Sequences and comprehensions

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We can write elegant and compact code with sequences. First we will see how to scan sequences with iterators, and then how to build them with comprehensions of lists.

What to do

1. Unzip exercises zip in a folder, you should obtain something like this:

sequences
    sequences1.ipynb
    sequences1-sol.ipynb
    sequences2-chal.ipynb
    jupman.py

WARNING: to correctly visualize the notebook, it MUST be in an unzipped folder !

2. open Jupyter Notebook from that folder. Two things should open, first a console and then a browser. The browser should show a file list: navigate the list and open the notebook sequences.ipynb

3. Go on reading the exercises file, sometimes you will find paragraphs marked Exercises which will ask to write Python commands in the following cells.

Shortcut keys:

• to execute Python code inside a Jupyter cell, press Control + Enter

• to execute Python code inside a Jupyter cell AND select next cell, press Shift + Enter

• to execute Python code inside a Jupyter cell AND a create a new cell aftwerwards, press Alt + Enter

• If the notebooks look stuck, try to select Kernel -> Restart

Iterables - lists

When dealing with loops with often talked about iterating sequences, but what does it exactly mean for a sequence to be iterable ? Concretely, it means we can call the functioniter on that sequence.

Let’s try for example with familiar lists:

iter(['a','b','c','d'])

<list_iterator at 0x7f8e886ef8d0>

We notice Python just created an object of type list_iterator.

NOTE: the list was not shown!

You can imagine an iterator as a sort of still machine, that each time is activated it produces an element from the sequence, one at a time

Typically, an iterator only knows its position inside the sequence, and can provide us with the sequence elements one by one if we keep asking with calls to the function next:

iterator = iter(['a','b','c','d'])

next(iterator)

'a'

next(iterator)

'b'

next(iterator)

'c'

next(iterator)

'd'

Note how the iterator has a state to keep track of where it is in the sequence (in other words, it’s stateful). The state is changed at each call of function next.

If we try asking more elements of the available ones, Python raises the exception StopIteration:

next(iterator)

---------------------------------------------------------------------------
StopIteration                             Traceback (most recent call last)
<ipython-input-65-4518bd5da67f> in <module>()
----> 1 next(iterator)

StopIteration:

V COMMANDMENT You shall never ever redefine next and iter system functions.

DO NOT use them as variables !!

iterables - range

We iterated a list, which is a completely materialized in memory sequence we scanned with the iterator object. There are also other peculiar sequences which are not materialized in memory, like for example range.

Previously we used range in for loops to obtain a sequence of numbers, but exactly, what is range doing? Let’s try calling it on its own:

range(4)

range(0, 4)

Maybe we expected a sequence of numbers, instead, Python is showing us an object of type range (with the lower range limit).

NOTE: No number sequence is currently present in memory

We only have a ‘still’ iterable object, which if we want can provide us with numbers

How can we ask for numbers?

We’ve seen we can use a for loop:

for x in range(4):
    print(x)

0
1
2
3

As an alternative, we can pass range to the function iter which produces an iterator.

WARNING: range is iterable but it is NOT an iterator !!

To obtain the iterator we must call the iter function on the range object

iterator = iter(range(4))

iter also produces a ‘still’ object, which hasn’t materialized numbers in memory yet:

iterator

<range_iterator at 0x7f8e88783030>

In order to ask we must use the function next:

next(iterator)

0

next(iterator)

1

next(iterator)

2

next(iterator)

3

Note the iterator has a state, which is changed at each next call to keep track of where it is in the sequence.

If we try asking for more elements than actually available, Python raises a StopIteration exception:

next(iterator)

---------------------------------------------------------------------------
StopIteration                             Traceback (most recent call last)
<ipython-input-65-4518bd5da67f> in <module>()
----> 1 next(iterator)

StopIteration:

Materializing a sequence

We said a range object does not physically materialize in memory all the numbers at the same time. We can get them one by one by only using the iterator. What if we wanted a list with all the numbers? In the tutorial on lists we’ve seen that by passing a sequence to function list, a new list is created with all the sequence elements. We talked generically about a sequence, but the more correct term would have been iterable.

If we pass any iterable object to list, then a new list will be built - we’ve seen range is iterable so let’s try:

list(range(4))

[0, 1, 2, 3]

Voilà ! Now the sequence is all physically present in memory.

