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I have a pandas df with various columns. One column - myCol - looks like this:

df

someCol   myCol
a         [{}]
b         [{'X': {'A': "value", 'B': "value"}}]
c         [{}, {}]
d         [{'X': {'A': "value", 'B': "value", 'C': "value"}}]

The maximum number of key-val pairs in X is unknown: some rows contain them all, some only contain a selection, and some are empty. I would like to replace myCol with actual columns, with as many columns as needed depending on the unique number of key-val pairs in X. So in this particular example, I would end up with:

df

someCol   A       B       C
a         N/A     N/A     N/A
b         value   value   N/A     
c         N/A     N/A     N/A
d         value   value   value

I am struggling in coming up with a general way to solve this, which is needed since I don't know how many 'additional' columns I will need in the end. Any ideas would be much appreciated.

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1 Answer 1

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Solution return first lists and dictionary with key X, then convert Nones to empty dicts and last pass to DataFrame constructor:

d = [{} if pd.isna(x) else x for x in df.pop('myCol').str[0].str.get('X')]
df = df.join(pd.DataFrame(d, index=df.index))
print (df)
  someCol      A      B      C
0       a    NaN    NaN    NaN
1       b  value  value    NaN
2       c    NaN    NaN    NaN
3       d  value  value  value
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3 Comments

Thank you. If the values are True or False instead of "value", how can I adapt your answer to that?
@CHRD - I think no change.
True, that didn't matter. Thanks alot!

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