如何将Pandas Dataframe转换为所需的Json格式

Question

start = datetime.datetime(2013, 1, 1)
end = datetime.datetime(2013, 01, 27)
f=web.get_data_yahoo('AAPL',start, end)
f['Adj Close'].to_json(date_format='iso',orient='split')

上面的代码给出了以下结果：

Out[85]: '{"name":"Adj Close","index":["2013-01-02T00:00:00","2013-01-03T00:00:0
0","2013-01-04T00:00:00","2013-01-07T00:00:00","2013-01-08T00:00:00","2013-01-09
T00:00:00","2013-01-10T00:00:00","2013-01-11T00:00:00","2013-01-14T00:00:00","20
13-01-15T00:00:00","2013-01-16T00:00:00","2013-01-17T00:00:00","2013-01-18T00:00
:00","2013-01-22T00:00:00","2013-01-23T00:00:00","2013-01-24T00:00:00","2013-01-
25T00:00:00"],"data":[535.58,528.82,514.09,511.06,512.44,504.43,510.68,507.55,48
9.45,474.01,493.69,490.36,487.75,492.4,501.41,439.46,429.1]}'

我想要的是：

'[{"index":"2013-01-02T00:00:00",value:535.58},{"index":"2013-01-04T00:00:00",value:528.82},...]'

这可能吗？ 我该怎么回事呢？

Answer 1

看起来这可能是to_json的一种有用的替代方法，目前，一种解决方法是将其读回python和munge：s

In [11]: s = f['Adj Close'].to_json(date_format='iso',orient='split')

In [12]: d = json.loads(s)  # import json

In [13]: [{"index": date, "value": val} for date, val in zip(d['index'], d['data'])]
Out[13]: 
[{'index': u'2013-01-02T00:00:00.000Z', 'value': 535.58},
 {'index': u'2013-01-03T00:00:00.000Z', 'value': 528.82},
 {'index': u'2013-01-04T00:00:00.000Z', 'value': 514.09},
 {'index': u'2013-01-07T00:00:00.000Z', 'value': 511.06},
 {'index': u'2013-01-08T00:00:00.000Z', 'value': 512.44},
 {'index': u'2013-01-09T00:00:00.000Z', 'value': 504.43},
 {'index': u'2013-01-10T00:00:00.000Z', 'value': 510.68},
 {'index': u'2013-01-11T00:00:00.000Z', 'value': 507.55},
 {'index': u'2013-01-14T00:00:00.000Z', 'value': 489.45},
 {'index': u'2013-01-15T00:00:00.000Z', 'value': 474.01},
 {'index': u'2013-01-16T00:00:00.000Z', 'value': 493.69},
 {'index': u'2013-01-17T00:00:00.000Z', 'value': 490.36},
 {'index': u'2013-01-18T00:00:00.000Z', 'value': 487.75},
 {'index': u'2013-01-22T00:00:00.000Z', 'value': 492.4},
 {'index': u'2013-01-23T00:00:00.000Z', 'value': 501.41},
 {'index': u'2013-01-24T00:00:00.000Z', 'value': 439.46},
 {'index': u'2013-01-25T00:00:00.000Z', 'value': 429.1}]

In [14]: json.dumps([{"index": date, "value": val} for date, val in zip(d['index'], d['data'])])
Out[14]: '[{"index": "2013-01-02T00:00:00.000Z", "value": 535.58}, {"index": "2013-01-03T00:00:00.000Z", "value": 528.82}, {"index": "2013-01-04T00:00:00.000Z", "value": 514.09}, {"index": "2013-01-07T00:00:00.000Z", "value": 511.06}, {"index": "2013-01-08T00:00:00.000Z", "value": 512.44}, {"index": "2013-01-09T00:00:00.000Z", "value": 504.43}, {"index": "2013-01-10T00:00:00.000Z", "value": 510.68}, {"index": "2013-01-11T00:00:00.000Z", "value": 507.55}, {"index": "2013-01-14T00:00:00.000Z", "value": 489.45}, {"index": "2013-01-15T00:00:00.000Z", "value": 474.01}, {"index": "2013-01-16T00:00:00.000Z", "value": 493.69}, {"index": "2013-01-17T00:00:00.000Z", "value": 490.36}, {"index": "2013-01-18T00:00:00.000Z", "value": 487.75}, {"index": "2013-01-22T00:00:00.000Z", "value": 492.4}, {"index": "2013-01-23T00:00:00.000Z", "value": 501.41}, {"index": "2013-01-24T00:00:00.000Z", "value": 439.46}, {"index": "2013-01-25T00:00:00.000Z", "value": 429.1}]'

显然，这违背了高效to_json功能的目的，但我认为值得将其作为功能请求添加 - 我认为这是一种相当标准的格式，我们只是忽略了它。

Answer 2

本文可以帮助您解决此问题。 你可以像这样写：

f['Adj Close'].to_json(orient="records")

在上面的文章中我们可以看到：

records : list like [{column -> value}, ... , {column -> value}]

我这样解决了问题。

Answer 3

它的工作原理与上面提到的一样，但我发现了一种更有趣的Pretty输出格式：

response = make_response(f['Adj Close'].to_json(orient='records'))
response.headers['Content-Type'] = 'application/json'
return response

这样你就可以获得JSON并使用标题，这样读者就能知道正确的格式。