如何从namedtuple实例列表创建pandas DataFrame(具有索引或多索引)? [英] How do I create pandas DataFrame (with index or multiindex) from list of namedtuple instances?

问题描述

简单的例子:

>>> from collections import namedtuple
>>> import pandas

>>> Price = namedtuple('Price', 'ticker date price')
>>> a = Price('GE', '2010-01-01', 30.00)
>>> b = Price('GE', '2010-01-02', 31.00)
>>> l = [a, b]
>>> df = pandas.DataFrame.from_records(l, index='ticker')
Traceback (most recent call last)
...
KeyError: 'ticker'

更恶劣的例子:

>>> df2 = pandas.DataFrame.from_records(l, index=['ticker', 'date'])
>>> df2

         0           1   2
ticker  GE  2010-01-01  30
date    GE  2010-01-02  31

现在它认为['ticker', 'date']是索引本身，而不是我想用作索引的列.

Now it thinks that ['ticker', 'date'] is the index itself, rather than the columns I want to use as the index.

有没有一种方法，而不必求助于中间的numpy ndarray或事后使用set_index?

Is there a way to do this without resorting to an intermediate numpy ndarray or using set_index after the fact?

推荐答案

要从命名元组获取系列，可以使用_fields属性:

To get a Series from a namedtuple you could use the _fields attribute:

In [11]: pd.Series(a, a._fields)
Out[11]:
ticker            GE
date      2010-01-01
price             30
dtype: object

类似地，您可以创建一个这样的DataFrame:

Similarly you can create a DataFrame like this:

In [12]: df = pd.DataFrame(l, columns=l[0]._fields)

In [13]: df
Out[13]:
  ticker        date  price
0     GE  2010-01-01     30
1     GE  2010-01-02     31

您必须 set_index 之后，但是您可以执行以下操作inplace:

In [14]: df.set_index(['ticker', 'date'], inplace=True)

In [15]: df
Out[15]:
                   price
ticker date
GE     2010-01-01     30
       2010-01-02     31

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