将 Python dict 转换为数据框 [英] Convert Python dict into a dataframe
问题描述
我有一个像下面这样的 Python 字典:
I have a Python dictionary like the following:
{u'2012-06-08': 388,
u'2012-06-09': 388,
u'2012-06-10': 388,
u'2012-06-11': 389,
u'2012-06-12': 389,
u'2012-06-13': 389,
u'2012-06-14': 389,
u'2012-06-15': 389,
u'2012-06-16': 389,
u'2012-06-17': 389,
u'2012-06-18': 390,
u'2012-06-19': 390,
u'2012-06-20': 390,
u'2012-06-21': 390,
u'2012-06-22': 390,
u'2012-06-23': 390,
u'2012-06-24': 390,
u'2012-06-25': 391,
u'2012-06-26': 391,
u'2012-06-27': 391,
u'2012-06-28': 391,
u'2012-06-29': 391,
u'2012-06-30': 391,
u'2012-07-01': 391,
u'2012-07-02': 392,
u'2012-07-03': 392,
u'2012-07-04': 392,
u'2012-07-05': 392,
u'2012-07-06': 392}
键是 Unicode 日期,值是整数.我想通过将日期及其相应的值作为两个单独的列将其转换为熊猫数据框.示例:col1:日期 col2:DateValue(日期仍然是Unicode,日期值仍然是整数)
The keys are Unicode dates and the values are integers. I would like to convert this into a pandas dataframe by having the dates and their corresponding values as two separate columns. Example: col1: Dates col2: DateValue (the dates are still Unicode and datevalues are still integers)
Date DateValue
0 2012-07-01 391
1 2012-07-02 392
2 2012-07-03 392
. 2012-07-04 392
. ... ...
. ... ...
在这方面的任何帮助将不胜感激.我无法在 pandas 文档中找到可以帮助我解决此问题的资源.
Any help in this direction would be much appreciated. I am unable to find resources on the pandas docs to help me with this.
我知道一种解决方案可能是将这个 dict 中的每个键值对转换为一个 dict,这样整个结构就变成了一个 dict 的 dict,然后我们可以将每一行单独添加到数据帧中.但我想知道是否有更简单和更直接的方法来做到这一点.
I know one solution might be to convert each key-value pair in this dict, into a dict so the entire structure becomes a dict of dicts, and then we can add each row individually to the dataframe. But I want to know if there is an easier way and a more direct way to do this.
到目前为止,我已经尝试将 dict 转换为系列对象,但这似乎并没有保持列之间的关系:
So far I have tried converting the dict into a series object but this doesn't seem to maintain the relationship between the columns:
s = Series(my_dict,index=my_dict.keys())
推荐答案
这里的错误是因为使用标量值调用 DataFrame 构造函数(它期望值是一个列表/dict/...即有多个列):
The error here, is since calling the DataFrame constructor with scalar values (where it expects values to be a list/dict/... i.e. have multiple columns):
pd.DataFrame(d)
ValueError: If using all scalar values, you must must pass an index
您可以从字典中获取项目(即键值对):
You could take the items from the dictionary (i.e. the key-value pairs):
In [11]: pd.DataFrame(d.items()) # or list(d.items()) in python 3
Out[11]:
0 1
0 2012-07-02 392
1 2012-07-06 392
2 2012-06-29 391
3 2012-06-28 391
...
In [12]: pd.DataFrame(d.items(), columns=['Date', 'DateValue'])
Out[12]:
Date DateValue
0 2012-07-02 392
1 2012-07-06 392
2 2012-06-29 391
但我认为传递 Series 构造函数更有意义:
But I think it makes more sense to pass the Series constructor:
In [21]: s = pd.Series(d, name='DateValue')
Out[21]:
2012-06-08 388
2012-06-09 388
2012-06-10 388
In [22]: s.index.name = 'Date'
In [23]: s.reset_index()
Out[23]:
Date DateValue
0 2012-06-08 388
1 2012-06-09 388
2 2012-06-10 388
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