Collection_Data_Types.rst 2.81 KB
 Bertrand NÉRON committed Jul 14, 2014 1 2 3 4 ``````.. sectnum:: :start: 4 `````` Bertrand NÉRON committed Jul 12, 2014 5 6 7 8 9 ``````.. _Collection_Data_types: ********************* Collection Data Types ********************* `````` Bertrand NÉRON committed Jul 14, 2014 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 `````` Exercices ========= 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)) 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()}``````