在 lambda 函数中解包的值太多

Question

我刚开始学习 Python。我正在使用 API 来构建 TFIDFs 模型，但是我遇到了一些无法解决的 lambda 函数错误。这是生成 TFIDF 的类的一部分：

class tfidf(ModelBuilder, Model):

    def __init__(self, max_ngram=1, normalize = True):
        self.max_ngram = max_ngram
        self.normalize = normalize

    def build(self, mentions, idfs):
        m = mentions\
            .map(lambda (target, (span, text)): (target, text))\  """error is triggered here  """
            .mapValues(lambda v: ngrams(v, self.max_ngram))\
            .flatMap(lambda (target, tokens): (((target, t), 1) for t in tokens))\
            .reduceByKey(add)\
            .map(lambda ((target, token), count): (token, (target, count)))\
            .leftOuterJoin(idfs)\

这是mentions 类的示例输出（这是导致tdfidf 类错误的输入）：

Out[24]:                                                                        
[{'_id': u'en.wikipedia.org/wiki/William_Cowper',
  'source': 'en.wikipedia.org/wiki/Beagle',
  'span': (165, 179),
  'text': u'References to the dog appear before the 19th century in works by such writers as William Shakespeare, John Webster, John Dryden, Thomas Tickell, Henry Fielding, and William Cowper, as well as in Alexander Pope\'s translation of Homer\'s "Iliad".'},
 {'_id': u"en.wikipedia.org/wiki/K-Run's_Park_Me_In_First",
  'source': 'en.wikipedia.org/wiki/Beagle',
  'span': (32, 62),
  'text': u" On 12 February 2008, a Beagle, K-Run's Park Me In First (Uno), won the Best In Show category at the Westminster Kennel Club show for the first time in the competition's history."},

错误信息是：

 .map(lambda (target, (span, text)): (target, text))\
ValueError: too many values to unpack

我试过了：.map(lambda ( src, target, span, text) : (target, text))\，因为我只需要在mentions\ 中导致相同错误的目标和文本。

一个简单且可编译的例子：

import math
import numpy


Data = [{'_id': '333981',

  'source': 'Apple',

  'span': (100, 119),

  'text': ' It is native to the northern Pacific.'}, {'_id': '27262',

  'source': 'Apple',

  'span': (4, 20),

  'text': ' Apples are yummy.'}]



m = map(lambda (ID, (span, text)) : (ID, text) , Data)

print(list(m))

我正在使用 python 2.7。任何帮助或指导将不胜感激。

非常感谢，

Answer 1

如果您想创建一个仅包含字段source 和text 的新字典数组，您可以使用

m = map(lambda item: {field: item.get(field) for field in ['source', 'text']}, Data)

如果要为键 source 和 text 创建内容元组数组：

m = map(lambda item: (item.get('source'), item.get('text')), Data)