WARNING: list consumes the iterator!

If you try calling twice list on the same iterator, you will get an empty list:

sequence = range(4)
iterator = iter(sequence)

new1 = list(iterator)

new1

[0, 1, 2, 3]

new2 = list(iterator)

new2

[]

What if we wanted to directly access a specific position in the sequence generated by the iterator? Let’s try extracting the character at index 2:

sequence = range(4)
iterator = iter(sequence)

iterator[2]
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-129-3c080cc9e700> in <module>()
      1 sequence = range(4)
      2 iterator = iter(sequence)
----> 3 iterator[3]

TypeError: 'range_iterator' object is not subscriptable

… sadly we get an error!

We are left with only two alternatives. Either:

1. First we convert to list and then use the squared brackets

2. We call next 4 times (remember indexes start from zero)

Option a) very often looks handy, but careful: converting an iterator into a list creates a NEW list in memory. If the list is very big and/or this operation is repeated many times, you risk occupying memory for nothing.

Let’s see the example in Python Tutor again:

# WARNING: FOR PYTHON TUTOR TO WORK, REMEMBER TO EXECUTE THIS CELL with Shift+Enter
#          (it's sufficient to execute it only once)

import jupman

sequence = range(4)
iterator = iter(sequence)
new1 = list(iterator)
new2 = list(iterator)

jupman.pytut()

Python Tutor visualization

QUESTION: Which object occupies more memory? a or b?

a = range(10)
b = range(10000000)

Show answer

ANSWER: they both occupy the same amount of memory.

QUESTION: Which object occupies more memory? a or b ?

a = list(range(10))
b = list(range(10000000))

Show answer

ANSWER: b occupies more (the list is materialized)

Questions - range

Look at the following expressions, and for each try guessing the result (or if it gives an error):

1. range(3)
   

2. range()
   

3. list(range(-3))
   

4. range(3,6)
   

5. list(range(5,4))
   

6. list(range(3,3))
   

7. range(3) + range(6)
   

8. list(range(3)) + list(range(6))
   

9. list(range(0,6,2))
   

10. list(range(9,6,-1))
    

reversed

reversed is a function which takes a sequence as parameter and PRODUCES a NEW iterator which allows to run through the sequence in reverse order.

WARNING: by calling reversed we directly obtain an iterator !

So you do not need to make further calls to iter as done with range!

Let’s have a better look with an example:

la = ['s','c','a','n']

reversed(la)

<list_reverseiterator at 0x7f8e886ad9d0>

We see reversed has produced an iterator as result (not a reversed list)

INFO: iterators occupy a small amount of memory

Creating an iterator from a sequence only creates a sort of pointer, it does not create new memory regions.

Furthermore , we see the original list associated to la was not changed:

print(la)

['s', 'c', 'a', 'n']

WARNING: the function reversed is different from reverse method

Note the final d! If we tried to call it as a method we would get an error:

>>> la.reversed()

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-182-c8d1eec57fdd> in <module>
----> 1 la.reversed()

AttributeError: 'list' object has no attribute 'reversed'

Iterating with next

How can we obtain a reversed list in memory? In other words, how can we actionate the iterator machine?

We can ask the iterator for one element at a time with the function next:

la = ['a','b','c']

iterator = reversed(la)

next(iterator)

'c'

next(iterator)

'b'

next(iterator)

'a'

Once the iterator is exhausted, by calling next again we will get an error:

next(iterator)

---------------------------------------------------------------------------
StopIteration                             Traceback (most recent call last)
<ipython-input-248-4518bd5da67f> in <module>
----> 1 next(iterator)

StopIteration:

Let’s try manually creating a destination list lb and adding elements we obtain one by one:

la = ['a','b','c']
iterator = reversed(la)
lb = []
lb.append(next(iterator))
lb.append(next(iterator))
lb.append(next(iterator))
print(lb)

jupman.pytut()

['c', 'b', 'a']

Python Tutor visualization

Exercise - sconcerto

Write some code which given a list of characters la, puts in a list lb all the characters at odd position taken from reversed list la.

