Python: Lists (Solutions)¶

Note

In some cases I will use the character \ at the end of a line.

Used in this way, \ tells Python that the command continues in the following line. Didn’t I use \, Python could think that the command is complete, giving an error message if the syntax is wrong.

You can ignore these \.

Operations¶

Solution:

list = []
print list, len(list)         # check

Solution:

list = range(5)
print list, len(list)         # check
print len(list)

Solution:

list = [0] * 100
print list, len(list)         # check

Solution:

list_1 = range(10)
list_2 = range(10, 20)

list_complete = list_1 + list_2
print list_complete

print list_complete == range(20)   # True

Solution:

list = ["I am", "a", "list"]
print list, len(list)         # check

print len(list[0])
print len(list[1])
print len(list[2])

Solution:

list = [0.0, "b", [3], [4, 5]]

print len(list)                # 4

print type(list[0])            # float

print list[1], len(list[1])   # "b", 1

print list[2], len(list[2])   # [3], 1

print list[-1], len(list[-1]) # [4, 5], 2

print "b" in list              # True

print 4 in list                # False
print 4 in list[-1]            # True

Solution: the first is a list of integers, the second a list of strings, the third is a string!:

print type(list_1)             # list
print type(list_2)             # list
print type(list_3)             # str

Solutions:

# an empty list
list = []
print len(list)                # 0
del list


# invalid syntax, Python gives an error message
list = [}


# a list that contains an empty list
list = [[]]
print len(list)                # 1
print len(list[0])             # 0
del list


# the following doesn't work because the list is not defined!
list.append(0)


# this works
list = []
list.append(0)
print list                     # [0]
del list


# this doesn't work because we forgot to put commas!
list = [1 2 3]


# this gives an error message because the list has only 3 elements!
list = range(3)
print list[3]


# Extract the last element
list = range(3)
print list[-1]
del list


# Extract the first two elements (list[2], the third,
# is excluded)
list = range(3)
sublist = list[0:2]
print list
del list


# Extract all the elements(list[3], not existent
# is excluded)
list = range(3)
sublist = list[0:3]
print list
del list


# Extract the first two elements (list[-1], the third,
# is excluded)
list = range(3)
sublist = list[0:-1]
print list
del list


# Insert in third position the string "two"
list = range(3)
list[2] = "two"
print list
del list


# this doesn't work, the list contains only three elements,
# there is no fourth position, and Python gives an error
list = range(3)
list[3] = "three"


# insert in third posizion the string "three"
list = range(3)
list[-1] = "three"
print list
del list


# the index has to be an integer, Python gives an error message
list = range(3)
list[1.2] = "one point two"


# substitute the second element of list (i.e. 1)
# with a list of two strings; this is can be done,
# since lists *can* contain other lists
list = range(3)
list[1] = ["protein1", "protein2"]
print list
del list

Solution:

matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9],
]

first_row = matrix[0]
print first_row

second_element_first_row = first_row[1]
# or
second_element_first_row = matrix[0][1]
print second_element_first_row

sum_first_row = matrix[0][0] + matrix[0][1] + matrix[0][2]
print sum_first_row

second_column = [matrix[0][1], matrix[1][1], matrix[2][1]]
print second_column

diagonal = [matrix[0][0], matrix[1][1], matrix[2][2]]
print diagonal

three_rows_together = matrix[0] + matrix[1] + matrix[2]
print three_rows_together

Methods¶

First, let’s create a list. For example, an empty list:
```
list = []
```
next, let’s add the required elements with append():
```
list.append(0)
list.append("text")
list.append([0, 1, 2, 3])
```

Solution:

# add one 3 at the end of the list
list = range(3)
list.append(3)
print list
del list

# add a list with a 3 at the end of the list
list = range(3)
list.append([3])
print list
del list

# add a 3 (the only element contained in the list [3])
# at the end of the list
list = range(3)
list.extend([3])
print list
del list

# doesn't work: extend() extends a list with the content of
# another list, but here 3 is *not* a list!
# Python gives an error message
list = range(3)
list.extend(3)

# replace the element in position 0, the first, with a 3
list = range(3)
list.insert(0, 3)
print list
del list

# insert a 3 at the end of list
list = range(3)
list.insert(3, 3)
print list
del list

# insert the list [3] at the end of list
list = range(3)
list.insert(3, [3])
print list
del list

# doesn't work: the first argument of insert() has to be an integer
# not a list! Python gives an error message
list = range(3)
list.insert([3], 3)

Solution:
```
list = []
list.append(range(10))
list.append(range(10, 20))
print list
```
Here we use append(), that inserts an element at the end of list. In this example we insert two lists, the results of range(10) and range(10, 20).

