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Commitc283dbd

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Reword the discussion on short ways to manipulate lists
 - Remove emphasis on map and filter, in favor of generator expressions. - Move list comprehensions and generator expressions under "Short ways to manipulate lists" rather than "Filtering a list". - Add some examples.
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‎docs/writing/style.rst

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@@ -521,7 +521,7 @@ Conventions
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Here are some conventions you should follow to make your code easier to read.
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Check if a variable equals a constant
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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You don't need to explicitly compare a value to True, or None, or 0 -- you can
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just add it to the if statement. See `Truth Value Testing
@@ -588,80 +588,104 @@ Short Ways to Manipulate Lists
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`List comprehensions
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<http://docs.python.org/tutorial/datastructures.html#list-comprehensions>`_
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provide a powerful, concise way to work with lists. Also, the:py:func:`map` and
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:py:func:`filter` functions can perform operations on lists using a different,
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more concise syntax.
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provide a powerful, concise way to work with lists.
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Filtering a list
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~~~~~~~~~~~~~~~~
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`Generator expressions
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<http://docs.python.org/tutorial/classes.html#generator-expressions>`_
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follow almost the same syntax as list comprehensions but return a generator
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instead of a list.
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**Bad**:
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Creating a new list requires more work and uses more memory. If you are just going
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to loop through the new list, prefer using an iterator instead.
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Never remove items from a list while you are iterating through it.
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**Bad**:
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..code-block::python
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# Filter elements greater than 4
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a= [3,4,5]
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for iin a:
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if i>4:
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a.remove(i)
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# needlessly allocates a list of all (gpa, name) entires in memory
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valedictorian=max([(student.gpa, student.name)for studentin graduates])
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Don't make multiple passes through the list.
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**Good**:
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..code-block::python
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while iin a:
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a.remove(i)
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valedictorian=max((student.gpa, student.name)for studentin graduates)
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Use list comprehensions when you really need to create a second list, for
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example if you need to use the result multiple times.
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If your logic is too complicated for a short list comprehension or generator
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expression, consider using a generator function instead of returning a list.
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**Good**:
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Python has a few standard ways of filtering lists.
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The approach you use depends on:
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..code-block::python
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defmake_batches(items,batch_size):
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"""
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>>>list(make_batches([1, 2, 3, 4, 5], batch_size=3))
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[[1, 2, 3], [4, 5]]
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"""
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current_batch= []
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for itemin items:
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current_batch.append(item)
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iflen(current_batch)== batch_size:
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yield current_batch
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current_batch= []
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yield current_batch
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* Python 2.x vs. 3.x
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* Lists vs. iterators
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* Possible side effects of modifying the original list
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Python 2.x vs. 3.x
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::::::::::::::::::
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Never use a list comprehension just for its side effects.
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Starting with Python 3.0, the:py:func:`filter` function returns an iterator instead of a list.
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Wrap it in:py:func:`list` if you truly need a list.
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**Bad**:
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..code-block::python
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list(filter(...))
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[print(x)for xin seqeunce]
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List comprehensions and generator expressions work the same in both 2.x and 3.x (except that comprehensions in 2.x "leak" variables into the enclosing namespace):
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**Good**:
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* Comprehensions create a new list object.
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* Generators iterate over the original list.
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..code-block::python
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The:py:func:`filter` function:
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for xin sequence:
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print(x)
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* in 2.x returns a list (use itertools.ifilter if you want an iterator)
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* in 3.x returns an iterator
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Lists vs. iterators
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:::::::::::::::::::
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Filtering a list
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~~~~~~~~~~~~~~~~
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**Bad**:
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Never remove items from a list while you are iterating through it.
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..code-block::python
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# Filter elements greater than 4
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a= [3,4,5]
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for iin a:
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if i>4:
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a.remove(i)
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Don't make multiple passes through the list.
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..code-block::python
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while iin a:
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a.remove(i)
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**Good**:
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Creating anewlistrequires more work and uses more memory. If you a just going to loop through the new list, consider using an iterator instead.
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Use a listcomprehension or generator expression.
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..code-block::python
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# comprehensions create a new list object
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filtered_values= [valuefor valuein sequenceif value!= x]
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# Or (2.x)
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filtered_values=filter(lambdai: i!= x, sequence)
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# generators don't create another list
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filtered_values= (valuefor valuein sequenceif value!= x)
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# Or (3.x)
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filtered_values=filter(lambdai: i!= x, sequence)
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# Or (2.x)
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filtered_values= itertools.ifilter(lambdai: i!= x, sequence)
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Possible side effects of modifying the original list
@@ -673,10 +697,6 @@ Modifying the original list can be risky if there are other variables referencin
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# replace the contents of the original list
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sequence[::]= [valuefor valuein sequenceif value!= x]
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# Or
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sequence[::]= (valuefor valuein sequenceif value!= x)
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# Or
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sequence[::]=filter(lambdavalue: value!= x, sequence)
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Modifying the values in a list
@@ -705,11 +725,6 @@ It's safer to create a new list object and leave the original alone.
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# assign the variable "a" to a new list without changing "b"
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a= [i+3for iin a]
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# Or (Python 2.x):
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a=map(lambdai: i+3, a)
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# Or (Python 3.x)
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a=list(map(lambdai: i+3, a))
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Use:py:func:`enumerate` keep a count of your place in the list.
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