Commit ef3842e6 by Bertrand NÉRON

### remove python2 specific code

parent 7e0b58e6
Pipeline #10138 passed with stages
in 21 seconds
 ... ... @@ -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 `_ In python there is a function to do that in ``itertools module``: `https://docs.python.org/3/library/itertools.html#itertools.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 `_ | see `https://docs.python.org/2/library/operator.html#operator.itemgetter `_ | see `https://docs.python.org/3/library/stdtypes.html#mutable-sequence-types `_ | see `https://docs.python.org/3/library/operator.html#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. ... ...
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