• use reversed and next

• DO NOT modify la

• DO NOT use negative indexes

• DO NOT use list

Example - given:

#      8    7    6    5    4    3    2    1    0
la = ['s', 'c', 'o', 'n', 'c', 'e', 'r', 't', 'o']
lb = []

After your code it must show:

>>> print(lb)
['t', 'e', 'n', 'c']
>>> print(la)
['s', 'c', 'o', 'n', 'c', 'e', 'r', 't', 'o']

We invite you to solve the problem in several ways:

WAY 1 - without cycle: Suppose the list length is fixed, and repeatedly call next without using a loop

WAY 2 - while: Suppose having a list of arbitrary length, and try generalizing previous code by using a while cycle, and calling next inside

• HINT 1: keep track of the position in which you are with a counter i

• HINT 2: you cannot call len on an iterator, so in the while conditions you will have to use the original list length

WAY 3 - for: this is the most elegant way. Suppose having a list of arbitrary length and use a loop like for x in reversed(la)

• HINT: you will still need to keep track of the position in which you are with an i counter

Show solution

# WAY 1: MANUAL

#      8    7    6    5    4    3    2    1    0
la = ['s', 'c', 'o', 'n', 'c', 'e', 'r', 't', 'o']
lb = []

# write here

iterator = reversed(la)

next(iterator)
lb.append(next(iterator))
next(iterator)
lb.append(next(iterator))
next(iterator)
lb.append(next(iterator))
next(iterator)
lb.append(next(iterator))
print(lb)

#jupman.pytut()

['t', 'e', 'n', 'c']

# WAY 1: MANUAL

#      8    7    6    5    4    3    2    1    0
la = ['s', 'c', 'o', 'n', 'c', 'e', 'r', 't', 'o']
lb = []

# write here

Show solution

# WAY 2: WHILE

#      8    7    6    5    4    3    2    1    0
la = ['s', 'c', 'o', 'n', 'c', 'e', 'r', 't', 'o']
lb = []

# write here

iterator = reversed(la)

i = 1
while i < len(la):
    if i % 2 == 1:
        next(iterator)
        lb.append(next(iterator))
    i += 2

print(lb)

['t', 'e', 'n', 'c']

# WAY 2: WHILE

#      8    7    6    5    4    3    2    1    0
la = ['s', 'c', 'o', 'n', 'c', 'e', 'r', 't', 'o']
lb = []

# write here

Show solution

# WAY 3: for

#      8    7    6    5    4    3    2    1    0
la = ['s', 'c', 'o', 'n', 'c', 'e', 'r', 't', 'o']
lb = []

# write here
i = 0

for x in reversed(la):
    if i % 2 == 1:
        lb.append(x)
    i += 1
print(lb)

['t', 'e', 'n', 'c']

# WAY 3: for

#      8    7    6    5    4    3    2    1    0
la = ['s', 'c', 'o', 'n', 'c', 'e', 'r', 't', 'o']
lb = []

# write here

Materializing an iterator

Luckily enough, we can obtain a list from an iterator with a less laborious method.

We’ve seen that when we want to create a new list from a sequence, we can use list as if it were a function. We can also do it in this case, interpreting the iterator as if it were a sequence:

la = ['s', 'c', 'a', 'n']
list( reversed(la) )

['n', 'a', 'c', 's']

Notice we generated a NEW list, the original one associated to la is always the same:

la

['s', 'c', 'a', 'n']

Let’s see what happens using Python Tutor (we created some extra variables to evidence relevant passages):

la = ['s', 'c', 'a', 'n']
iterator = reversed(la)
new = list(iterator)
print("la is",la)
print("new is",new)

jupman.pytut()

la is ['s', 'c', 'a', 'n']
new is ['n', 'a', 'c', 's']

Python Tutor visualization

QUESTION Which effect is the following code producing?

la = ['b','r','i','d','g','e']
lb = list(reversed(reversed(la)))

sorted

The function sorted takes as parameter a sequence and returns a NEW sorted list.

WARNING: sorted returns a LIST, not an iterator!

sorted(['g','a','e','d','b'])

['a', 'b', 'd', 'e', 'g']

WARNING: sorted is a function different from sort method !

