python - numpy "Mean of empty slice."警告

Question

更新(真正的错误)

我错误地识别了错误的来源。这是我的完整功能(抱歉，如果某些行晦涩难懂...)

def removeLines(input,CRVAL1,CDELT1): #Masks out the Balmer lines from the spectrum
    #Numbers 4060, 4150, 4300, 4375, 4800, and 4950 obtained from fit_RVs.pro.
    #Other numbers obtained from the Balmer absorption series lines
    
    for i in range(0,len(lineWindows),2):
        left = toIndex(lineWindows[i],CRVAL1,CDELT1)
        right = toIndex(lineWindows[i+1],CRVAL1,CDELT1)
        
        print "left = ", left
        print "right = ", right
        print "20 from right =\n", input[right:right+20]
        print "mean of 20 = ", numpy.mean(input[right:right+20])
        
        #Find the averages on the left and right sides
        left_avg = numpy.mean(input[left-20:left])
        right_avg = numpy.mean(input[right:right+20])   #<--- NOT here
        
        print "right_avg = ", right_avg
        
        #Find the slope between the averages
        slope = (left_avg - right_avg)/(left - right)
        
        #Find the y-intercept of the line conjoining the averages
        bval = ((left_avg - slope*left) + (right_avg - slope*right)) / 2
        
        for j in range(left,right):     #Redefine the data to follow the line conjoining
            input[j] = slope*j + bval   #the sides of the peaks

    left = int(input[0])
    left_avg = int(input[0])
    right = toIndex(lineWindows[0],CRVAL1,CDELT1)
    right_avg = numpy.mean(input[right:right+20])   #<---- THIS IS WHERE IT IS!
    slope = (left_avg - right_avg)/(left - right)
    bval = ((left_avg - slope*left) + (right_avg - slope*right)) / 2
    
    for i in range(left, right):
        input[i] = slope*i + bval
    return input

我调查了这个问题并找到了答案，答案发布在下面(不在这篇文章中)。

---

错误(愚蠢的假错误)

#left  = An index in the data (on the 'left' side)
#right = An index in the data (on the 'right' side)
#input = The data array

print "left = ", left
print "right = ", right
print "20 from right =\n", input[right:right+20]
print "mean of 20 = ", numpy.mean(input[right:right+20])

#Find the averages on the left and right sides
left_avg = numpy.mean(input[left-20:left])
right_avg = numpy.mean(input[right:right+20])

产生输出

left =  1333
right =  1490
20 from right =
[ 0.14138737  0.14085886  0.14038289  0.14045525  0.14078836  0.14083192
  0.14072289  0.14082283  0.14058594  0.13977806  0.13955595  0.13998236
  0.1400764   0.1399636   0.14025062  0.14074247  0.14094831  0.14078569
  0.14001536  0.13895717]
mean of 20 =  0.140395
Traceback (most recent call last):
...
  File "getRVs.py", line 201, in removeLines
    right_avg = numpy.mean(input[right:right+20])
  File "C:\Users\MyName\Anaconda\lib\site-packages\numpy\core\fromnumeric.py", line 2735, in mean
    out=out, keepdims=keepdims)
  File "C:\Users\MyName\Anaconda\lib\site-packages\numpy\core\_methods.py", line 59, in _mean
    warnings.warn("Mean of empty slice.", RuntimeWarning)
RuntimeWarning: Mean of empty slice.

看起来 numpy.mean 在我打印时运行正确，但在我将其分配给一个值时却不同。任何反馈将不胜感激。感谢您花时间阅读我的问题。

---

简要说明

简而言之，我正在编写一段代码来处理科学数据，部分代码涉及取大约 20 个值的平均值。

#left  = An index in the data (on the 'left' side)
#right = An index in the data (on the 'right' side)
#input = The data array

#Find the averages on the left and right sides
left_avg = numpy.mean(input[left-20:left])
right_avg = numpy.mean(input[right:right+20])

这段代码返回一个 numpy 的“空切片的平均值”。警告并烦人地将其打印在我宝贵的输出中!我决定尝试追踪警告的来源，如 here 所示，例如，所以我放置了

import warnings
warnings.simplefilter("error")

在我的代码顶部，然后返回以下片段的 Traceback:

  File "getRVs.py", line 201, in removeLines
    right_avg = numpy.mean(input[right:right+20])
  File "C:\Users\MyName\Anaconda\lib\site-packages\numpy\core\fromnumeric.py", line 2735, in mean
    out=out, keepdims=keepdims)
  File "C:\Users\MyName\Anaconda\lib\site-packages\numpy\core\_methods.py", line 59, in _mean
    warnings.warn("Mean of empty slice.", RuntimeWarning)
RuntimeWarning: Mean of empty slice.

我省略了大约 2/3 的 Traceback，因为它通过了大约 5 个难以解释的函数，这些函数不影响数据的可读性或大小。

所以我决定打印出整个操作，看看 right_avg 是否真的在尝试空切片的 numpy.mean ... em>真的很奇怪。

Answer 1

我无法重现您的错误。您使用的是最新的 numpy 版本吗？但是，您可以通过使用关键字 ignore 来抑制警告(请参阅 https://docs.python.org/2/library/warnings.html#temporarily-suppressing-warnings)

此错误通常意味着向函数传递了一个空列表。

>>> a = []

>>> import numpy
>>> numpy.mean(a)
/shahlab/pipelines/apps_centos6/Python-2.7.10/lib/python2.7/site-packages/numpy/core/_methods.py:59: RuntimeWarning: Mean of empty slice.
  warnings.warn("Mean of empty slice.", RuntimeWarning)
/shahlab/pipelines/apps_centos6/Python-2.7.10/lib/python2.7/site-packages/numpy/core/_methods.py:71: RuntimeWarning: invalid value encountered in double_scalars
  ret = ret.dtype.type(ret / rcount)
nan
>>> print numpy.mean(a)
nan

>>> import warnings
>>> warnings.simplefilter("ignore")
>>> numpy.mean(a)
nan

>>> a=[ 0.14138737, 0.14085886, 0.14038289, 0.14045525, 0.14078836, 0.14083192, 0.14072289, 0.14082283, 0.14058594, 0.13977806, 0.13955595, 0.13998236, 0.1400764,  0.1399636,  0.14025062, 0.14074247, 0.14094831, 0.14078569, 0.14001536, 0.13895717]
>>> numpy.mean(a)
0.140394615
>>> x = numpy.mean(a)
>>> print x
0.140394615
>>> numpy.__version__
'1.9.2'

希望对您有所帮助。