Question

我在numpy中有成千上万的val。我想通过平均相邻值来减小其大小。例如：

a = [2,3,4,8,9,10]
#average down to 2 values here
a = [3,9]
#it averaged 2,3,4 and 8,9,10 together

所以，基本上，我在数组中有n个元素，我想告诉它平均值为X个值，并且平均值如上所述。

有没有办法用numpy做到这一点（已经将它用于其他事情，所以我想坚持下去。）

Answer 1

正如评论中所提到的，你想要的可能是：

group = 3
a = a.reshape(-1, group).mean(axis=1)

Answer 2

对我来说，看起来像一个简单的非重叠移动窗口，如何：

In [3]:

import numpy as np
a = np.array([2,3,4,8,9,10])
window_sz = 3
a[:len(a)/window_sz*window_sz].reshape(-1,window_sz).mean(1) 
#you want to be sure your array can be reshaped properly, so the [:len(a)/window_sz*window_sz] part
Out[3]:
array([ 3.,  9.])

Answer 3

试试这个：

n_averaged_elements = 3
averaged_array = []
a = np.array([ 2,  3,  4,  8,  9, 10])
for i in range(0, len(a), n_averaged_elements):
   slice_from_index = i
   slice_to_index = slice_from_index + n_averaged_elements
   averaged_array.append(np.mean(a[slice_from_index:slice_to_index]))

>>>> averaged_array
>>>> [3.0, 9.0]

Answer 4

在此示例中，我假设a是需要平均的一维numpy数组。在下面提供的方法中，我们首先找到该数组a的长度的因素。然后，我们选择适当的因子作为平均数组的步长。

这是代码。

import numpy as np
from functools import reduce

''' Function to find factors of a given number 'n' '''
def factors(n):    
    return list(set(reduce(list.__add__, 
        ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))

a = [2,3,4,8,9,10]  #Given array.

'''fac: list of factors of length of a. 
   In this example, len(a) = 6. So, fac = [1, 2, 3, 6] '''
fac = factors(len(a)) 

'''step: choose an appropriate step size from the list 'fac'.
   In this example, we choose one of the middle numbers in fac 
   (3). '''   
step = fac[int( len(fac)/3 )+1]

'''avg: initialize an empty array. '''
avg = np.array([])
for i in range(0, len(a), step):
    avg = np.append( avg, np.mean(a[i:i+step]) ) #append averaged values to `avg`

print avg  #Prints the final result

[3.0, 9.0]

通过使用numpy平均相邻值来减小数组大小

4 个答案: