I create an OrderedDict:
from collections import OrderedDict
od = OrderedDict([((2, 9), 0.5218),
((2, 0), 0.3647),
((3, 15), 0.3640),
((3, 8), 0.3323),
((2, 28), 0.3310),
((2, 15), 0.3281),
((2, 10), 0.2938),
((3, 9), 0.2719)])
Then I feed that into the pandas DataFrame constructor:
import pandas as pd
df = pd.DataFrame({'values': od})
the result is this:
instead it should give this:
What is going on here that I don't understand?
PS: I am not looking for an alternative way to solving the problem (though you are welcome to post it if you think it would help the community). All I want is to understand why this here doesn't work. Is it a bug, or is there some logic to it? This is also not a duplicate of this link , because i am using specifically an OrderedDict and not a normal dict.
If you want to get the DataFrame in the same order as your dictionary you can
df = pd.DataFrame(od.values(), index=od.keys(), columns=['values'])
Output
values
2 9 0.5218
0 0.3647
3 15 0.3640
8 0.3323
2 28 0.3310
15 0.3281
10 0.2938
3 9 0.2719
The only mention of OrderedDict
in the frame source code is for an example of df.to_dict()
, so not useful here.
It seems that even though you are passing an ordered structure, it is being parsed and re-ordered by default once you wrap it in a common dictionary {'values': od}
and pandas takes its index from the OrderedDict.
This behavior seems to be overruled if you build your dictionary with the column labels as well (à la json).
od = OrderedDict([
((2, 9), {'values':0.5218}),
((2, 0), {'values':0.3647}),
((3, 15), {'values':0.3640}),
((3, 8), {'values':0.3323}),
((2, 28), {'values':0.3310}),
((2, 15), {'values':0.3281}),
((2, 10), {'values':0.2938}),
((3, 9), {'values':0.2719})
])
df = pd.DataFrame(od).T
print(df)
values
2 9 0.5218
0 0.3647
3 15 0.3640
8 0.3323
2 28 0.3310
15 0.3281
10 0.2938
3 9 0.2719
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