我错误地识别了错误的来源。这是我的全部功能（对不起，如果有些线路模糊不清......）
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
我已经调查了这个问题并找到了答案，该答案发布在下面（不在这篇文章中）。
#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])
产生了输出
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.
当我打印它时，numpy.mean似乎正确运行，但是当我将它分配给一个值时，它会有所不同。任何反馈都将非常感激。感谢您抽出宝贵时间阅读我的问题。
简而言之，我正在编写处理科学数据的代码，部分代码涉及大约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])
此代码返回一个numpy“空切片的平均值”。警告并恼人地将它打印在我宝贵的输出中！我决定尝试跟踪警告的来源，例如here，所以我放置了
import warnings
warnings.simplefilter("error")
在我的代码顶部，然后返回以下剪切的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，因为它通过大约5个难以解释的函数，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是否真的在尝试一个空切片的numpy.mean ......那就是当事情变得非常奇怪时
答案 0 :(得分：4)
left和right边）太靠近边缘数据阵列。
def removeLines(input,CRVAL1,CDELT1): #Masks out the Balmer lines from the spectrum
for i in range(0,len(lineWindows),2):
left = toIndex(lineWindows[i],CRVAL1,CDELT1)
right = toIndex(lineWindows[i+1],CRVAL1,CDELT1)
#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])
#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 = 0
left_avg = int(input[0])
if toIndex(lineWindows[0],CRVAL1,CDELT1) < 0: right = 0
else: right = toIndex(lineWindows[0],CRVAL1,CDELT1)
right_avg = numpy.mean(input[right:right+20])
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
只需更改此
即可right = toIndex(lineWindows[0],CRVAL1,CDELT1) #Error occurs where right = -10
right_avg = numpy.mean(input[right:right+20]) #Index of -10? Yeah, right.
到这个
if toIndex(lineWindows[0],CRVAL1,CDELT1) < 0: right = 0 #Index 0, much better!
else: right = toIndex(lineWindows[0],CRVAL1,CDELT1) #Leave it alone if it isn't a problem.
right_avg = numpy.mean(input[right:right+20])
此外，您对left = int(input[0])的完全错误，因此我将其更改为left = 0。 谁知道这个草率，草率的代码会产生什么其他简单错误？请稍等一下，然后发布到Stack Overflow吧！
答案 1 :(得分：1)
我无法重现您的错误。你使用的是最新的numpy版本吗？ 但是，您可以通过使用关键字ignore（请参阅https://docs.python.org/2/library/warnings.html#temporarily-suppressing-warnings）
来取消警告此错误通常表示将空列表传递给函数。
>>> 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'
希望有所帮助。