What is the difference between flatten and ravel functions in numpy?
Question:
import numpy as np
y = np.array(((1,2,3),(4,5,6),(7,8,9)))
OUTPUT:
print(y.flatten())
[1 2 3 4 5 6 7 8 9]
print(y.ravel())
[1 2 3 4 5 6 7 8 9]
Both function return the same list.
Then what is the need of two different functions performing same job.
Answers:
The current API is that:
flatten
always returns a copy.ravel
returns a view of the original array whenever possible. This isn’t visible in the printed output, but if you modify the array returned by ravel, it may modify the entries in the original array. If you modify the entries in an array returned from flatten this will never happen. ravel will often be faster since no memory is copied, but you have to be more careful about modifying the array it returns.reshape((1,))
gets a view whenever the strides of the array allow it even if that means you don’t always get a contiguous array.
As explained here a key difference is that:

flatten
is a method of an ndarray object and hence can only be called for true numpy arrays. 
ravel
is a librarylevel function and hence can be called on any object that can successfully be parsed.
For example ravel
will work on a list of ndarrays, while flatten
is not available for that type of object.
@IanH also points out important differences with memory handling in his answer.
Here is the correct namespace for the functions:
Both functions return flattened 1D arrays pointing to the new memory structures.
import numpy
a = numpy.array([[1,2],[3,4]])
r = numpy.ravel(a)
f = numpy.ndarray.flatten(a)
print(id(a))
print(id(r))
print(id(f))
print(r)
print(f)
print("nbase r:", r.base)
print("nbase f:", f.base)
returns
140541099429760
140541099471056
140541099473216
[1 2 3 4]
[1 2 3 4]
base r: [[1 2]
[3 4]]
base f: None
In the upper example:
 the memory locations of the results are different,
 the results look the same
 flatten would return a copy
 ravel would return a view.
How we check if something is a copy?
Using the .base
attribute of the ndarray
. If it’s a view, the base will be the original array; if it is a copy, the base will be None
.
Check if a2
is copy of a1
import numpy
a1 = numpy.array([[1,2],[3,4]])
a2 = a1.copy()
id(a2.base), id(a1.base)
Out:
(140735713795296, 140735713795296)