Python IndexError: invalid index to scalar variable

Question

Why do i get the indexError?

I have tried to change the slicing to 2001 but i did not help

inputs = training_data[:-1] #EVERYTHING EXCEPT last values
outputs = training_data[-1] #last value

training_inputs = inputs[:2000]
training_outputs = outputs[:2000]
testing_inputs = inputs[2000:]
testing_outputs = outputs[2000:]

IndexError: invalid index to scalar variable.

outputs = training_data[-1] will have a single value so you can not apply slicing on it — Sociopath
– Sociopath, Commented Sep 18, 2019 at 8:35
Share these details please: shape of training and test arrays. — CypherX
– CypherX, Commented Sep 18, 2019 at 8:37
Share these details please: shape of training and test arrays and inputs. Perhaps you want to use: inputs = training_data[:, :-1] and outputs = training_data[:,-1]. — CypherX
– CypherX, Commented Sep 18, 2019 at 8:45

Sai prateek · Accepted Answer · 2019-09-18 08:42:43Z

2

This problem occurs when You are trying to index into a scalar non-iterable value.

>> data = [3, 6, 9]
>> result = data[0] # gives you result=3
>> print(result[0]) # Error

answered Sep 18, 2019 at 8:42

Sai prateek

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1 Comment

enthu Over a year ago

what is the solution?

Massifox · Accepted Answer · 2019-09-18 08:43:32Z

1

Because output is not a list, and therefore it is not possible to perform slice operations. If you want output to be a list you can use this trick:

outputs = training_data[-1:]

answered Sep 18, 2019 at 8:43

Massifox

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