utils.py 6.65 KB
 Mathieu Giraud committed Feb 21, 2015 1 2 `````` import collections `````` Mathieu Giraud committed Jul 30, 2015 3 ``````import defs `````` 4 ``````import sys `````` Mathieu Giraud committed Feb 21, 2015 5 6 7 8 9 10 11 12 13 14 15 16 `````` #### Utilities on dictionaries def ordered(d, key=None): '''sorts a dictionary into an OrderedDict''' return collections.OrderedDict([(k, d[k]) for k in sorted(d, key=key)]) def concatenate_with_padding(d, d1, d1_size, d2, d2_size, `````` Mathieu Giraud committed Apr 24, 2015 17 `````` ignore_keys=None): `````` Mathieu Giraud committed Feb 21, 2015 18 19 20 21 22 23 24 25 26 27 `````` '''Concatenate two dictionaries d1 and d2 into d The dictionaries d1 and d2 store several values that are lists with d1_size and d2_size elements, and the resulting dictionary will store values that are lists with size d1_size + d2_size elements. Pads with lists [0, ... 0] data that appear in either only d1 or only d2. The values that are not lists are ignored (but this should not happen). >>> d = {} >>> d1 = { 'a': [1, 2], 'b': [11, 22], 'z':17 } >>> d2 = { 'a': [3, 4, 5], 'c': [333, 444, 555] } >>> concatenate_with_padding(d, d1, 2, d2, 5, ['z']) `````` Mathieu Giraud committed Feb 25, 2015 28 29 30 31 32 33 `````` >>> d['a'] [1, 2, 3, 4, 5] >>> d['b'] [11, 22, 0, 0, 0, 0, 0] >>> d['c'] [0, 0, 333, 444, 555] `````` Mathieu Giraud committed Feb 21, 2015 34 35 36 37 `````` ''' t1=[] t2=[] `````` Mathieu Giraud committed Apr 24, 2015 38 39 40 41 `````` if ignore_keys == None: ignore_keys = [] `````` Mathieu Giraud committed Feb 21, 2015 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 `````` for i in range(d1_size): t1.append(0) for i in range(d2_size): t2.append(0) for key in d1: if key in ignore_keys: continue if type(d1[key]) is not list: continue d[key] = d1[key] if key not in d2 : d[key] += t2 for key in d2: if key in ignore_keys: continue if type(d2[key]) is not list: continue if key not in d : d[key] = t1 + d2[key] else : d[key] = d[key] + d2[key] class AccessedDict(dict): '''Dictionary providing a .not_accessed_keys() method Note that access with .get(key) are not tracked. >>> d = AccessedDict({1:11, 2:22, 3: 33, 4: 44}) >>> d[1], d[3] (11, 33) >>> list(d.not_accessed_keys()) [2, 4] ''' def __init__(self, *args, **kwargs): dict.__init__(self, *args, **kwargs) self.accessed_keys = [] def __getitem__(self, key): self.accessed_keys.append(key) return dict.__getitem__(self, key) def not_accessed_keys(self): for key in self.keys(): if key in self.accessed_keys: continue yield key ###### Utilities on strings def common_substring(l): '''Return the longest common substring among the strings in the list >>> common_substring(['abcdfffff', 'ghhhhhhhhhbcd']) 'bcd' >>> common_substring(['abcdfffff', 'ghhhhhhhhh']) '' >>> common_substring(['b-abc-123', 'tuvwxyz-abc-321', 'tuvwxyz-abc-456', 'd-abc-789']) '-abc-' ''' table = [] for s in l: # adds in table all substrings of s - duplicate substrings in s are added only once table += set(s[j:k] for j in range(len(s)) for k in range(j+1, len(s)+1)) # sort substrings by length (descending) `````` Mathieu Giraud committed Feb 25, 2015 119 `````` table = sorted(table, key=lambda x:-len(x)) `````` Mathieu Giraud committed Feb 21, 2015 120 121 122 123 124 125 126 `````` # get the position of duplicates and get the first one (longest) duplicates=[i for i, x in enumerate(table) if table.count(x) == len(l)] if len(duplicates) > 0: return table[duplicates[0]] else: return "" `````` Mikaël Salson committed Apr 21, 2016 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 ``````def get_common_suffpref(l, max_length, order): ''' Get common prefixes