Note the final ed! If we tried to call it with a different method we would get an error:

>>> la.sorted()

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-182-c8d1eec57fdd> in <module>
----> 1 la.reversed()

AttributeError: 'list' object has no attribute 'sorted'

Exercise - reversort

✪ Given a list of names, write some code to produce a list sorted in reverse

There are at least a couple of ways to do it in a single line of code, find them both

• INPUT: ['Maria','Paolo','Giovanni','Alessia','Greta']

• OUTPUT: ['Paolo', 'Maria', 'Greta', 'Giovanni', 'Alessia']

Show solution

# write here

list(sorted(['Maria','Paolo','Giovanni','Alessia','Greta'], reverse=True))

# or
#list(reversed(sorted(['Maria','Paolo','Giovanni','Alessia','Greta'])))

['Paolo', 'Maria', 'Greta', 'Giovanni', 'Alessia']

# write here

zip

Suppose we have two lists paintings and years, with rispectively names of famous paintings and the dates in which they were painted:

paintings = ["The Mona Lisa", "The Birth of Venus", "Sunflowers"]
years = [1503, 1482, 1888]

We want to produce a new list which contains some tuples which associate each painting with the year it was made:

[('The Mona Lisa', 1503),
 ('The Birth of Venus', 1482),
 ('Sunflowers', 1888)]

There are various ways to do it but certainly the most elegant is by using the function zip which produces an iterator:

zip(paintings, years)

<zip at 0x7f8e88550c80>

Even if you don’t see written ‘iterator’ in the object name, we can still use it as such with next:

iterator = zip(paintings, years)
next(iterator)

('The Mona Lisa', 1503)

next(iterator)

('The Birth of Venus', 1482)

next(iterator)

('Sunflowers', 1888)

As done previously, we can convert everything to a list with list:

paintings = ["The Mona Lisa", "The Birth of Venus", "Sunflowers"]
years = [1503, 1482, 1888]

list(zip(paintings,years))

[('The Mona Lisa', 1503), ('The Birth of Venus', 1482), ('Sunflowers', 1888)]

If the lists have different length, the sequence produced by zip will be as long as the shortest input sequence:

list(zip([1,2,3], ['a','b','c','d','e']))

[(1, 'a'), (2, 'b'), (3, 'c')]

If we will, we can pass an arbitrary number of sequences - for example, by passing three of them we will obtain triplets of values:

songs = ['Imagine', 'Hey Jude', 'Satisfaction', 'Yesterday' ]
authors = ['John Lennon','The Beatles', 'The Rolling Stones', 'The Beatles']
years = [1971, 1968, 1965, 1965]
list(zip(songs, authors, years))

[('Imagine', 'John Lennon', 1971),
 ('Hey Jude', 'The Beatles', 1968),
 ('Satisfaction', 'The Rolling Stones', 1965),
 ('Yesterday', 'The Beatles', 1965)]

Exercise - ladder

Given a number n, create a list of tuples that for each integer number \(x\) such that \(0 \leq x \leq n\) associates the number \(n - x\)

• INPUT: n=5

• OUTPUT: [(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)]

Show solution

n = 5
# write here
list(zip(range(n), reversed(range(n))))

[(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)]

n = 5
# write here

List comprehensions

List comprehensions are handy when you need to generate a NEW list by executing the same operation on all the elements of a sequence. Comprehensions start and end with square brackets [ ] so theit syntax reminds lists, but inside they contain a special for to loop inside a sequence:

numbers = [2,5,3,4]

doubled = [x*2 for x in numbers]

doubled

[4, 10, 6, 8]

Note the variable numbers is still associated to the original list:

numbers

[2, 5, 3, 4]

What happened ? We wrote the name of a variable x we just invented, and we told Python to go through the list numbers: at each iteration, the variable x is associated to a different value of the list numbers. This value can be reused in the expression we wrote on left of the for, which in this case is x*2

As name for the variable we used x, but we could have used any other name, for example this code is equivalent to the previous one:

numbers = [2,5,3,4]

doubled = [number * 2 for number in numbers]

doubled

[4, 10, 6, 8]

On the left of the for we can write any expression which produces a value, for example here we write x + 1 to increment all the numbers of the original list:

numbers = [2,5,3,4]

augmented = [x + 1 for x in numbers]

augmented

[3, 6, 4, 5]

QUESTION: What is this code going to produce? If we visualize it in Python Tutor, will la and lb point to different objects?

la = [7,5,6,9]
lb = [x for x in la]

Show answer

ANSWER: When [x for x in la] is executed, during the first iteration x is valued 7, during the second 5, during the third one 6 and so on and so forth. In the expression on the left of the for we put only x, so as expression result we will get the same identical number taken from the original string.

