5   Collection Data Types

5.1   Exercises

5.1.1   Exercise

Draw the representation in memory of the following expressions.
what is the data type of each object?
x = [1, 2, 3, 4]
y = x[1]
y = 3.14
x[1] = 'foo'
set
x = [1, 2, 3, 4]
x += [5, 6]
set
>>> x = [1, 2, 3, 4]
>>> id(x)
139950507563632
>>> x += [5,6]
>>> id(x)
139950507563632

With mutable object like list when we mutate the object the state of the object is modified. But the reference to the object is still unchanged. So in this example we have two ways to access to the list [1,2] if we modify the state of the list itself. but not the references to this object, then the 2 variables x and y still reference the list containing [1,2,3,4].

compare with the exercise on string and integers:

Since list are mutable, when += is used the original list object is modified, so no rebinding of x is necessary. We can observe this using id() which give the memory adress of an object. This adress does not change after the += operation.

Note

even the results is the same there is a subtelty to use augmented operator. in a operator= b python looks up a ’s value only once, so it is potentially faster than the a = a operator b.

compare

x = 3
y = x
y += 3
x = ?
y = ?
augmented_assignment

and

x = [1,2]
y = x
y += [3,4]
x = ?
y = ?
list extend

5.1.2   Exercise

wihout using python shell, what is the results of the following statements:

Note

sum is a function which return the sum of each elements of a list.

x = [1, 2, 3, 4]
x[3] = -4 # what is the value of x now ?
y = sum(x)/len(x) #what is the value of y ? why ?

y = 0.5

Warning

In python2 the result is

y = 0

because sum(x) is an integer, len(x) is also an integer so in python2.x the result is an integer, all the digits after the periods are discarded.

5.1.3   Exercise

Draw the representation in memory of the following expressions.

x = [1, ['a','b','c'], 3, 4]
y = x[1]
y[2] = 'z'
# what is the value of x ?
set
_images/spacer.png

When we execute y = x[1], we create y wich reference the list ['a', 'b', 'c']. This list has 2 references on it: y and x[1] .

set
_images/spacer.png

This object is a list so it is a mutable object. So we can access and modify it by the two ways y or x[1]

x = [1, ['a','b','z'], 3, 4]

5.1.4   Exercise

from the list l = [1, 2, 3, 4, 5, 6, 7, 8, 9] generate 2 lists l1 containing all odd values, and l2 all even values.:

l = [1, 2, 3, 4, 5, 6, 7, 8, 9]
l1 = l[::2]
l2 = l[1::2]

5.1.5   Exercise

generate a list containing all codons.

5.1.5.1   pseudocode:

function all_codons()
all_codons <- empty list
let varying the first base
for each first base let varying the second base
for each combination first base, second base let varying the third base
add the concatenation base 1 base 2 base 3 to all_codons
return all_codons

5.1.5.2   first implementation:

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def all_codons():
    all_codons = []
    alphabet = 'acgt'
    for base_1 in alphabet:
        for base_2 in alphabet:
            for base_3 in alphabet:
                codon = base_1 + base_2 + base_3
                all_codons.append(codon)
    return all_codons
python -i codons.py
>>> codons = all_codons()

codons.py .

5.1.5.3   second implementation:

Mathematically speaking the generation of all codons can be the cartesian product between 3 vectors ‘acgt’. In python there is a function to do that in itertools module: https://docs.python.org/3/library/itertools.html#itertools.product

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from itertools import product


def all_codons():
    """
    :return: the 64 codons
    :rtype: list of string
    """
    alphabet = 'acgt'
    all_codons = [''.join(codon) for codon in product(alphabet, repeat=3)]
    return all_codons
python -i codons.py
>>> codons = all_codons()

codons_itertools.py .

5.1.6   Exercise

From a list return a new list without any duplicate, regardless of the order of items. For example:

>>> l = [5,2,3,2,2,3,5,1]
>>> uniqify(l)
>>> [1,2,3,5] #is one of the solutions

5.1.6.1   pseudocode:

function uniqify(l)
uniq <- empty list
for each element of l
add element if is not in uniq
return uniq

5.1.6.2   implementation:

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def uniqify(l):
    uniq = []
    for item in l:
       if item not in uniq:
          uniq.append(item)
    return (uniq)
>>> l=[1,2,3,2,3,4,5,1,2,3,3,2,7,8,9]
>>> uniqify(l)
[1, 2, 3, 4, 5, 7, 8, 9]

codons_itertools.py .

5.1.6.3   second implementation:

The problem with the first implementation come from the line 4. Remember that the membership operator uses a linear search for list, which can be slow for very large collections. If we plan to use uniqify with large list we should find a better algorithm. In the specification we can read that uniqify can work regardless the order of the resulting list. So we can use the specifycity of set

>>> list(set(l))

5.1.7   Exercise

We need to compute the occurrence of all kmers of a given length present in a sequence.