or common suffixes. Maximal length of the prefixes/suffixes is max_length. Order is either 1 (common prefix) or -1 (common suffix). >>> get_common_suffpref(['ablkjsdflkj', 'ablmlksdf', 'alkjr'], 5, 1) 'a' >>> get_common_suffpref(['ablkjsdflkj', 'ablmlksdf', 'lkjr'], 5, 1) '' >>> get_common_suffpref(['ablkjsdflkj', 'ablmlksdf', 'lkjr'], 5, -1) '' >>> get_common_suffpref(['ablkjsdflkj', 'ablmlklkj', 'lkjrlkj'], 5, -1) 'lkj' ''' common_string = 0 for i in range(max_length): index = i if order == -1: index = i+1 if all(map(lambda x: x[order*index] == l[0][order*index], l)): common_string += 1 else: break if order==1 or common_string == 0: return l[0][:common_string] else: return l[0][-common_string:] `````` Mathieu Giraud committed Feb 21, 2015 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 `````` def interesting_substrings(l, target_length=6, substring_replacement='-'): '''Return a list with intersting substrings. Now it removes common prefixes and suffixes, and then the longest common substring. But it stops removing once all lengths are at most 'target_length'. >>> interesting_substrings(['ec-3--bla', 'ec-512-bla', 'ec-47-bla'], target_length=0) ['3-', '512', '47'] >>> interesting_substrings(['ec-A-seq-1-bla', 'ec-B-seq-512-bla', 'ec-C-seq-21-bla'], target_length=0, substring_replacement='') ['A1', 'B512', 'C21'] >>> interesting_substrings(['ec-A-seq-1-bla', 'ec-B-seq-512-bla', 'ec-C-seq-21-bla'], target_length=0) ['A-1', 'B-512', 'C-21'] >>> interesting_substrings(['ec-A-seq-1-bla', 'ec-B-seq-512-bla', 'ec-C-seq-21-bla'], target_length=9) ['A-seq-1', 'B-seq-512', 'C-seq-21'] ''' if not l: return {} if max(map (len, l)) <= target_length: return l min_length = min(map (len, l)) ### Remove prefixes `````` Mikaël Salson committed Apr 21, 2016 184 `````` common_prefix = get_common_suffpref(l, min_length, 1) `````` Mathieu Giraud committed Feb 21, 2015 185 `````` `````` Mikaël Salson committed Apr 21, 2016 186 `````` substrings = [x[len(common_prefix):] for x in l] `````` Mathieu Giraud committed Feb 21, 2015 187 188 189 190 191 192 `````` if max(map (len, substrings)) <= target_length: return substrings ### Remove suffixes `````` Mikaël Salson committed Apr 21, 2016 193 `````` common_suffix = get_common_suffpref(l, min_length - len(common_prefix), -1) `````` Mathieu Giraud committed Feb 21, 2015 194 `````` `````` Mikaël Salson committed Apr 21, 2016 195 `````` substrings = [x[len(common_prefix):-len(common_suffix)] for x in l] `````` Mathieu Giraud committed Feb 21, 2015 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 `````` if max(map (len, substrings)) <= target_length: return substrings ### Remove the longest common substring #Have to replace '' by '_' if the removal have place between 2 substrings common = common_substring(substrings) if common: substrings = [s.replace(common, substring_replacement) for s in substrings] return substrings # ### Build dict # substrings = {} # for x in l: # substrings[x] = x[common_prefix:-(common_suffix+1)] # return substrings `````` Mathieu Giraud committed Jul 30, 2015 217 218 219 220 221 222 223 224 225 226 `````` ######### class VidjilJson(): def check_version(self, filepath): '''Check vidjil_json_version''' if "vidjil_json_version" in self.d: `````` 227 228 229 `````` if self.d["vidjil_json_version"] < defs.VIDJIL_JSON_VERSION: sys.stderr.write("! Reading file with old .json version %s\n" % self.d["vidjil_json_version"]) if self.d["vidjil_json_version"] >= defs.VIDJIL_JSON_VERSION_REQUIRED: `````` Mathieu Giraud committed Jul 30, 2015 230 231 232 `````` return raise IOError ("File '%s' is too old -- please regenerate it with a newer version of Vidjil" % filepath) ``````