4

为了说明我的意思，这里有一个例子

messages = [
  ('Ricky',  'Steve',  'SMS'),
  ('Steve',  'Karl',   'SMS'),
  ('Karl',   'Nora',   'Email')
]

我想将此列表和组的定义转换为整数列表和查找字典，以便组中的每个元素都有一个唯一的 id。该 id 应该像这样映射到查找表中的元素

messages_int, lookup_table = create_lookup_list(
              messages, ('person', 'person', 'medium'))

print messages_int
[ (0, 1, 0),
  (1, 2, 0),
  (2, 3, 1) ]

print lookup_table
{ 'person': ['Ricky', 'Steve', 'Karl', 'Nora'],
  'medium': ['SMS', 'Email']
}

我想知道这个问题是否有一个优雅和pythonic的解决方案。

我也愿意接受比create_lookup_listetc更好的术语

4

7 回答 7

3

defaultdict combined with the itertools.count().next method is a good way to assign identifiers to unique items. Here's an example of how to apply this in your case:

from itertools import count
from collections import defaultdict

def create_lookup_list(data, domains):
    domain_keys = defaultdict(lambda:defaultdict(count().next))
    out = []
    for row in data:
        out.append(tuple(domain_keys[dom][val] for val, dom in zip(row, domains)))
    lookup_table = dict((k, sorted(d, key=d.get)) for k, d in domain_keys.items())
    return out, lookup_table

Edit: note that count().next becomes count().__next__ or lambda: next(count()) in Python 3.

于 2009-09-09T22:04:41.107 回答
2

我的长度和复杂度大致相同：

import collections

def create_lookup_list(messages, labels):

    # Collect all the values
    lookup = collections.defaultdict(set)
    for msg in messages:
        for l, v in zip(labels, msg):
            lookup[l].add(v)

    # Make the value sets lists
    for k, v in lookup.items():
        lookup[k] = list(v)

    # Make the lookup_list
    lookup_list = []
    for msg in messages:
        lookup_list.append([lookup[l].index(v) for l, v in zip(labels, msg)])

    return lookup_list, lookup
于 2009-09-09T20:17:44.517 回答
2

在 Otto 的回答（或任何其他人的字符串->id 字典）中，我会替换（如果你对速度的痴迷是你的事）：

# create the lookup table
lookup_dict = {}
for group in indices:
    lookup_dict[group] = sorted(indices[group].keys(),
            lambda e1, e2: indices[group][e1]-indices[group][e2])

经过

# k2i must map keys to consecutive ints [0,len(k2i)-1)
def inverse_indices(k2i):
    inv=[0]*len(k2i)
    for k,i in k2i.iteritems():
        inv[i]=k
    return inv

lookup_table = dict((g,inverse_indices(gi)) for g,gi in indices.iteritems())

这更好，因为直接分配给逆数组中的每个项目比排序更快。

于 2009-09-09T21:01:32.223 回答
1

这是我自己的解决方案 - 我怀疑这是最好的

def create_lookup_list(input_list, groups):
    # use a dictionary for the indices so that the index lookup 
    # is fast (not necessarily a requirement)
    indices = dict((group, {}) for group in groups) 
    output = []

    # assign indices by iterating through the list
    for row in input_list:
        newrow = []
        for group, element in zip(groups, row):
            if element in indices[group]:
                index = indices[group][element]
            else:
                index = indices[group][element] = len(indices[group])
            newrow.append(index)
        output.append(newrow)

    # create the lookup table
    lookup_dict = {}
    for group in indices:
        lookup_dict[group] = sorted(indices[group].keys(),
                lambda e1, e2: indices[group][e1]-indices[group][e2])

    return output, lookup_dict
于 2009-09-09T19:53:48.830 回答
1

这更简单，更直接。

from collections import defaultdict

def create_lookup_list( messages, schema ):
    def mapped_rows( messages ):
        for row in messages:
            newRow= []
            for col, value in zip(schema,row):
                if value not in lookups[col]:
                    lookups[col].append(value)
                code= lookups[col].index(value)
                newRow.append(code)
            yield newRow
    lookups = defaultdict(list)
    return list( mapped_rows(messages) ), dict(lookups)

如果查找是正确的字典，而不是列表，则可以进一步简化。
使您的“查找表”具有以下结构

{ 'person': {'Ricky':0, 'Steve':1, 'Karl':2, 'Nora':3},
  'medium': {'SMS':0, 'Email':1}
}

并且可以进一步降低复杂性。

您可以将查找的这个工作副本转换为它的逆，如下所示：

>>> lookups = { 'person': {'Ricky':0, 'Steve':1, 'Karl':2, 'Nora':3},
      'medium': {'SMS':0, 'Email':1}
    }
>>> dict( ( d, dict( (v,k) for k,v in lookups[d].items() ) ) for d in lookups )
{'person': {0: 'Ricky', 1: 'Steve', 2: 'Karl', 3: 'Nora'}, 'medium': {0: 'SMS', 1: 'Email'}}
于 2009-09-09T20:26:21.993 回答
0

这是我的解决方案，不是更好 - 它只是不同:)

def create_lookup_list(data, keys):
  encoded = []
  table = dict([(key, []) for key in keys])

  for record in data:
      msg_int = []
      for key, value in zip(keys, record):
          if value not in table[key]:
              table[key].append(value)
          msg_int.append(table[key].index(value))  
      encoded.append(tuple(msg_int))

  return encoded, table
于 2009-09-09T20:20:25.530 回答
0

这是我的，内部函数让我将索引元组编写为生成器。

def create_lookup_list( data, format):
    table = {}
    indices = []
    def get_index( item, form ):
        row = table.setdefault( form, [] )
        try:
            return row.index( item )
        except ValueError:
            n = len( row )
            row.append( item )
            return n
    for row in data:
        indices.append( tuple( get_index( item, form ) for item, form in zip( row, format ) ))

    return table, indices
于 2009-09-09T20:55:53.953 回答