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I am following the Pandas tutorials
The tutorials are written using python 2.7 and I am doing them in python 3.4
Here is my version details.
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
I create the zip as per the tutorial
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)
But after creation the first error shows that I cannot see the contents of the zip.
In [17]: BabyDataSet
Out[17]: <zip at 0x4f28848>
In [18]: print(BabyDataSet)
<zip object at 0x0000000004F28848>
Then when I go to create the dataframe I get this iterator error.
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]:
Is this a python 3 gotcha where I need to do it differently? Or other?
326
Python zip() FunctionThe zip() function returns a zip object, which is an iterator of tuples where the first item in each passed iterator is paired together, and then the second item in each passed iterator are paired together etc.
zip() can receive multiple iterables as input. It returns an iterator that can generate tuples with paired elements from each argument.
The zip() function combines the contents of two or more iterables. zip() returns a zip object. This is an iterator of tuples where all the values you have passed as arguments are stored as pairs. Python's zip() function takes an iterable—such as a list, tuple, set, or dictionary—as an argument.
12.5 Lists and tupleszip is a built-in function that takes two or more sequences and “zips” them into a list of tuples where each tuple contains one element from each sequence. In Python 3, zip returns an iterator of tuples, but for most purposes, an iterator behaves like a list.
You need to change this line:
BabyDataSet = zip(names,births)
to:
BabyDataSet = list(zip(names,births))
This is because zip now returns an iterator in python 3, hence your error message. For more details see: http://www.diveintopython3.net/porting-code-to-python-3-with-2to3.html#zip and https://docs.python.org/3/library/functions.html#zip
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