I have an array in numpy: [0 1 2 3]
When I try to get the first element with a[0], it throws: IndexError: too many indices for array
If I use for x in a: print(x) Python throws TypeError: iteration over a 0-d array
Attempting to make a list with list(a) throws TypeError: 'numpy.uint8' object is not iterable
How do I convert this list of numbers into a standard list?
It would help to know how you created a so that we could try to reproduce the errors.
import numpy as np
a = np.array([0, 1, 2, 3])
print(a[0]) # 0
for x in a:
print(x) # 0 1 2 3
# you can call list() to convert to a python list
print(list(a))
# you can also call the built-in numpy array method
print(a.tolist())
a = np.array(a)
print(a, type(a)) # [0 1 2 3] <class 'numpy.ndarray'>
print(a[0]) # 0
print(a[1]) # 1
All of these operations succeed. Python 3.7.1, Numpy 1.15.4
Try to check the type of the array, using a=array('[0 1 2 3]', dtype='<U9') I was able to replicate the firts two errors. If that is the case try the following
b=list(str(a))
newArray=[]
for val in b:
try:
newArray.append(int(val))
except ValueError:
pass
Python will not change automatically the value of a string to int or floatas some others languajes. Hope it helps
list(a) is essentially [x for x in a], so should produce the same error, not one referencing a 'uint8' object.
a?dtype. And just to be suretypeandshape. The errors are not consistent with the initial display. Also tryrepr(arr).