파이썬은 스크립트 언어지만.. 

module.py를 import 하였을때 module.pyc로 컴파일후 본문내 불러와 진다. 


module.py을 변경하여도 컴파일러가 알아챌만한 큰 변화가 없으면, module.pyc을 매번 컴파일 하지 않는다는 문제. 물론 매번 컴파일 한다면 속도면에서 큰 제약이 있겠다. 


그래서 module을 강제 reload하고 싶을때는, 


import moduleA as ma
import importlib
importlib.reload(ma)



이렇게 reload를 해주면 변경된 moduleA.py가 반영된다. 

https://www.quora.com/How-can-we-apply-early-stopping-for-Generative-Adversarial-Network-GAN-to-prevent-overfitting


Ian Goodfellow
Ian Goodfellow, Lead author of the Deep Learning textbook: http://www.deeplearningbook.org

Overfitting isn’t much of a problem for GANs: https://arxiv.org/pdf/1611.04273...

Because we aren’t able to find the Nash equilibrium very well, they generall underfit.


좋은 설명이 있어서 가져옴.. 


https://stackoverflow.com/questions/22053050/difference-between-numpy-array-shape-r-1-and-r



1. The meaning of shapes in NumPy

You write, "I know literally it's list of numbers and list of lists where all list contains only a number" but that's a bit of an unhelpful way to think about it.

The best way to think about NumPy arrays is that they consist of two parts, a data buffer which is just a block of raw elements, and a view which describes how to interpret the data buffer.

For example, if we create an array of 12 integers:

>>> a = numpy.arange(12)
>>> a
array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11])

Then a consists of a data buffer, arranged something like this:

┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
  0   1   2   3   4   5   6   7   8   9  10  11 
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

and a view which describes how to interpret the data:

>>> a.flags
  C_CONTIGUOUS : True
  F_CONTIGUOUS : True
  OWNDATA : True
  WRITEABLE : True
  ALIGNED : True
  UPDATEIFCOPY : False
>>> a.dtype
dtype('int64')
>>> a.itemsize
8
>>> a.strides
(8,)
>>> a.shape
(12,)

Here the shape (12,) means the array is indexed by a single index which runs from 0 to 11. Conceptually, if we label this single index i, the array a looks like this:

i= 0    1    2    3    4    5    6    7    8    9   10   11
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
  0   1   2   3   4   5   6   7   8   9  10  11 
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

If we reshape an array, this doesn't change the data buffer. Instead, it creates a new view that describes a different way to interpret the data. So after:

>>> b = a.reshape((3, 4))

the array b has the same data buffer as a, but now it is indexed by two indices which run from 0 to 2 and 0 to 3 respectively. If we label the two indices i and j, the array b looks like this:

i= 0    0    0    0    1    1    1    1    2    2    2    2
j= 0    1    2    3    0    1    2    3    0    1    2    3
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
  0   1   2   3   4   5   6   7   8   9  10  11 
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

which means that:

>>> b[2,1]
9

You can see that the second index changes quickly and the first index changes slowly. If you prefer this to be the other way round, you can specify the order parameter:

>>> c = a.reshape((3, 4), order='F')

which results in an array indexed like this:

i= 0    1    2    0    1    2    0    1    2    0    1    2
j= 0    0    0    1    1    1    2    2    2    3    3    3
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
  0   1   2   3   4   5   6   7   8   9  10  11 
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

which means that:

>>> c[2,1]
5

It should now be clear what it means for an array to have a shape with one or more dimensions of size 1. After:

>>> d = a.reshape((12, 1))

the array d is indexed by two indices, the first of which runs from 0 to 11, and the second index is always 0:

i= 0    1    2    3    4    5    6    7    8    9   10   11
j= 0    0    0    0    0    0    0    0    0    0    0    0
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
  0   1   2   3   4   5   6   7   8   9  10  11 
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

and so:

>>> d[10,0]
10

A dimension of length 1 is "free" (in some sense), so there's nothing stopping you from going to town:

>>> e = a.reshape((1, 2, 1, 6, 1))

giving an array indexed like this:

i= 0    0    0    0    0    0    0    0    0    0    0    0
j= 0    0    0    0    0    0    1    1    1    1    1    1
k= 0    0    0    0    0    0    0    0    0    0    0    0
l= 0    1    2    3    4    5    0    1    2    3    4    5
m= 0    0    0    0    0    0    0    0    0    0    0    0
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
  0   1   2   3   4   5   6   7   8   9  10  11 
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

and so:

>>> e[0,1,0,0,0]
6

See the NumPy internals documentation for more details about how arrays are implemented.

2. What to do?

Since numpy.reshape just creates a new view, you shouldn't be scared about using it whenever necessary. It's the right tool to use when you want to index an array in a different way.

However, in a long computation it's usually possible to arrange to construct arrays with the "right" shape in the first place, and so minimize the number of reshapes and transposes. But without seeing the actual context that led to the need for a reshape, it's hard to say what should be changed.

The example in your question is:

numpy.dot(M[:,0], numpy.ones((1, R)))

but this is not realistic. First, this expression:

M[:,0].sum()

computes the result more simply. Second, is there really something special about column 0? Perhaps what you actually need is:

M.sum(axis=0)


Tag // Numpy, Python