I have a pandas dataframe as follows, I want to convert it to a dictionary format with 2 keys as shown:

    id              name                        energy             fibre    
0   11005   4-Grain Flakes                          1404            11.5    
1   35146   4-Grain Flakes, Gluten Free             1569             6.1    
2   32570   4-Grain Flakes, Riihikosken Vehnämylly  1443            11.2     

I am expecting the result to be of

 nutritionValues = {
  ('4-Grain Flakes', 'id'): 11005,
  ('4-Grain Flakes', 'energy'): 1404,
  ('4-Grain Flakes', 'fibre'):  11.5,
  ('4-Grain Flakes, Gluten Free', 'id'): 11005,
  ('4-Grain Flakes, Gluten Free', 'energy'): 1569,
  ('4-Grain Flakes, Gluten Free', 'fibre'):  6.1,
  ('4-Grain Flakes, Riihikosken Vehnämylly', 'id'): 32570,
  ('4-Grain Flakes, Riihikosken Vehnämylly', 'energy'): 1443,
  ('4-Grain Flakes, Riihikosken Vehnämylly', 'fibre'):  11.2}

foods, fiber = multidict({
  '4-Grain Flakes': 11.5,
  '4-Grain Flakes, Gluten Free':   6.1,
  '4-Grain Flakes, Riihikosken Vehnämylly':   11.2})

How can I achieve this?

  • $\begingroup$ Have you tried the DataFrame's to_dict method? $\endgroup$
    – bradS
    Commented Apr 18, 2019 at 11:10
  • $\begingroup$ yes, but I couldnot achieve the multi-key dictionary I require as mentioned above. df = df.set_index(["item1", "item2"]) # Columns for dict keys df_dict = df.to_dict("index") # Turn into dict It gives error $\endgroup$
    – KHAN irfan
    Commented Apr 18, 2019 at 11:12
  • $\begingroup$ @bradS can you create the dictionary? $\endgroup$
    – KHAN irfan
    Commented Apr 18, 2019 at 12:10
  • $\begingroup$ Does this help: stackoverflow.com/questions/52192177/… ? $\endgroup$
    – bradS
    Commented Apr 18, 2019 at 13:48
  • 1
    $\begingroup$ Why do you want to do this? There might be a nicer data structure to use. The original dataframe is already very useful as it is, in my opinion. If you want to check combination of name and other variables, you could e.g. use the DataFrame.groupby() method to then do something for each value of name. $\endgroup$
    – n1k31t4
    Commented Apr 19, 2019 at 0:12

1 Answer 1


In order to be able to create a dictionary from your dataframe, such that the keys are tuples of combinations (according to your example output), my idea would be to use a Pandas MultiIndex. This will then generate a dictionary of the form you want.

First I just recreate your example dataframe (would be nice if you provide this code in the future!):

import pandas as pd

# Create the example dataframe
df = pd.DataFrame(["4-Grain Flakes", "4-Grain Flakes, Gluten Free", "4-Grain Flakes, Riihikosken Vehnämylly"])
df["id"] = [11005, 35146, 32570]
df["energy"] = [1404, 1569, 1443]
df["fibre"] = [11.5, 6.1, 11.2]
df.columns = ["name"] + list(df.columns[1:])

                                     name     id  energy  fibre
0                          4-Grain Flakes  11005    1404   11.5
1             4-Grain Flakes, Gluten Free  35146    1569    6.1
2  4-Grain Flakes, Riihikosken Vehnämylly  32570    1443   11.2

Now we can create the combinations of each value in "name" with each of the other column names. I will use lists, within a list comprehension, where I bundle up the values together into tuples. We end with a list of tuples:

names = df.name.tolist()
others = list(df.columns)
others.remove("name")         # We don't want "name" to be included

index_tuples = [(name, other) for name in names for other in others]

We can create the MultiIndex from this list of tuples as follows:

multi_ix = pd.MultiIndex.from_tuples(index_tuples)

Now we can create a new dataframe using out multi_ix. To populate this dataframe, notice that we simple need to row-wise values from columns ["id", "energy", "fibre"]. We can do this easily by extracting as an n * 3 NumPy array (using the values attribute of the dataframe) and then flattening the matrix, using NumPy's ravel method:

df1 = pd.DataFrame(df[others].values.ravel(), index=multi_ix, columns=["data"])


4-Grain Flakes                         id      11005.0
                                       energy   1404.0
                                       fibre      11.5
4-Grain Flakes, Gluten Free            id      35146.0
                                       energy   1569.0
                                       fibre       6.1
4-Grain Flakes, Riihikosken Vehnämylly id      32570.0
                                       energy   1443.0
                                       fibre      11.2

Now we can simply use to to_dict() method of the datframe to create the dictionary you are looking for:

nutritionValues = df1.to_dict()["data"]


{('4-Grain Flakes', 'energy'): 1404.0,
 ('4-Grain Flakes', 'fibre'): 11.5,
 ('4-Grain Flakes', 'id'): 11005.0,
 ('4-Grain Flakes, Gluten Free', 'energy'): 1569.0,
 ('4-Grain Flakes, Gluten Free', 'fibre'): 6.1,
 ('4-Grain Flakes, Gluten Free', 'id'): 35146.0,
 ('4-Grain Flakes, Riihikosken Vehnämylly', 'energy'): 1443.0,
 ('4-Grain Flakes, Riihikosken Vehnämylly', 'fibre'): 11.2,
 ('4-Grain Flakes, Riihikosken Vehnämylly', 'id'): 32570.0}

It is also possible to get your final example of a multidict, directly from the multi-indexed dataframe. You need to just use multi-index slicing:

fibre_df = final_df.loc[(slice(None), ["fibre"]), :]

4-Grain Flakes                         fibre  11.5
4-Grain Flakes, Gluten Free            fibre   6.1
4-Grain Flakes, Riihikosken Vehnämylly fibre  11.2

You can then generate a dictionary as before:

d = final_df.loc[(slice(None), ["fibre"]), :].to_dict()[0]

{('4-Grain Flakes', 'fibre'): 11.5,
 ('4-Grain Flakes, Gluten Free', 'fibre'): 6.1,
 ('4-Grain Flakes, Riihikosken Vehnämylly', 'fibre'): 11.2}

And you can drop the "fibre" value from the tuple-keys with a simple dictionary comprehension:

final_dict = {k[0]: v for k, v in d.items()}

{'4-Grain Flakes': 11.5,
 '4-Grain Flakes, Gluten Free': 6.1,
 '4-Grain Flakes, Riihikosken Vehnämylly': 11.2}

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.