1

如何计算列表中省略特殊值（-999）的值的平均值？

import numpy as np
A = [4,5,7,8,-999]
M = np.mean(A)

任何想法 ？？？

4

4 回答 4

3 
>>> import numpy as np
>>> a = np.array([1,2,3,4,5])
>>> np.mean(a)
3.0
>>> np.mean(a[a<=3])
2.0
>>> np.mean(a[a!=4])
2.75

对于 OP 案例：

np.mean(A[A!=-999])

表现

让我们用 Python 生成器测试三个片段：plainnp.mean和masked_array“naive”解决方案。数组有 1000000 个值。

from timeit import timeit
setup = 'import numpy as np; a=np.arange(0, 1000000)'
snippets = [
    'assert np.mean(a[a!=999999]) == 499999.0',
    'm=np.ma.masked_array(a,a==999999); assert np.ma.mean(m) == 499999.0',
    'assert sum(x for x in a if x != 999999)/999999 == 499999'
]
timings = [timeit(x, setup=setup, number=10) for x in snippets]
print('\n'.join(str(x) for x in timings))

结果：

0.0840559005737
0.0890350341797
10.4104599953

简单np.mean且masked_array时间很接近，而“幼稚”的解决方案要慢 100 倍以上。

于 2013-11-07T11:39:01.183 回答
3

在 numpy 中，您可以使用掩码数组的意思：

import numpy as np
A = np.array([4,5,7,8,-999])
mask_A = A == -999
ma_A = np.ma.masked_array(A, mask_A)
print np.ma.mean(ma_A)

结果是：

6.0
于 2013-11-07T11:41:58.727 回答
1

我不知道麻木。但这会起作用

A = [4,5,7,8,-999]
A = [item for item in A if item != -999]
print sum(A)/float(len(A))

输出

6.0

编辑：

要找到所有子列表的均值，

A = [[4,5,7,8,-999],[3,8,5,7,-999]]
M = [sum(z)/float(len(z)) for z in [[x for x in y if x != -999] for y in A]]
print M

输出

[6.0, 5.75]
于 2013-11-07T11:24:10.410 回答
0 
from numpy import*
A = [4,5,7,8,-999]
result = mean(A[A!=-999])
print (result)
于 2013-11-12T14:17:47.923 回答