[Tutor] weird lambda expression -- can someone help me understand how this works

[Michael Crawford]

It answered it.  I had forgotten that you could pass functions around in python.

Thanks,
Mike


On Dec 13, 2013, at 9:31 PM, Amit Saha <amitsaha.in at gmail.com> wrote:

> On Sat, Dec 14, 2013 at 12:29 PM, Amit Saha <amitsaha.in at gmail.com> wrote:
>> On Sat, Dec 14, 2013 at 12:14 PM, Michael Crawford <dalupus at gmail.com> wrote:
>>> I found this piece of code on github
>>> 
>>> https://gist.github.com/kljensen/5452382
>>> 
>>> def one_hot_dataframe(data, cols, replace=False):
>>>    """ Takes a dataframe and a list of columns that need to be encoded.
>>>        Returns a 3-tuple comprising the data, the vectorized data,
>>>        and the fitted vectorizor.
>>>    """
>>>    vec = DictVectorizer()
>>>    mkdict = lambda row: dict((col, row[col]) for col in cols)
>>> #<<<<<<<<<<<<<<<<<<
>>>    vecData = pandas.DataFrame(vec.fit_transform(data[cols].apply(mkdict,
>>> axis=1)).toarray())
>>>    vecData.columns = vec.get_feature_names()
>>>    vecData.index = data.index
>>>    if replace is True:
>>>        data = data.drop(cols, axis=1)
>>>        data = data.join(vecData)
>>>    return (data, vecData, vec)
>>> 
>>> I don't understand how that lambda expression works.
>>> For starters where did row come from?
>>> How did it know it was working on data?
>> 
>> Consider this simple example:
>> 
>>>>> l = lambda x: x**2
>>>>> apply(l, (3,))
>> 9
>> 
>> A lambda is an anonymous function. So, when you use apply(), the
>> lambda, l gets the value 3 in x and then returns x**2 which is 9 in
>> this case.
> 
> Argh, no sorry, that doesn't answer your question. Sorry, my bad. I
> should have read your query properly.
> 
> 
> -- 
> http://echorand.me

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