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.. sectnum:: 
   :start: 4


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.. _Collection_Data_types:

*********************
Collection Data Types
*********************
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Exercices
=========

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Exercice
--------

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| Draw the representation in memory of the following expressions.
| what is the data type of each object?
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::   

   x = [1, 2, 3, 4]
   y = x[1]
   y = 3.14
   x[1] = 'foo'
   
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.. figure:: _static/figs/list_1.png
   :width: 400px
   :alt: set
   :figclass: align-center
   
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Exercice
--------

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

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.
In python3 we will obtain the expected result (see :ref:``) 
   
   
Exercice
--------

How to compute safely the average of a list? ::

   float(sum(l))/float(len(l)

exercise
--------

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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 ?
   
.. figure:: _static/figs/list_2-1.png
   :width: 400px
   :alt: set
   :figclass: align-center
   

.. container:: clearer

    .. image :: _static/figs/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]`` .
   
   
.. figure:: _static/figs/list_2-2.png
   :width: 400px
   :alt: set
   :figclass: align-center
 
   
.. container:: clearer

    .. image :: _static/figs/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]
    
    
exercise
--------
   
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generate a list containing all codons. ::

   codons = []
      for a in 'acgt':
         for b in 'acgt':
            for c in 'acgt':
               codon = a + b + c
               codons.append(codon)
               
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exercice
--------

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 

solution ::

   >>> list(set(l))

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exercice
--------

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?

::

   dna_1 = dna_1.replace('\n', '')
   dans_2 = dna_2.replace('\n', '')
   
   enzymes = [ecor1, ecor5, bamh1, hind3, taq1, not1, sau3a1, hae3, sma1]
   digest_1 = []
   for enz in enzymes:
      pos = dna_1.find(enz.sequence)
      if pos != -1:
         digest_1.append(enz)

with this first algorithm we find if an enzyme cut the dna but we cannot find all cuts in the dna for an enzyme.
If we find a cutting site, we must search again starting at the first nucleotid after the begining of the match 
until the end of the the dna, for this we use the start parameter of the find function, and so on. 
As we don't know how many loop we need to scan the dna until the end we use a ``while`` loop testing for the presence of a cutting site.::  

   digest_1 = []
   for enz in enzymes:
      pos = dna_1.find(enz.sequence)
      while pos != -1:
         digest_1.append(enz)
         pos = dna_1.find(enz.sequence, pos + 1)
         
   digest_2 = []
   for enz in enzymes:
      pos = dna_2.find(enz.sequence)
      while pos != -1:
         digest_2.append(enz)
         pos = dna_2.find(enz.sequence, pos + 1)  
                
   cut_dna_1 = set(digest_1)
   cut_dna_2 = set(digest_2)
   cut_dna_1_not_dna_2 = cut_dna_1 - cut_dna_2
         
If we want also the position, for instance to compute the fragments of dna. ::

   digest_1 = []
   for enz in enzymes:
      pos = dna_1.find(enz.sequence)
      while pos != -1:
         digest_1.append((enz, pos))
         pos = dna_1.find(enz.sequence, pos + 1)
    
   from operator import itemgetter
   digest_1.sort(key=itemgetter(1))
   [(e.name, d) for e, d in digest_1]
   
   digest_2 = []
   for enz in enzymes:
      pos = dna_2.find(enz.sequence)
      while pos != -1:
         digest_2.append((enz, pos))
         pos = dna_2.find(enz.sequence, pos + 1)
           
   cut_dna_1 = set([e.name for e in digest_1])
   cut_dna_2 = set([e.name for e in digest_2])
   cut_dna_1_not_dna_2 = cut_dna_1 - cut_dna_2
   
   

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exercice
--------
From a list return a new list without any duplicate, but keeping the order of items. 
For example: ::

   >>> l = [5,2,3,2,2,3,5,1]
   >>> uniqify_with_order(l)
   >>> [5,2,3,1]  

solution ::

   >>> uniq = []
   >>> for item in l:
   >>>   if item not in uniq:
   >>>      uniq.append(item)

solution ::

   >>> uniq_items = set()
   >>> l_uniq = [x for x in l if x not in uniq_items and not uniq_items.add(x)]
   
exercice
--------

list and count occurences of every 3mers in the following sequence ::

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

and finally print the results one 3mer and it's occurence per line. 

write first the pseudocode, then implement it.

bonus:
print the kmer by incresing occurences.

solution ::

   s = s.replace('\n', '')
   kmers = {}
   for i in range(len(s) - 3):
      kmer = s[i:i+3]
      kmers[kmer] = kmers.get(kmer, 0) + 1

   for kmer, occurence in kmers.items():
      print kmer, " = ", occurence

solution bonus ::

   list_of_kmers = kmers.items()  
   from operator import itemgetter
   list_of_kmers.sort(key=itemgetter(1)) 
   for kmer, occurence in list_of_kmers:
      print kmer, " = ", occurence

 solution bonus ::

   list_of_kmers = kmers.items()      
   list_of_kmers.sort(key = lambda kmer: kmer[1])
   for kmer, occurence in list_of_kmers:
      print kmer, " = ", occurence   
      
      
exercice
--------

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