【发布时间】:2014-11-25 02:12:34
【问题描述】:
我在关注Pandas tutorials
教程是使用 python 2.7 编写的,我是在 python 3.4 中编写的
这是我的版本详情。
In [11]: print('Python version ' + sys.version)
Python version 3.4.1 |Anaconda 2.0.1 (64-bit)| (default, Jun 11 2014, 17:27:11)
[MSC v.1600 64 bit (AMD64)]
In [12]: print('Pandas version ' + pd.__version__)
Pandas version 0.14.1
我按照教程创建 zip
In [13]: names = ['Bob','Jessica','Mary','John','Mel']
In [14]: births = [968, 155, 77, 578, 973]
In [15]: zip?
Type: type
String form: <class 'zip'>
Namespace: Python builtin
Init definition: zip(self, *args, **kwargs)
Docstring:
zip(iter1 [,iter2 [...]]) --> zip object
Return a zip object whose .__next__() method returns a tuple where
the i-th element comes from the i-th iterable argument. The .__next__()
method continues until the shortest iterable in the argument sequence
is exhausted and then it raises StopIteration.
In [16]: BabyDataSet = zip(names,births)
但创建后第一个错误显示我看不到 zip 的内容。
In [17]: BabyDataSet
Out[17]: <zip at 0x4f28848>
In [18]: print(BabyDataSet)
<zip object at 0x0000000004F28848>
然后当我去创建数据框时,我得到了这个迭代器错误。
In [21]: df = pd.DataFrame(data = BabyDataSet, columns=['Names', 'Births'])
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-21-636a49c94b6e> in <module>()
----> 1 df = pd.DataFrame(data = BabyDataSet, columns=['Names', 'Births'])
c:\Users\Sayth\Anaconda3\lib\site-packages\pandas\core\frame.py in __init__(self
, data, index, columns, dtype, copy)
255 copy=copy)
256 elif isinstance(data, collections.Iterator):
--> 257 raise TypeError("data argument can't be an iterator")
258 else:
259 try:
TypeError: data argument can't be an iterator
In [22]:
这是 python 3 的陷阱,我需要以不同的方式做吗?还是其他?
【问题讨论】:
-
将您的行改为:
BabyDataSet = dict(zip(names,births)) -
zip在 python 3 中更改它现在返回一个迭代器:diveintopython3.net/porting-code-to-python-3-with-2to3.html#zip 所以你必须将它传递给你的情况下的 dict 构造函数
标签: python python-3.x pandas