This is my code,
l1=[[1,2,3],[4,5,6],[7,8,9]]
suppose, If I want to find the Index of an Item, I can easily find it like this
l1.index([7,8,9])
--output
2
type(l1)
--output
list
But, When I try to convert the list into Numpy.array and then convert it back to list and then if I try to find the index of an element, I am getting an error
l1=list(np.array([[1,2,3],[4,5,6],[7,8,9]]))
l1.index([7,8,9])
Why am I getting this error and how to resolve this?
You need to use tolist() function in order to get the proper conversion from np.array to Python built-in list:
import numpy as np
l1=[[1,2,3],[4,5,6],[7,8,9]]
print(l1.index([7,8,9]))
l1=np.array([[1,2,3],[4,5,6],[7,8,9]]).tolist()
print(l1.index([7,8,9]))
Output:
2
2
list. The thing is that list() is a built-in Python implementation that converts objects to lists. But tolist() is an specific function made by NumPy developers that convert an np.array object to a Python list.list() method can fail at the conversion, meanwhile tolist() is assured not to fail because it only converts from one type of object. Please take a look at both Python and Numpy documentation to learn more about both functions.In [4]: l1=[[1,2,3],[4,5,6],[7,8,9]]
In [5]: arr = np.array(l1)
In [6]: arr
Out[6]:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
tolist performs a thorough conversion to python base structures - not just lists, but also the numeric elements.
In [7]: arr.tolist()
Out[7]: [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
list just converts the first dimension:
In [8]: list(arr)
Out[8]: [array([1, 2, 3]), array([4, 5, 6]), array([7, 8, 9])]
It's equivalent to this list comprehension:
In [9]: [a for a in arr]
Out[9]: [array([1, 2, 3]), array([4, 5, 6]), array([7, 8, 9])]
tolist is also faster.
As for your error
In [10]: list(arr).index([7,8,9])
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
Input In [10], in <cell line: 1>()
----> 1 list(arr).index([7,8,9])
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
you have to understand how index works, and how == works for arrays.
index iterates through the elements of the list and applies an equality test, first with id and then with ==.
for the list of lists:
In [14]: [sublist==[7,8,9] for sublist in l1]
Out[14]: [False, False, True]
but for the list of arrays:
In [15]: [sublist==[7,8,9] for sublist in list(arr)]
Out[15]:
[array([False, False, False]),
array([False, False, False]),
array([ True, True, True])]
For lists == tests the whole lists, and returns a simple True/False. For arrays, the result is an element-wise test, producing multiple True/False. That's what the "ambiguity" error is all about.
l1. It is a list, but is it the list of lists that you expect?