The code will produce a NEW list [7,5,6,9] and it will be associated to the variable lb.

la = [7,5,6,9]
lb = [x for x in la]

jupman.pytut()

Python Tutor visualization

List comprehensions on strings

QUESTION: What is this code going to produce?

[x for x in 'question']

Show answer

ANSWER: It will produce ['q', 'u', 'e', 's', 't', 'i', 'o', 'n']

Since question is a string, if we interpret it as a sequence each element of it is a character, so during the first iteration x is valued 'q', during the second 'u', during the third 'e' and so on and so forth. In the expression on the left of the for we put only x, so as expression result we will obtain the same identical character taken from the original string.

Let’s now suppose to have a list of animals and we want to produce another one with the same names as uppercase. We can do it in a compact way with a list comprehension like this:

animals = ['dogs', 'cats', 'squirrels', 'elks']

new_list = [animal.upper() for animal in animals]

new_list

['DOGS', 'CATS', 'SQUIRRELS', 'ELKS']

In the left part reserved to the expression we used the method .upper() on the string variable animal. We know strings are immutable, so we’re sure the method call produces a NEW string. Let’s see what happened with Python Tutor:

animals = ['dogs', 'cats', 'squirrels', 'elks']

new_list = [animal.upper() for animal in animals]

jupman.pytut()

Python Tutor visualization

✪ EXERCISE: Try writing here a list comprehension to put all characters as lowercase (.lower() method)

Show solution

animals = ['doGS', 'caTS', 'SQUIrreLs', 'ELks']

# write here

[animal.lower() for animal in animals]

['dogs', 'cats', 'squirrels', 'elks']

animals = ['doGS', 'caTS', 'SQUIrreLs', 'ELks']

# write here

Questions - List comprehensions

Look at the following code fragments, and for each try guessing the result it produces (or if it gives an error):

1. [x for [4,2,5]]
   

2. x for x in range(3)
   

3. [x for y in 'cartoccio']
   

4. [for x in 'zappa']
   

5. [for [3,4,5]]
   

6. [k + 1 for k in 'bozza']
   

7. [k + 1 for k in range(5)]
   

8. [k > 3 for k in range(7)]
   

9. [s + s for s in ['lam','pa','da']]
   

10. la = ['x','z','z']
    [x for x in la] + [y for y in la]
    

11. [x.split('-') for x in ['a-b', 'c-d', 'e-f']]
    

12. ['@'.join(x) for x in [['a','b.com'],['c','d.org'],['e','f.net'] ]]
    

13. ['z' for y in 'borgo'].count('z') == len('borgo')
    

14. m = [['a','b'],['c','d'],['e','f'] ]
    la = [x.pop() for x in m]   # not advisable - why ?
    print(' m:', m)
    print('la:',la)
    

Exercises - list comprehension

Exercise - Bubble bubble

✪ Given a list of strings, produce a sequence with all the strings replicated 4 times

• INPUT: ['chewing','gum','bubble']

• OUTPUT: ['chewingchewingchewingchewing', 'gumgumgumgum', 'bubblebubblebubblebubble']

Show solution

import math

bubble_bubble = ['chewing','gum','bubble']

# write here
[x*4 for x in bubble_bubble]

['chewingchewingchewingchewing', 'gumgumgumgum', 'bubblebubblebubblebubble']

import math

bubble_bubble = ['chewing','gum','bubble']

# write here

Exercise - root

✪ Given a list of numbers, produce a list with the square root of the input numbers

• INPUT: [16,25,81]

• OUTPUT: [4.0, 5.0, 9.0]

Show solution

import math

# write here
[math.sqrt(x) for x in [16,25,81]]

[4.0, 5.0, 9.0]

import math

# write here

Exercise - When The Telephone Rings

✪ Given a list of strings, produce a list with the first characters of each string

• INPUT: ['When','The','Telephone','Rings']

• OUTPUT: ['W', 'T', 'T', 'R']

Show solution

# write here
[x[0] for x in ['When','The','Telephone','Rings']]

['W', 'T', 'T', 'R']

