匹配pyspark数据帧在 pandas 中的索引

问题描述

我有以下pyspark数据框(testDF=ldamodel.describeTopics().select("termIndices").toPandas())

I have following pyspark dataframe (testDF=ldamodel.describeTopics().select("termIndices").toPandas())

topic|    termIndices|         termWeights|
+-----+---------------+--------------------+
|    0|    [6, 118, 5]|[0.01205522104545...|
|    1|   [0, 55, 100]|[0.00125521761966...|

我有以下单词列表

['one',
 'peopl',
 'govern',
 'think',
 'econom',
 'rate',
 'tax',
 'polici',
 'year',
 'like',
........]

我正在尝试将vocablist匹配到termIndices匹配到termWeights.

I am trying to match vocablist to termIndices to termWeights.

到目前为止，我有以下内容:

So far I have following:

for i in testDF.items():
    for j in i[1]:
        for m in j:
            t=vocablist[m],m
            print(t)

其结果为:

('tax', 6)
('insur', 118)
('rate', 5)
('peopl', 1)
('health', 84)
('incom', 38)
('think', 3)
('one', 0)
('social', 162)
.......

但是我想要类似

('tax', 6, 0.012055221045453202)
('insur', 118, 0.001255217619666775)
('rate', 5, 0.0032220995010401187)

('peopl', 1,0.008342115226031033)
('health', 84,0.0008332053105123403)
('incom', 38, ......)

任何帮助将不胜感激.

Any help will be appreciated.

推荐答案

我建议您将那些lists分散在termIndices和termWeights列中.完成此操作后，您实际上可以map为其术语名称建立索引，同时使术语权重与每个术语索引保持一致.以下是说明:

I would recommend you spread those lists in the columns termIndices and termWeights downward. Once you've done that, then you can actually map indices to their term names while having the term weights aligned with each term index. The following is an illustration:

df = pd.DataFrame(data={'topic': [0, 1],
                        'termIndices': [[6, 118, 5],
                                        [0, 55, 100]],
                        'termWeights': [[0.012055221045453202, 0.012055221045453202, 0.012055221045453202],
                                        [0.00125521761966, 0.00125521761966, 0.00125521761966]]})

dff = df.apply(lambda s: s.apply(pd.Series).stack().reset_index(drop=True, level=1))

vocablist = ['one', 'peopl', 'govern', 'think', 'econom', 'rate', 'tax', 'polici', 'year', 'like'] * 50

dff['termNames'] = dff.termIndices.map(vocablist.__getitem__)

dff[['termNames', 'termIndices', 'termWeights']].values.tolist()

我希望这会有所帮助.

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