Raspberry Pi: The Essential Guide on Starting Your Own Raspberry Pi 3 Projects With Ingenious Tips & Tricks! by Jared Hendrix

Author:Jared Hendrix [Hendrix, Jared]
Language: eng
Format: epub
ISBN: 9781539972242
Barnesnoble:
Publisher: CreateSpace Publishing
Published: 2016-11-09T00:00:00+00:00

>>> lst = [1, 2, 3]

>>> def foo2():

... lst += [5] # ... but this bombs!

...

>>> foo2()

Traceback (most recent call last):

File "<stdin>", line 1, in <module>

File "<stdin>", line 2, in foo

UnboundLocalError: local variable 'lst' referenced before assignment

This code would cause foo1 to run properly, but foo2 to mess up. The reason this happened in this previous coding example is the same as in the first, and in this case, foo1 does not connect to an assignment to 1st, but foo2 is. It is important to note, 1st += [5] really just means lst = lst + [5], this is an attempt to assign a value to 1st, but the value given to it is based on 1st itself, which Python assumes is local scope, but it has not really been defined.

Changing a list and iterating over it – Take issue in the following should be obvious:

>>> odd = lambda x : bool(x % 2)

>>> numbers = [n for n in range(10)]

>>> for i in range(len(numbers)):

... if odd(numbers[i]):

... del numbers[i] # BAD: Deleting item from a list while iterating over it

...

Traceback (most recent call last):

File "<stdin>", line 2, in <module>

IndexError: list index out of range

Any experienced software developer knows that deleting an item from an array or list while iterating over it will cause a problem. The coding above contains a more blatant mistake, but not all of these will be this easy to see, and it can still trip up even advanced developers.

Python uses many different programming paradigms that simplify and streamline code. An added benefit of simpler code is that it generally creates less opportunity for mistakes. One of the paradigms is the list comprehension paradigm, which is very useful for avoiding this very issue. Here is an example of an alternative code that works properly:

>>> odd = lambda x : bool(x % 2)

>>> numbers = [n for n in range(10)]

>>> numbers[:] = [n for n in numbers if not odd(n)] # ahh, the beauty of it all

>>> numbers

[0, 2, 4, 6, 8]

Not fully understanding how Python binds closures – Look at the following code:

>>> def create_multipliers():

... return [lambda x : i * x for i in range(5)]

>>> for multiplier in create_multipliers():

... print multiplier(2)

...

You would anticipate the following ouput:

0

2

4

6

8

This is actually very different from what you would actually get with that code:

8

8

8

8

8

This occurs because of the late binding behavior of Python, the values of variables used in closures are looked up when the inner function is called. In the code each time the returned functions are called, the value of i is found in the surrounding scope when it is called. Some people think that the solution to this is more of a hack, but there is a solution:

>>> def create_multipliers():

... return [lambda x, i=i : i * x for i in range(5)]

...

>>> for multiplier in create_multipliers():

... print multiplier(2)

...

0

2

4

6

8

The solution is using the default arguments that create anonymous functions in your favor, to achieve your preferred results. Some people do not agree with this type of solution, but it can be helpful to understand how it works anyway.

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