Clearly, len(list) is 2, since we added only 2 elements.

On the other hand:
```
list = []
list.extend(range(10))
list.extend(range(10, 20))
print list
```
here we use extend(), that extends a list with another list. Here the final list has 20 elements, as we can see with:
```
print len(list)
```

Solution:

list = [0, 0, 0, 0]
list.remove(0)
print list

only the first occurrence of 0 is removed!

Solution:

list = [1, 2, 3, 4, 5]

# invert the order of the elements of list
list.reverse()
print list

# order the elements of list
list.sort()
print list

After the two operations,``list`` gets back to the initial value.

On the other hand:

list = [1, 2, 3, 4, 5]
list.reverse().sort()

cannot be done, since the result of list.reverse() is None:

list = [1, 2, 3, 4, 5]
result = list.reverse()
print result

being result not a list,``sort()`` cannot be applied. Python gives an error message.

Let’s try this:

list = range(10)
inverse_list = list.reverse()

print list                         # modified!
print inverse_list                 # None!

the code doesn’t work: reverse() modifies list e returns None! Moreover, this code modifies directly``list``, and we don’t want that.

First, let’s create a copy of list, next we can order the copy:

list = range(10)
inverse_list = list[:]            # *not* inverse_list = list
inverse_list.reverse()
print list                         # unvaried
print inverse_list                 # inverted

On the other hand, this code:

list = range(10)
inverse_list = list
inverse_list.reverse()
print list                         # modified!
print inverse_list                # inverted

doesn’t work as we want: inverse_list doesn’t contain a copy of list, but a reference to the same object referred by list.

As a consequence, when we invert inverse_list we also invert list.

As before:

motifs = [
    "KSYK",
    "SVALVV"
    "GVTGI",
    "VGSSLAEVLKLPD",
]
sorted_motifs = motifs.sort()

the code doesn’t work: reverse() modifies motifs e returns None! First, let’s create a copy of motifs, next we can order the copy:

sorted_motifs = motifs[:]
fsorted_motifs.sort()
print motifs                    # unvaried
print sorted_motifs            # ordered

String-List Methods¶

Solution:

text = """The Wellcome Trust Sanger Institute
is a world leader in genome research.""""

words = text.split()
print len(words)

Solution:

table = [
    "protein | database | domain | start | end",
    "YNL275W | Pfam | PF00955 | 236 | 498",
    "YHR065C | SMART | SM00490 | 335 | 416",
    "YKL053C-A | Pfam | PF05254 | 5 | 72",
    "YOR349W | PANTHER | 353 | 414",
]

first_row = table[0]
almost_column_titles = first_row.split("|")
almost_column_titles
# ["protein ", " database ", ...]

# unfortunately, column titles contain superfluous spaces
# to remove them, we can change the argument of split()

column_titles = first_row.split(" | ")
print column_titles
# ["protein", "database", ...]

We could also use strip() together with a list comprehension on almost_column_titles, but it’s not necessary.

Solution:

words = ["word_1", "word_2", "word_3"]

print " ".join(words)

print ",".join(words)

print " e ".join(words)

print "".join(words)

backslash = r"\"
print backslash.join(words)

Solution:

verses = [
    "Taci. Su le soglie"
    "del bosco non odo"
    "parole che dici"
    "umane; ma odo"
    "parole piu' nuove"
    "che parlano gocciole e foglie"
    "lontane."
]

poem = "\n".join(verses)

List Comprehension¶

Solutions:

Solution:

list_plus_three = [number + 3 for number in list]

print list_plus_three             # check

Solution:

odds = [number for number in list
                 if (number % 2 == 1)]

Solution:

opposites = [-number for number in list]

Solution:

inverses = [1.0 / number for number in list
                 if number != 0]

Solution:
```
first_and_last = [list[0], list[-1]]
```

Solution:

from_second_to_penultimate = list[1:-1]

Solution:

list_odds = [number for number in list
                 if (number % 2 == 1)]
number_odds = len(list_odds)
print number_odds

or:

number_odds = len([number for number in list
                      if (number % 2 == 1)])

Solution:

list_divided_by_5 = [float(number) / 5
                      for number in list]

Solution:

list_multiples_5_divided = [float(number) / 5.0)
                           for number in list
                           if (number % 5 == 0)]

Solution:

list_of_strings = [str(number) for number in list]

Solution:

# As before, but iterating on `list_of_strings`
# rather than directly on `list`
number_odds = len([string for string in list_of_strings
                      if (int(string) % 5 == 0)])

Solution:
```
text = " ".join([str(number) for number in list])
```
Notice that if we forget to write``str(number)``, join() doesn’t work.