# write here

Exercise - don’t worry

✪ Given a list of strings, produce a list with the lengths of all the lists

• INPUT: ["don't", 'worry','and', 'be','happy']

• OUTPUT: [5, 5, 3, 2, 5]

Show solution

# write here
[len(x) for x in ["don't", 'worry','and', 'be','happy']]

[5, 5, 3, 2, 5]

# write here

Exercise - greater than 3

✪ Given a list of numbers, produce a list with True if the corresponding element is greater than 3, False otherwise

• INPUT: [4,1,0,5,0,9,1]

• OUTPUT: [True, False, False, True, False, True, False]

Show solution

# write here
[x > 3 for x in [4,1,0,5,0,9,1]]

[True, False, False, True, False, True, False]

# write here

Exercise - even

✪ Given a list of numbers, produce a list with True if the corresponding element is even

• INPUT: [3,2,4,1,5,3,2,9]

• OUTPUT: [False, True, True, False, False, False, True, False]

Show solution

# write here
[x % 2 == 0 for x in [3,2,4,1,5,3,2,9]]

[False, True, True, False, False, False, True, False]

# write here

Exercise - both ends

✪ Given a list of strings having at least two characters each, produce a list of strings with the first and last characters of each

• INPUT: ['departing', 'for', 'the', 'battlefront']

• OUTPUT: ['dg', 'fr', 'te', 'bt']

Show solution

# write here
[x[0] + x[-1] for x in ['departing', 'for', 'the', 'battlefront']]

['dg', 'fr', 'te', 'bt']

# write here

Exercise - dashes

✪ Given a list of lists of characters, produce a list of strings with characters separated by dashes

• INPUT: [['a','b'],['c','d','e'], ['f','g']]

• OUTPUT: ['a-b', 'c-d-e', 'f-g']

Show solution

# write here
['-'.join(x) for x in [['a','b'],['c','d','e'], ['f','g']]]

['a-b', 'c-d-e', 'f-g']

# write here

Exercise - lollosa

✪ Given a string s, produce a list of tuples having for each character the number of occurrences of that character in the string

• INPUT: s = 'lollosa'

• OUTPUT: [('l', 3), ('o', 2), ('l', 3), ('l', 3), ('o', 2), ('s', 1), ('a', 1)]

Show solution

s = 'lollosa'
# write here
[(car, s.count(car)) for car in s]

[('l', 3), ('o', 2), ('l', 3), ('l', 3), ('o', 2), ('s', 1), ('a', 1)]

s = 'lollosa'
# write here

Exercise - dog cat

✪ Given a list of strings of at least two characters each, produce a list with the strings without intial and final characters

• INPUT: ['donkey','eagle','ox', 'dog' ]

• OUTPUT: ['onke', 'agl', '', 'o']

Show solution

# write here
[x[1:-1] for x in ['donkey','eagle','ox', 'dog' ]]

['onke', 'agl', '', 'o']

# write here

Exercise - smurfs

✪ Given some names produce a list with the names sorted alphabetically and all in uppercase

• INPUT: ['Brainy', 'Hefty', 'Smurfette', 'Clumsy']

• OUTPUT: ['BRAINY', 'CLUMSY', 'HEFTY', 'SMURFETTE']

Show solution

# write here
[x.upper() for x in sorted(['Brainy', 'Hefty', 'Smurfette', 'Clumsy'])]

['BRAINY', 'CLUMSY', 'HEFTY', 'SMURFETTE']

# write here

Exercise - precious metals

✪ Given two lists values and metals produce a list containing all the couples value-metal as tuples

INPUT:

values = [10,25,50]
metals = ['silver','gold','platinum']

OUTPUT: [(10, 'silver'), (25, 'gold'), (50, 'platinum')]

Show solution

values = [10,25,50]
metals = ['silver','gold','platinum']

# write here
list(zip(values, metals))

[(10, 'silver'), (25, 'gold'), (50, 'platinum')]

values = [10,25,50]
metals = ['silver','gold','platinum']

# write here

Filtered list comprehensions

During the construction of a list comprehension we can filter the elements taken from the sequence by using an if. For example, the following expression takes from the sequence only numbers greater than 5:

[x for x in [7,4,8,2,9] if x > 5]

[7, 8, 9]

After the if we can put any expression which reuses the variable on which we are iterating, for example if we are iterating a string we can keep only the uppercase characters:

[x for x in 'The World Goes Round' if x.isupper()]

['T', 'W', 'G', 'R']

WARNING: else is not supported

For example, writing this generates an error:

[x for x in [7,4,8,2,9] if x > 5 else x + 1]   # WRONG!