Below we propose 2 algorithms.

5.1.7.1   pseudo code 1

function get_kmer_occurences(seq, kmer_len)
all_kmers <- generate all possible kmer of kmer_len
occurences <- 0
for each kmer in all_kmers
count occurence of kmer
store occurence

5.1.7.2   pseudo code 2

function get_kmer_occurences(seq, kmer_len)
all_kmers <- empty
from i = 0 to sequence length - kmer_len
kmer <- kmer startin at pos i im sequence
increase by of occurence of kmer

Note

Computer scientists typically measure an algorithm’s efficiency in terms of its worst-case running time, which is the largest amount of time an algorithm can take given the most difficult input of a fixed size. The advantage to considering the worst case running time is that we are guaranteed that our algorithm will never behave worse than our worst-case estimate.

Big-O notation compactly describes the running time of an algorithm. For example, if your algorithm for sorting an array of n numbers takes roughly n2 operations for the most difficult dataset, then we say that the running time of your algorithm is O(n2). In reality, depending on your implementation, it may be use any number of operations, such as 1.5n2, n2 + n + 2, or 0.5n2 + 1; all these algorithms are O(n2) because big-O notation only cares about the term that grows the fastest with respect to the size of the input. This is because as n grows very large, the difference in behavior between two O(n2) functions, like 999 · n2 and n2 + 3n + 9999999, is negligible when compared to the behavior of functions from different classes, say O(n2) and O(n6). Of course, we would prefer an algorithm requiring 1/2 · n2 steps to an algorithm requiring 1000 · n2 steps.

When we write that the running time of an algorithm is O(n2), we technically mean that it does not grow faster than a function with a leading term of c · n2, for some constant c. Formally, a function f(n) is Big-O of function g(n), or O(g(n)), when f(n) <= c · g(n) for some constant c and sufficiently large n.

For more on Big-O notation, see A http://rob-bell.net/2009/06/a-beginners-guide-to-big-o-notation/Beginner’s.

Compare the pseudocode of each of them and implement the fastest one.

"""gtcagaccttcctcctcagaagctcacagaaaaacacgctttctgaaagattccacactcaatgccaaaatataccacag
   gaaaattttgcaaggctcacggatttccagtgcaccactggctaaccaagtaggagcacctcttctactgccatgaaagg
   aaaccttcaaaccctaccactgagccattaactaccatcctgtttaagatctgaaaaacatgaagactgtattgctcctg
   atttgtcttctaggatctgctttcaccactccaaccgatccattgaactaccaatttggggcccatggacagaaaactgc
   agagaagcataaatatactcattctgaaatgccagaggaagagaacacagggtttgtaaacaaaggtgatgtgctgtctg
   gccacaggaccataaaagcagaggtaccggtactggatacacagaaggatgagccctgggcttccagaagacaaggacaa
   ggtgatggtgagcatcaaacaaaaaacagcctgaggagcattaacttccttactctgcacagtaatccagggttggcttc
   tgataaccaggaaagcaactctggcagcagcagggaacagcacagctctgagcaccaccagcccaggaggcacaggaaac
   acggcaacatggctggccagtgggctctgagaggagaaagtccagtggatgctcttggtctggttcgtgagcgcaacaca"""

In the first alogrithm.

we first compute all kmers we generate 4kmer length
then we count the occurence of each kmer in the sequence
so for each kmer we read all the sequence so the algorith is in O( 4kmer length * sequence length)
In the secon algorithm we read the sequence only once
So the algorithm is in O(sequence length)

Compute the 6 mers occurences of the sequence above, and print each 6mer and it’s occurence one per line.

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def get_kmer_occurences(seq, kmer_len):
    """
    return a list of tuple 
    each tuple contains a kmers present in seq and its occurence
    """
    kmers = {}
    stop = len(seq) - kmer_len
    for i in range(stop + 1):
        kmer = s[i : i + kmer_len]
        kmers[kmer] = kmers.get(kmer, 0) + 1
    return kmers.items()
>>> s = """"gtcagaccttcctcctcagaagctcacagaaaaacacgctttctgaaagattccacactcaatgccaaaatataccacag
... gaaaattttgcaaggctcacggatttccagtgcaccactggctaaccaagtaggagcacctcttctactgccatgaaagg
... aaaccttcaaaccctaccactgagccattaactaccatcctgtttaagatctgaaaaacatgaagactgtattgctcctg
... atttgtcttctaggatctgctttcaccactccaaccgatccattgaactaccaatttggggcccatggacagaaaactgc
... agagaagcataaatatactcattctgaaatgccagaggaagagaacacagggtttgtaaacaaaggtgatgtgctgtctg
... gccacaggaccataaaagcagaggtaccggtactggatacacagaaggatgagccctgggcttccagaagacaaggacaa
... ggtgatggtgagcatcaaacaaaaaacagcctgaggagcattaacttccttactctgcacagtaatccagggttggcttc
... tgataaccaggaaagcaactctggcagcagcagggaacagcacagctctgagcaccaccagcccaggaggcacaggaaac
... acggcaacatggctggccagtgggctctgagaggagaaagtccagtggatgctcttggtctggttcgtgagcgcaacaca"""
>>> s = s.replace('\n', '')
>>> kmers = get_kmer_occurences(s, 6)
>>> for kmer in kmers:
>>>   print kmer[0], '..', kmer[1]
gcagag .. 2
aacttc .. 1
gcaact .. 1
aaatat .. 2

kmer.py .

