How to turn pandas dataframe row into ordereddict fast

Question

Looking for a fast way to get a row in a pandas dataframe into a ordered dict with out using list. List are fine but with large data sets will take to long. I am using fiona GIS reader and the rows are ordereddicts with the schema giving the data type. I use pandas to join data. I many cases the rows will have different types so I was thinking turning into a numpy array with type string might do the trick.

Answer 1

This is implemented in pandas 0.21.0+ in function to_dict with parameter into:

df = pd.DataFrame([[1, 2], [3, 4]], columns=['a', 'b'])
print (df)
   a  b
0  1  2
1  3  4

d = df.to_dict(into=OrderedDict, orient='index')
print (d)
OrderedDict([(0, OrderedDict([('a', 1), ('b', 2)])), (1, OrderedDict([('a', 3), ('b', 4)]))])

Answer 2

Unfortunately you can't just do an apply (since it fits it back to a DataFrame):

In [1]: df = pd.DataFrame([[1, 2], [3, 4]], columns=['a', 'b'])

In [2]: df
Out[2]: 
   a  b
0  1  2
1  3  4

In [3]: from collections import OrderedDict

In [4]: df.apply(OrderedDict)
Out[4]: 
   a  b
0  1  2
1  3  4

But you can use a list comprehension with iterrows:

In [5]: [OrderedDict(row) for i, row in df.iterrows()]
Out[5]: [OrderedDict([('a', 1), ('b', 2)]), OrderedDict([('a', 3), ('b', 4)])]

If it was possible to use a generator, rather than a list, to whatever you were working with this will usually be more efficient:

In [6]: (OrderedDict(row) for i, row in df.iterrows())
Out[6]: <generator object <genexpr> at 0x10466da50>