[2680 rows x 3 columns] Pearson相关系数: 0.14628706843368472 Pearson p值: 2.759812589889411e-14 Spearman秩相关系数: 0.10814107230151733 Spearman p值: 2.0101039940091e-08 Kendall秩相关系数: 0.0731082564031467 Kendall p值: 1.4088940969025172e-08
时间: 2024-03-30 21:34:32 浏览: 16
根据程序输出的结果,沪深300指数和标普500指数之间的Pearson相关系数为0.146,Spearman秩相关系数为0.108,Kendall秩相关系数为0.073,这些系数都表明两个指数之间存在一定程度的正相关关系,但程度不是很强。p值都非常小,表明这些相关系数的显著性非常高,即两个指数之间的相关性不是由于偶然因素导致的。
相关问题
xx = np.array(x[0:5, 0: x.shape[1] - 1])
This line of code creates a new NumPy array called "xx" by selecting the first 5 rows and all columns except for the last column of an existing array "x".
The syntax "x[0:5, 0: x.shape[1] - 1]" specifies the range of rows and columns to select. The first part "0:5" selects the first 5 rows, while the second part "0: x.shape[1] - 1" selects all columns except for the last one. The expression "x.shape[1] - 1" calculates the index of the last column of the array "x".
Overall, this line of code is useful for selecting a subset of data from a larger array for further analysis or processing.
Give a positive integer n not greater than 9 and output a “Snake matrix” with n rows and n columns. For example: Input: 4 output:
Sure, I can help you with that! Here's a Python code that generates a Snake matrix with n rows and n columns for a positive integer n not greater than 9:
```
n = int(input("Enter a positive integer not greater than 9: "))
matrix = [[0] * n for i in range(n)]
num = 1
for i in range(n):
if i % 2 == 0:
for j in range(n):
matrix[i][j] = num
num += 1
else:
for j in range(n-1, -1, -1):
matrix[i][j] = num
num += 1
for row in matrix:
print(row)
```
This code first prompts the user to enter a positive integer not greater than 9, creates an empty matrix of size n x n, and then fills in the matrix in a "snake" pattern by iterating over each row and column. Finally, it prints the matrix to the console.
For example, if the user enters 4, the output will be:
```
[1, 2, 3, 4]
[8, 7, 6, 5]
[9, 10, 11, 12]
[16, 15, 14, 13]
```
I hope this helps! Let me know if you have any other questions.