File "<ipython-input-74-9ba5c135c58c>", line 1
    [x for x in [7,4,8,2,9] if x > 5 else x + 1]
                                        ^
SyntaxError: invalid syntax

Questions - filtered list comprehensions

Look at the following code fragments, and for each try guessing the result it produces (or if it gives an error):

1. [x for x in range(100) if False]
   

2. [x for x in range(3) if True]
   

3. [x for x in range(6) if x > 3 else 55]
   

4. [x for x in range(6) if x % 2 == 0]
   

5. [x for x in {'a','b','c'}]  # careful about ordering
   

6. [x for x in [[5], [2,3], [4,2,3], [4]] if len(x) > 2]
   

7. [(x,x) for x in 'xyxyxxy' if x != 'x' ]
   

8. [x for x in ['abCdEFg'] if x.upper() == x]
   

9. la = [1,2,3,4,5]
   [x for x in la if x > la[len(la)//2]]
   

Exercises - filtered list comprehensions

Exercise - savannah

Given a list of strings, produce a list with only the strings of length greater than 6:

• INPUT: ['zebra', 'leopard', 'giraffe', 'gnu', 'rhinoceros', 'lion']

• OUTPUT: ['leopard', 'giraffe', 'rhinoceros']

Show solution

# write here
[x for x in ['zebra', 'leopard', 'giraffe', 'gnu', 'rhinoceros', 'lion'] if len(x) > 6]

['leopard', 'giraffe', 'rhinoceros']

# write here

Exercise - puZZled

Given a list of strings, produce a list with only the strings which contain at least a 'z'. The selected strings must be transformed so to place the Z in uppercase.

• INPUT: ['puzzled', 'park','Aztec', 'run', 'mask', 'zodiac']

• OUTPUT: ['puZZled', 'AZtec', 'Zodiac']

[x.replace('z','Z') for x in ['puzzled', 'park','Aztec', 'run', 'mask', 'zodiac'] if 'z' in  x]

Exercise - Data science

Produce a string with the words of the input string alternated uppercase / lowercase

• INPUT:

phrase = """Data science is an interdisciplinary field
that uses scientific methods, processes, algorithms and systems
to extract knowledge and insights from noisy, structured
and unstructured data, and apply knowledge and actionable insights
from data across a broad range of application domains."""

• OUTPUT (only one line):

  DATA science IS an INTERDISCIPLINARY field THAT uses SCIENTIFIC methods, PROCESSES, algorithms AND systems TO extract KNOWLEDGE and INSIGHTS from NOISY, structured AND unstructured DATA, and APPLY knowledge AND actionable INSIGHTS from DATA across A broad RANGE of APPLICATION domains.
  

✪✪✪ WRITE ONLY ONE code line

✪✪✪✪ USE ONLY ONE list comprehension

Show solution

phrase = """Data science is an interdisciplinary field
that uses scientific methods, processes, algorithms and systems
to extract knowledge and insights from noisy, structured
and unstructured data, and apply knowledge and actionable insights
from data across a broad range of application domains."""

# write here

print(' '.join(
    [t[0].upper() + ' ' + t[1] for t in  zip(phrase.split()[::2], phrase.split()[1::2])]))

# or
#print(' '.join([phrase.split()[i].upper() + ' ' + phrase.split()[i + 1] for i in range(0,len(phrase.split())-1,2)]))

DATA science IS an INTERDISCIPLINARY field THAT uses SCIENTIFIC methods, PROCESSES, algorithms AND systems TO extract KNOWLEDGE and INSIGHTS from NOISY, structured AND unstructured DATA, and APPLY knowledge AND actionable INSIGHTS from DATA across A broad RANGE of APPLICATION domains.

phrase = """Data science is an interdisciplinary field
that uses scientific methods, processes, algorithms and systems
to extract knowledge and insights from noisy, structured
and unstructured data, and apply knowledge and actionable insights
from data across a broad range of application domains."""

# write here

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