File size: 1,598 Bytes
113d67d | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 | def load_vocab():
human_vocab = {' ': 0, '.': 1, '/': 2, '0': 3, '1': 4, '2': 5, '3': 6, '4': 7, '5': 8, '6': 9, '7': 10, '8': 11, '9': 12, 'a': 13, 'b': 14, 'c': 15, 'd': 16, 'e': 17, 'f': 18, 'g': 19, 'h': 20, 'i': 21, 'j': 22, 'l': 23, 'm': 24, 'n': 25, 'o': 26, 'p': 27, 'r': 28, 's': 29, 't': 30, 'u': 31, 'v': 32, 'w': 33, 'y': 34, '<unk>': 35, '<pad>': 36}
machine_vocab = {'-': 0, '0': 1, '1': 2, '2': 3, '3': 4, '4': 5, '5': 6, '6': 7, '7': 8, '8': 9, '9': 10}
inv_machine_vocab = {0: '-', 1: '0', 2: '1', 3: '2', 4: '3', 5: '4', 6: '5', 7: '6', 8: '7', 9: '8', 10: '9'}
return human_vocab, machine_vocab, inv_machine_vocab
def string_to_int(string, length, vocab):
"""
Converts all strings in the vocabulary into a list of integers representing the positions of the
input string's characters in the "vocab"
Arguments:
string -- input string, e.g. 'Wed 10 Jul 2007'
length -- the number of time steps you'd like, determines if the output will be padded or cut
vocab -- vocabulary, dictionary used to index every character of your "string"
Returns:
rep -- list of integers (or '<unk>') (size = length) representing the position of the string's character in the vocabulary
"""
#make lower to standardize
string = string.lower()
string = string.replace(',','')
if len(string) > length:
string = string[:length]
rep = list(map(lambda x: vocab.get(x, '<unk>'), string))
if len(string) < length:
rep += [vocab['<pad>']] * (length - len(string))
return rep |