Solutions:

# forward
list_1 = [1, 2, 3]
list_2 = [str(x) for x in list_1]

# backward
list_2 = ["1", "2", "3"]
list_1 = [int(x) for x in list_2]

# forward
list_1 = ["name", "surname", "age"]
list_2 = [[x] for x in list_1]

# backward
list_2 = [["name"], ["surname"], ["age"]]
list_1 = [l[0] for l in list_2]

# forward
list_1 = ["ACTC", "TTTGGG", "CT"]
list_2 = [[x.lower(), len(x)] for x in list_1]

# backward
list_2 = [["actc", 4], ["tttgggcc", 6], ["ct", 2]]
list_1 = [l[0].upper() for l in list_2]

Solution:
1. [x for x in list]: creates a copy of list.
2. [y for y in list]: creates a copy of list (the same as before).
3. [y for x in list]: invalid. (If x represents an element of the list, what is y?)
4. ["x" for x in list]: creates a list full of strings "x" as long as list. The result will be: ["x", "x", ..., "x"].
5. [str(x) for x in list]: for each int x in list, x is converted to a string str(x) and included in the resulting list. The result will be: ["0", "1", ..., "9"].
6. [x for str(x) in list]: invalid: the transformation str(...) is in the wrong place!
7. [x + 1 for x in list]: for each int x in list, adds 1 with x + 1 and includes the result in the resulting list. The result will be: [1, 2, ..., 10].
8. [x + 1 for x in list if x == 2]: for each int x in list checks whether it is equal to 2. In the positive case x + 1 is included in the resulting list, otherwise it’s excluded. The result will be’: [3].

Solution:

clusters = """\
>Cluster 0
0 >YLR106C at 100.00%
>Cluster 50
0 >YPL082C at 100.00%
>Cluster 54
0 >YHL009W-A at 90.80%
1 >YHL009W-B at 100.00%
2 >YJL113W at 98.77%
3 >YJL114W at 97.35%
>Cluster 52
0 >YBR208C at 100.00%
"""

rows = clusters.split("\n")

In order to get cluster names, we have to keep only lines starting with ">", and for each of them apply split() in order to get the second element (the nam of the cluster):

cluster_names = [row.split()[1] for row in rows
                if row.startswith(">")]

In order to get protein names, we have to keep only lines not starting with ">", and for each of them apply split() and keep the second element (also removing ">" from the name of the protein):

proteins = [row.split()[1].lstrip(">") for row in rows
            if not row.startswith(">")]

In order to get protein-percentage pairs, we have to keep only lines not starting with ">". On each of them, we apply split() and keep the second element (protein name) and the last (percentage):

protein_percentage_pairs = \
    [[row.split()[1].lstrip(">"), row.split()[-1].rstrip("%")]
     for row in rows if not row.startswith(">")]

Annotated version:

protein_percentage_pairs = \
    [[row.split()[1].lstrip(">"), row.split()[-1].rstrip("%")]
#     ^^^^^^^^^^^^^^^^^^^^^^^^^^^  ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
#       protein name, as before            percentage
#    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
#                     protein-percentage pair
     for row in rows if not row.startswith(">")]

Solution:

matrix = [range(0,3), range(3,6), range(6,9)]


# extract first row
first_row = matrix[0]


# extract first column
first_column = [matrix[0][i] for i in range(3)]


# invert row order
upside_down = matrix[:]
upside_down.reverse()
# or
upside_down = [matrix[2-i] for i in range(3)]


# invert column order
palindrome = []
# append the first line
palindrome.append([matrix[0][2-i] for i in range(3)])
# append the second line
palindrome.append([matrix[1][2-i] for i in range(3)])
# append the third line
palindrome.append([matrix[2][2-i] for i in range(3)])

# or in a single step -- it's complicated and you can ignore it!!!
palindrome = [[row[2-i] for i in range(3)]
               for row in matrix]


# we can re-create matrix with a single list comprehension
matrix_again = [range(i, i+3) for i in range(9)
                    if i % 3 == 0]

Solutions:

list = range(100)

squares = [number**2 for number in list]

difference_of_squares = \
    [squares[i+1] - squares[i]
     for i in range(len(squares) - 1)]

Solutions:

mouse_genes = ["Fus", "Tdp43", "Sod1", "Ighmbp2", "Srsf2"]

sorted_mouse_genes = mouse_genes[:]
sorted_mouse_genes.sort()
print sorted_mouse_genes

human_genes = [gene.upper() for gene in sorted_mouse_genes]