What does 'index 0 is out of bounds for axis 0 with size 0' mean?

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I ran a code that I wrote and I am getting this message:

'index 0 is out of bounds for axis 0 with size 0'

index 0 means the first value in the array, but I can't figure out what axis 0 and size 0 mean.

The 'data' is a text file with lots of numbers in two columns.

x = np.linspace(1735.0,1775.0,100)
column1 = (data[0,0:-1]+data[0,1:])/2.0
column2 = data[1,1:]
x_column1 = np.zeros(x.size+2)
x_column1[1:-1] = x
x_column1[0] = x[0]+x[0]-x[1]
x_column1[-1] = x[-1]+x[-1]-x[-2]
experiment = np.zeros_like(x)
for i in range(np.size(x_edges)-2):
    indexes = np.flatnonzero(np.logical_and((column1>=x_column1[i]),(column1<x_column1[i+1])))
    temp_column2 = column2[indexes]
    temp_column2[0] -= column2[indexes[0]]*(x_column1[i]-column1[indexes[0]-1])/(column1[indexes[0]]-column1[indexes[0]-1])
    temp_column2[-1] -= column2[indexes[-1]]*(column1[indexes[-1]+1]-x_column1[i+1])/(column1[indexes[-1]+1]-column1[indexes[-1]])
    experiment[i] = np.sum(temp_column2)   
return experiment
4

There are 4 answers

0
hpaulj On

In numpy, index and dimension numbering starts with 0. So axis 0 means the 1st dimension. Also in numpy a dimension can have length (size) 0. The simplest case is:

In [435]: x = np.zeros((0,), int)
In [436]: x
Out[436]: array([], dtype=int32)
In [437]: x[0]
...
IndexError: index 0 is out of bounds for axis 0 with size 0

I also get it if x = np.zeros((0,5), int), a 2d array with 0 rows, and 5 columns.

So someplace in your code you are creating an array with a size 0 first axis.

When asking about errors, it is expected that you tell us where the error occurs.

Also when debugging problems like this, the first thing you should do is print the shape (and maybe the dtype) of the suspected variables.

Applied to pandas

Resolving the error:

  1. Use a try-except block
  2. Verify the size of the array is not 0
    • if x.size != 0:
0
Dean Smith On

I encountered this error and found that it was my data type causing the error. The type was an object, after converting it to an int or float the issue was solved.

I used the following code:

df = df.astype({"column": new_data_type,
                "example": float})
0
Daniel Abud On

Essentially it means you don't have the index you are trying to reference. For example:

df = pd.DataFrame()
df['this']=np.nan
df['my']=np.nan
df['data']=np.nan
df['data'][0]=5 #I haven't yet assigned how long df[data] should be!
print(df)

will give me the error you are referring to, because I haven't told Pandas how long my dataframe is. Whereas if I do the exact same code but I DO assign an index length, I don't get an error:

df = pd.DataFrame(index=[0,1,2,3,4])
df['this']=np.nan
df['is']=np.nan
df['my']=np.nan
df['data']=np.nan
df['data'][0]=5 #since I've properly labelled my index, I don't run into this problem!
print(df)

Hope that answers your question!

0
kmario23 On

This is an IndexError in python, which means that we're trying to access an index which isn't there in the tensor. Below is a very simple example to understand this error.

# create an empty array of dimension `0`
In [14]: arr = np.array([], dtype=np.int64) 

# check its shape      
In [15]: arr.shape  
Out[15]: (0,)

with this array arr in place, if we now try to assign any value to some index, for example to the index 0 as in the case below

In [16]: arr[0] = 23     

Then, we will get an IndexError, as below:


IndexError                                Traceback (most recent call last)
<ipython-input-16-0891244a3c59> in <module>
----> 1 arr[0] = 23

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

The reason is that we are trying to access an index (here at 0th position), which is not there (i.e. it doesn't exist because we have an array of size 0).

In [19]: arr.size * arr.itemsize  
Out[19]: 0

So, in essence, such an array is useless and cannot be used for storing anything. Thus, in your code, you've to follow the traceback and look for the place where you're creating an array/tensor of size 0 and fix that.