Keras function parameters not valid Python?

Question

I am working through a course on deep learning with Python and ran into this line:

hidden = Dense(2)(visible)

What does the second parameter do? Is this a python language feature I've missed?

jkr · Accepted Answer · 2020-07-11 23:03:13Z

2

I remember asking the same thing when I was learning Keras.

hidden = Dense(2)(visible)

You can rewrite this in a more verbose way as below:

dense_layer = Dense(2)
hidden = dense_layer(visible)

As you can see from the above, the first line creates an instance of the Dense layer, and then you can call that layer on a tensor. This adds the Dense operation to a graph of operations.

edited Jul 11, 2020 at 23:03

answered Jul 11, 2020 at 22:52

jkr

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2 Comments

jdbennet Over a year ago

Ah. I was wondering if it returned a function. Thank you.

jkr Over a year ago

It does not return a function. It creates a class, and that class implements the __call__ method, which allows you to call it like a function. Also if this answered your question, please accept it by clicking the green check mark.

ismetguzelgun · Accepted Answer · 2020-07-11 15:03:30Z

0

Visible is not a language specific feature or something. Your code should not have only this line but before that you probably define a variable named visible like below.

from keras.layers import Input
from keras.layers import Dense
visible = Input(shape=(2,))
hidden = Dense(2)(visible)

Here are some examples:

source1 source2

answered Jul 11, 2020 at 15:03

ismetguzelgun

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