更新（真正的错误）
我误认为错误来自于其他地方。这是我的整个函数（如果有些行不清楚和混乱，很抱歉...）
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，下面是发布的答案（不在此帖子中）。
错误（愚蠢的虚假错误）
#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])
生成输出
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.
似乎当我打印
numpy.mean时，它可以正常运行，但是当我将其赋给一个值时，结果不同。非常感谢您提供的任何反馈。谢谢您抽出时间阅读我的问题。
简要解释
简而言之，我正在编写一段代码来处理科学数据，其中一部分代码涉及对大约20个值取平均。
#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，并且会在输出中烦人地打印出来！我决定尝试追踪警告的来源，例如在这里看到的那样，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，因为它经过了约5个难以解释的函数，这些函数不会影响数据的可读性或大小。
所以我决定打印整个操作，看看right_avg是否真的尝试对一个空切片进行numpy.mean...就在那时，事情变得非常奇怪。
numpy.mean(input[right:right+20])时，input[right:right+20]是空的。在第一次和第二次调用之间必须有代码更改了input的值。 - unutbuinput的值正在被更改，但是在调用之前甚至打印相同代码时没有错误。您认为这与我如何分配left_avg有关吗？ - boofright_avg！我已经更新了主线程，以避免浪费大家的时间。真尴尬... - boofright = toIndex(lineWindows[0],CRVAL1,CDELT1)的值。如果它太大，那么切片input[right:right+20]可能为空。 - unutbu