如何修复python中的“index is of bounds”错误，使用numpy？

Question

I'm trying to build a function that removes any negative numbers out of an array, but gets a error message. What do I need to change?

This is for python 3 using the numpy library The code I've tried so far is:

def drop_negative_numbers(a):
    b = a
    for i in range(a.size): 
        if a[i] < 0:
            b = np.delete(b, a[i])       
    return b

I'm trying to make this assert work:

a = np.array([1, 2, -3, 4, -5])
b = drop_negative_numbers(a)
npt.assert_equal(b, [1, 2, 4])

But i get this error message: IndexError: index -5 is out of bounds for axis 0 with size 4

Answer 1

The mistake is in the assignment b = a . b is then not a copy, just a(nother) view into a : any chances in b are reflected in a . If you use b = a.copy() (or b = a[:] ) in your function, your code will work.

With NumPy, however, it's clearer and faster to think vector(array)-wise: use, for example,

a = np.array([1, 2, -3, 4, -5])
b = a[a >= 0]

and you don't need your function anymore.

Answer 2

In your for loop you iterate over the array a number of times, if you then remove -3, you only have 4 items in the array. You then get to -5 and you try to remove that item, but that would be item 5 (4, when you start on 0) which means that your index is invalid as there isn't an item 5 in your array any more.

You want to make a copy of the array at the start, then iterate over one list and remove from another

Answer 3

Look in detail at what happens with each delete:

In [43]: a                                                                      
Out[43]: array([ 1,  2, -3,  4, -5])
In [44]: b=a                                                                    
In [45]: a[2]                                                                   
Out[45]: -3
In [46]: b=np.delete(b, a[2])                                                   
In [47]: b                                                                      
Out[47]: array([ 1,  2,  4, -5])
In [48]: a[4]                                                                   
Out[48]: -5
In [49]: b=np.delete(b, a[4])                                                   
---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-49-a326bbe5d5c9> in <module>
----> 1 b=np.delete(b, a[4])

/usr/local/lib/python3.6/dist-packages/numpy/lib/function_base.py in delete(arr, obj, axis)
   4374             raise IndexError(
   4375                 "index %i is out of bounds for axis %i with "
-> 4376                 "size %i" % (obj, axis, N))
   4377         if (obj < 0):
   4378             obj += N

IndexError: index -5 is out of bounds for axis 0 with size 4

delete makes a new array each time. That makes it slower than similar list methods. But it does mean that a does not get changed.

np.delete takes an index, not a value (it's not the same as the list del ). a[2] just happens to work because a[-3] is [-3] . a[4] is -5 which isn't a valid index in the size 4 b .

If we specify the index rather than value, the deletes work correctly:

In [51]: b=a                                                                    
In [52]: b=np.delete(b, 2)                                                      
In [53]: b                                                                      
Out[53]: array([ 1,  2,  4, -5])
In [54]: b=np.delete(b, 3)                                                      
In [55]: b                                                                      
Out[55]: array([1, 2, 4])

But a repeated delete is slow; better to delete all negative numbers at once:

In [56]: a                                                                      
Out[56]: array([ 1,  2, -3,  4, -5])
In [57]: np.delete(a,[2,4])                                                     
Out[57]: array([1, 2, 4])

We can use where to get the indices:

In [65]: a<0                                                                    
Out[65]: array([False, False,  True, False,  True])
In [66]: np.where(a<0)                                                          
Out[66]: (array([2, 4]),)
In [67]: np.delete(a, np.where(a<0))                                            
Out[67]: array([1, 2, 4])

But we can use the mask directly:

In [68]: a[~(a<0)]                                                              
Out[68]: array([1, 2, 4])

for a list, a simple list comprehension works nicely:

In [69]: b = a.tolist()                                                         
In [70]: [i for i in b if not i<0]                                              
Out[70]: [1, 2, 4]

Another well known trick with lists, is to iterate from the end, so that any deletions won't mess up the remaining list.