5.1.7.3   bonus:

Print the kmers by ordered by occurences.

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import collections
from operator import itemgetter

def get_kmer_occurences(seq, kmer_len):
    """
    return a list of tuple 
    each tuple contains a kmers present in seq and its occurence
    """
    kmers = collections.defaultdict(int)
    stop = len(seq) - kmer_len
    for i in range(stop + 1):
        kmer = s[i : i + kmer_len]
        kmers[kmer] += 1
    kmers = kmers.items()
    kmers.sort(key = itemgetter(1), reverse =True)
    return kmers
        
>>> s = """"gtcagaccttcctcctcagaagctcacagaaaaacacgctttctgaaagattccacactcaatgccaaaatataccacag
... gaaaattttgcaaggctcacggatttccagtgcaccactggctaaccaagtaggagcacctcttctactgccatgaaagg
... aaaccttcaaaccctaccactgagccattaactaccatcctgtttaagatctgaaaaacatgaagactgtattgctcctg
... atttgtcttctaggatctgctttcaccactccaaccgatccattgaactaccaatttggggcccatggacagaaaactgc
... agagaagcataaatatactcattctgaaatgccagaggaagagaacacagggtttgtaaacaaaggtgatgtgctgtctg
... gccacaggaccataaaagcagaggtaccggtactggatacacagaaggatgagccctgggcttccagaagacaaggacaa
... ggtgatggtgagcatcaaacaaaaaacagcctgaggagcattaacttccttactctgcacagtaatccagggttggcttc
... tgataaccaggaaagcaactctggcagcagcagggaacagcacagctctgagcaccaccagcccaggaggcacaggaaac
... acggcaacatggctggccagtgggctctgagaggagaaagtccagtggatgctcttggtctggttcgtgagcgcaacaca"""
>>> s = s.replace('\n', '')
>>> kmers = get_kmer_occurences(s, 6)
>>> for kmer, occ in kmers:
>>>   print kmer, '..', occ
cacagg .. 4
aggaaa .. 4
ttctga .. 3
ccagtg .. 3

kmer_2.py .

5.1.8   Exercise

Write a function which take a sequence as parameter and return it’s reversed complement.
Write the pseudocode before to propose an implementation.

5.1.8.1   pseudocode:

function reverse_comp(sequence)
complement <- establish a correpondance and each base and its complement
rev_seq <- revert the sequence
rev_comp <- empty
for each nt of rev_seq
concatenate nt complement to rev_comp
return rev_comp
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def rev_comp(seq):
    """
    return the reverse complement of seq
    the sequence must be in lower case
    """
    complement = {'a' : 't',
                  'c' : 'g',
                  'g' : 'c',
                  't' : 'a'}
    rev_seq = seq[::-1]
    rev_comp = ''
    for nt in rev_seq:
        rev_comp += complement[nt]
    return rev_comp
::
>>> from rev_comp import rev_comp
>>>
>>> seq = 'acggcaacatggctggccagtgggctctgagaggagaaagtccagtggatgctcttggtctggttcgtgagcgcaacaca'
>>> print rev_comp(seq)
tgtgttgcgctcacgaaccagaccaagagcatccactggactttctcctctcagagcccactggccagccatgttgccgt

rev_comp.py .

5.1.8.2   other solution

python provide an interresting method for our problem. The translate method work on string and need a parameter which is a object that can do the correspondance between characters in old string a the new one. maketrans is a function in module string that allow us to build this object. maketrans take 2 arguments, two strings, the first string contains the characters to change, the second string the corresponding characters in the new string. Thus the two strings must have the same lenght. The correspondance between the characters to change and their new values is made in funtion of thier position. the first character of the first string will be replaced by the first character of the second string, the second character of the first string will be replaced by the second character of the second string, on so on. So we can write the reverse complement without loop.

