相关数据帧转换为字典{key =（sample_x，sample_y），value = correlation}

输入：

输出：

DataFrame.to_dict（）不会为此

1 个答案: