remove python2 specific code

source/Collection_Data_Types.rst

@@ -110,19 +110,16 @@ wihout using python shell, what is the results of the following statements:
    x[3] = -4 # what is the value of x now ?
    y = sum(x)/len(x) #what is the value of y ? why ?
    
-   y = 0
+   y = 0.5
+.. warning::
 
-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:``) 
-   
-   
-Exercise
---------
+    In python2 the result is ::
+
+        y = 0
 
-How to compute safely the average of a list? ::
+    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.
 
-   float(sum(l)) / float(len(l))
 
 Exercise
 --------
@@ -206,9 +203,9 @@ first implementation:
 second implementation:
 """"""""""""""""""""""
 
-Mathematically speaking the generation of all codons can be the cartesiens product 
+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/2/library/itertools.html#itertools.product <product>`_
+In python there is a function to do that in ``itertools module``: `https://docs.python.org/3/library/itertools.html#itertools.product <product>`_
 
 
 .. literalinclude:: _static/code/codons_itertools.py
@@ -274,7 +271,7 @@ So we can use the specifycity of set ::
 Exercise
 --------
 
-We need to compute the occurence of all kmers of a given lenght present in a sequence.
+We need to compute the occurrence of all kmers of a given length present in a sequence.
 
 Below we propose 2 algorithms. 
 
@@ -378,8 +375,8 @@ bonus:
 
 Print the kmers by ordered by occurences.
 
-| see `https://docs.python.org/2/library/stdtypes.html#mutable-sequence-types <sort>`_
-| see `https://docs.python.org/2/library/operator.html#operator.itemgetter <operator.itemgetter>`_
+| see `https://docs.python.org/3/library/stdtypes.html#mutable-sequence-types <sort>`_
+| see `https://docs.python.org/3/library/operator.html#operator.itemgetter <operator.itemgetter>`_
 
 
 .. literalinclude:: _static/code/kmer_2.py
@@ -475,7 +472,7 @@ Exercise
 let the following enzymes collection: ::
  
    import collections
-   RestrictEnzyme = collections.namedtuple("RestrictEnzyme", "name comment sequence cut end")
+   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")
@@ -504,7 +501,7 @@ and the 2 dna fragments: ::
 
 #. Write a function *seq_one_line* which take a multi lines sequence and return a sequence in one line.
 #. Write a function *enz_filter* which take a sequence and a list of enzymes and return a new list containing 
-   the enzymes which are a binding site in the sequence
+   the enzymes which have a binding site in the sequence
 #. 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
@@ -532,7 +529,7 @@ with this algorithm we find if an enzyme cut the dna but we cannot find all cuts
    for enz in enzymes:
       print enz.name, dna_1.count(enz.sequence)
 
-the latter algorithm display the number of occurence of each enzyme, But we cannot determine the position of every sites.
+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.