Python 如何循环相关排序列表？

下面是查找相关矩阵并对其排序的简单代码，但如何通过获取列对名称来循环它呢

import pandas as pd
import numpy as np

d = {
    'x1': [1, 4, 4, 5, 6], 
    'x2': [0, 0, 8, 2, 4], 
    'x3': [2, 8, 8, 10, 12], 
    'x4': [-1, -4, -4, -4, -5]
}
df = pd.DataFrame(data=d)
print(df)
print('---')
print(df.corr())
print('---')

corr_matrix = df.corr().abs()
sol = (corr_matrix.where(np.triu(np.ones(corr_matrix.shape), k=1).astype(np.bool)).stack().sort_values(ascending=False))
print(sol)
print('---')

for s in sol:
    print(s)
    # how to print column 1 and 2 pair names with this "s" corr?

结果:

   x1  x2  x3  x4
0   1   0   2  -1
1   4   0   8  -4
2   4   8   8  -4
3   5   2  10  -4
4   6   4  12  -5
---
          x1        x2        x3        x4
x1  1.000000  0.399298  1.000000 -0.969248
x2  0.399298  1.000000  0.399298 -0.472866
x3  1.000000  0.399298  1.000000 -0.969248
x4 -0.969248 -0.472866 -0.969248  1.000000
---
x1  x3    1.000000
x3  x4    0.969248
x1  x4    0.969248
x2  x4    0.472866
    x3    0.399298
x1  x2    0.399298
dtype: float64
---
1.0
0.9692476431690819
0.9692476431690819
0.4728662437434603
0.39929785312496247
0.39929785312496247

x1, x3, 1.000000
x3, x4, 0.969248
x1, x4, 0.969248
x2, x4, 0.472866
x1, x2, 0.399298

'x1', 'x3', 1.0
'x3', 'x4', 0.9692476431690819
'x1', 'x4', 0.9692476431690819
'x2', 'x4', 0.4728662437434603
'x2', 'x3', 0.39929785312496247
'x1', 'x2', 0.39929785312496247

我所期望的是：

for (column1, column2, s) in sol:
    print(column1 + ',' + column2 + ',' + str(s))

结果:

   x1  x2  x3  x4
0   1   0   2  -1
1   4   0   8  -4
2   4   8   8  -4
3   5   2  10  -4
4   6   4  12  -5
---
          x1        x2        x3        x4
x1  1.000000  0.399298  1.000000 -0.969248
x2  0.399298  1.000000  0.399298 -0.472866
x3  1.000000  0.399298  1.000000 -0.969248
x4 -0.969248 -0.472866 -0.969248  1.000000
---
x1  x3    1.000000
x3  x4    0.969248
x1  x4    0.969248
x2  x4    0.472866
    x3    0.399298
x1  x2    0.399298
dtype: float64
---
1.0
0.9692476431690819
0.9692476431690819
0.4728662437434603
0.39929785312496247
0.39929785312496247

x1, x3, 1.000000
x3, x4, 0.969248
x1, x4, 0.969248
x2, x4, 0.472866
x1, x2, 0.399298

'x1', 'x3', 1.0
'x3', 'x4', 0.9692476431690819
'x1', 'x4', 0.9692476431690819
'x2', 'x4', 0.4728662437434603
'x2', 'x3', 0.39929785312496247
'x1', 'x2', 0.39929785312496247

您可以使用以命名对的形式迭代数据帧行：

pairs = sol.reset_index().itertuples(index=False, name=None)
print('\n'.join(str(p).strip('()') for p in pairs))

或者也可以使用：

结果:

   x1  x2  x3  x4
0   1   0   2  -1
1   4   0   8  -4
2   4   8   8  -4
3   5   2  10  -4
4   6   4  12  -5
---
          x1        x2        x3        x4
x1  1.000000  0.399298  1.000000 -0.969248
x2  0.399298  1.000000  0.399298 -0.472866
x3  1.000000  0.399298  1.000000 -0.969248
x4 -0.969248 -0.472866 -0.969248  1.000000
---
x1  x3    1.000000
x3  x4    0.969248
x1  x4    0.969248
x2  x4    0.472866
    x3    0.399298
x1  x2    0.399298
dtype: float64
---
1.0
0.9692476431690819
0.9692476431690819
0.4728662437434603
0.39929785312496247
0.39929785312496247

x1, x3, 1.000000
x3, x4, 0.969248
x1, x4, 0.969248
x2, x4, 0.472866
x1, x2, 0.399298

'x1', 'x3', 1.0
'x3', 'x4', 0.9692476431690819
'x1', 'x4', 0.9692476431690819
'x2', 'x4', 0.4728662437434603
'x2', 'x3', 0.39929785312496247
'x1', 'x2', 0.39929785312496247

您可以使用以命名对的形式迭代数据帧行：

pairs = sol.reset_index().itertuples(index=False, name=None)
print('\n'.join(str(p).strip('()') for p in pairs))

或者也可以使用：

结果:

   x1  x2  x3  x4
0   1   0   2  -1
1   4   0   8  -4
2   4   8   8  -4
3   5   2  10  -4
4   6   4  12  -5
---
          x1        x2        x3        x4
x1  1.000000  0.399298  1.000000 -0.969248
x2  0.399298  1.000000  0.399298 -0.472866
x3  1.000000  0.399298  1.000000 -0.969248
x4 -0.969248 -0.472866 -0.969248  1.000000
---
x1  x3    1.000000
x3  x4    0.969248
x1  x4    0.969248
x2  x4    0.472866
    x3    0.399298
x1  x2    0.399298
dtype: float64
---
1.0
0.9692476431690819
0.9692476431690819
0.4728662437434603
0.39929785312496247
0.39929785312496247

x1, x3, 1.000000
x3, x4, 0.969248
x1, x4, 0.969248
x2, x4, 0.472866
x1, x2, 0.399298

'x1', 'x3', 1.0
'x3', 'x4', 0.9692476431690819
'x1', 'x4', 0.9692476431690819
'x2', 'x4', 0.4728662437434603
'x2', 'x3', 0.39929785312496247
'x1', 'x2', 0.39929785312496247

---

这就是你想要的吗：

print(sol.reset_index())

  level_0 level_1         0
0      x1      x3  1.000000
1      x3      x4  0.969248
2      x1      x4  0.969248
3      x2      x4  0.472866
4      x2      x3  0.399298
5      x1      x2  0.399298

---

这就是你想要的吗：

print(sol.reset_index())

  level_0 level_1         0
0      x1      x3  1.000000
1      x3      x4  0.969248
2      x1      x4  0.969248
3      x2      x4  0.472866
4      x2      x3  0.399298
5      x1      x2  0.399298

---

如果接近，您可以通过

（第1列，第2列）

循环解包多索引值：

与

f-string

s类似的解决方案：

for ((column1, column2), s) in sol.items():
    print( f"{column1},{column2},{s}")
    

---

如果接近，您可以通过

（第1列，第2列）

循环解包多索引值：

与

f-string

s类似的解决方案：

for ((column1, column2), s) in sol.items():
    print( f"{column1},{column2},{s}")