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import string

def rev_comp(seq):
    """
    return the reverse complement of seq
    the case is respect but if the sequence mix upper and lower case the function will failed
    """
    reverse = seq[::-1]
    direct = 'acgtATCG'
    comp = 'tgcaTGCA'
    table = string.maketrans(direct, comp)
    rev_comp = reverse.translate(table)
    return rev_comp
::
>>> from rev_comp2 import rev_comp
>>>
>>> seq = 'acggcaacatggctggccagtgggctctgagaggagaaagtccagtggatgctcttggtctggttcgtgagcgcaacaca'
>>> print rev_comp(seq)
tgtgttgcgctcacgaaccagaccaagagcatccactggactttctcctctcagagcccactggccagccatgttgccgt

rev_comp2.py .

5.1.9   Exercise

let the following enzymes collection:

import collections
RestrictEnzyme = collections.namedtuple("RestrictEnzyme", ("name", "comment", "sequence", "cut", "end"))

ecor1 = RestrictEnzyme("EcoRI", "Ecoli restriction enzime I", "gaattc", 1, "sticky")
ecor5 = RestrictEnzyme("EcoRV", "Ecoli restriction enzime V", "gatatc", 3, "blunt")
bamh1 = RestrictEnzyme("BamHI", "type II restriction endonuclease from Bacillus amyloliquefaciens ", "ggatcc", 1, "sticky")
hind3 = RestrictEnzyme("HindIII", "type II site-specific nuclease from Haemophilus influenzae", "aagctt", 1 , "sticky")
taq1 = RestrictEnzyme("TaqI", "Thermus aquaticus", "tcga", 1 , "sticky")
not1 = RestrictEnzyme("NotI", "Nocardia otitidis", "gcggccgc", 2 , "sticky")
sau3a1 = RestrictEnzyme("Sau3aI", "Staphylococcus aureus", "gatc", 0 , "sticky")
hae3 = RestrictEnzyme("HaeIII", "Haemophilus aegyptius", "ggcc", 2 , "blunt")
sma1 =  RestrictEnzyme("SmaI", "Serratia marcescens", "cccggg", 3 , "blunt")

and the 2 dna fragments:

dna_1 = """tcgcgcaacgtcgcctacatctcaagattcagcgccgagatccccgggggttgagcgatccccgtcagttggcgtgaattcag
cagcagcgcaccccgggcgtagaattccagttgcagataatagctgatttagttaacttggatcacagaagcttccaga
ccaccgtatggatcccaacgcactgttacggatccaattcgtacgtttggggtgatttgattcccgctgcctgccagg"""

dna_2 = """gagcatgagcggaattctgcatagcgcaagaatgcggccgcttagagcgatgctgccctaaactctatgcagcgggcgtgagg
attcagtggcttcagaattcctcccgggagaagctgaatagtgaaacgattgaggtgttgtggtgaaccgagtaag
agcagcttaaatcggagagaattccatttactggccagggtaagagttttggtaaatatatagtgatatctggcttg"""
which enzymes cut the dna_1 ?
the dna_2 ?
the dna_1 but not the dna_2?
  1. Write a function seq_one_line which take a multi lines sequence and return a sequence in one line.
  2. Write a function enz_filter which take a sequence and a list of enzymes and return a new list containing the enzymes which have a binding site in the sequence
  3. use the functions above to compute the enzymes which cut the dna_1 apply the same functions to compute the enzymes which cut the dna_2 compute the difference between the enzymes which cut the dna_1 and enzymes which cut the dna_2
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def one_line(seq):
    return seq.replace('\n', '')

def enz_filter(enzymes, dna):
    cuting_enz = []
    for enz in enzymes:
        if enz.sequence in dna: 
            cuting_enz.append(enz)
    return cuting_enz

::

from enzyme_1 import *

enzymes = [ecor1, ecor5, bamh1, hind3, taq1, not1, sau3a1, hae3, sma1] dna_1 = one_line(dna_1) dans_2 = one_line(dna_2) enz_1 = enz_filter(enzymes, dna_1) enz_2 = enz_filter(enzymes, dna_2) enz1_only = set(enz_1) - set(enz_2)

enzymes_1.py .

with this algorithm we find if an enzyme cut the dna but we cannot find all cuts in the dna for an enzyme.

enzymes = [ecor1, ecor5, bamh1, hind3, taq1, not1, sau3a1, hae3, sma1]
digest_1 = []
for enz in enzymes:
   print enz.name, dna_1.count(enz.sequence)

the latter algorithm display the number of occurrence of each enzyme, But we cannot determine the position of every sites. We will see how to do this later.

5.1.10   Exercise

given the following dict :

d = {1 : 'a', 2 : 'b', 3 : 'c' , 4 : 'd'}

We want obtain a new dict with the keys and the values inverted so we will obtain:

inverted_d  {'a': 1, 'c': 3, 'b': 2, 'd': 4}

solution

inverted_d = {}
for key in d.keys():
    inverted_d[d[key]] = key

solution

inverted_d = {}
for key, value in d.items():
    inverted_d[value] = key

solution

inverted_d = {v : k for k, v in d.items()}