Question

我想知道是否有人可以给我提示如何计算Python中的协方差;我不想使用numpy中的任何东西。我只想学习如何手动执行此操作并使用for循环练习。

基本上，我想计算协方差：

X = [1,2]
Y = [1,2,3]
P = [[0.25,0.25,0.0], [0.0, 0.25, 0.25]]

Mean of X: 1.5
Mean of Y: 2

这些值取自：https://onlinecourses.science.psu.edu/stat414/node/109

结果应为0.25。

我一直在嵌套for循环中循环遍历X，Y和P，但是不知道我可以用来组合它的其他方法。

我基本上想要做这个计算：

(1-1.5)(1-2)(0.25) + (1-1.5)(2-2)(0.25) +  ..... + (2-1.5)(3-2)(0.25)

Answer 1

要计算协方差，您需要类似下面的内容，它具有嵌套循环，遍历每个列表，并使用协方差公式累积协方差。

# let's get the mean of `X` (add all the vals in `X` and divide by
# the length
x_mean = float(sum(X)) / len(X)

# now, let's get the mean for `Y`
y_mean = float(sum(Y)) / len(Y)

# initialize the covariance to 0 so we can add it up
cov = 0

# we'll use a nested loop structure -- the outer loop can be through `Y`
# or `X`, it doesn't matter in this case
# we'll use python's `enumerate`, which lets us iterate through the `list`
# using a `tuple` that contains (the_current_index, the_current_element),
# or in `C`/`Java` terms, `(i, arr[i])`
for y_idx,y in enumerate(Y):
    for x_idx,x in enumerate(X):

        # the covariance is defined by the following equation
        # you don't need to loop through `P` -- the outer list
        # contains 2 elements, which is the size of `X`, and
        # the inner list contains 3 elements, which is the size of `Y`
        cov += (x - x_mean) * (y - y_mean) * P[x_idx][y_idx]

print cov # => 0.25

Answer 2

itertools中的product函数在这里也可以提供帮助，可以与enumerate结合使用，以返回P所需的索引，如下所示：

from itertools import product

X = [1, 2]
Y = [1, 2, 3]
P = [[0.25,0.25,0.0], [0.0, 0.25, 0.25]]

mean_x = float(sum(X) / len(X))
mean_y = float(sum(Y) / len(Y))

print sum((x[1] - mean_x) * (y[1] - mean_y) * P[x[0]][y[0]] for x, y in product(enumerate(X), enumerate(Y)))

给出结果：

0.25

在Python中计算协方差

2 个答案: