python – What does index 0 is out of bounds for axis 0 with size 0 mean?
Table of Contents
python – What does index 0 is out of bounds for axis 0 with size 0 mean?
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
- The same error can occur for those using
pandas
, when sending aSeries
orDataFrame
to anumpy.array
, as with the following:
Resolving the error:
- Use a
try-except
block - Verify the size of the array is not 0
if x.size != 0:
Essentially it means you dont 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 havent yet assigned how long df[data] should be!
print(df)
will give me the error you are referring to, because I havent told Pandas how long my dataframe is. Whereas if I do the exact same code but I DO assign an index length, I dont 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 Ive properly labelled my index, I dont run into this problem!
print(df)
Hope that answers your question!
python – What does index 0 is out of bounds for axis 0 with size 0 mean?
This is an IndexError
in python, which means that were trying to access an index which isnt 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 doesnt 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, youve to follow the traceback and look for the place where youre creating an array/tensor of size 0
and fix that.