Efficient way to set all empty lists in column to None in Pandas

In pandas, it is possible to do the following:

<class 'pandas.core.frame.DataFrame'>
Int64Index: 99919 entries, 0 to 99918
Data columns (total 47 columns):
 #   Column                        Non-Null Count  Dtype         
---  ------                        --------------  -----         
 0   reason                        99919 non-null  object   

Most of these entries contain [], and some contain an actual object, e.g. [{'a':'x', 'b':'y', 'c':'z'}]

I want to efficiently (without using df.apply) set all entries that contain only the empty list to e.g. None.

I thought that I might be able to do something like the following:

df[df['reasons'].str.len() == 0]['reasons'] = None

but this does not work since it is setting it on a copy of the dataframe (see https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy)

Usually in the case where I set something on df[row][column], I can instead do:

df.loc[:, (row, column)]

to set it on the actual dataframe, but that does not work in this case since df['reasons'].str.len() == 0 returns a series which is unhashable and not a valid argument for loc.

Is there any way to do this without using apply?

>Solution :

The syntax should be:

df.loc[df['reasons'].str.len() == 0, 'reasons'] = None

The correct use of loc is loc[row, col], not loc[:, (row, col)]

df.loc[:, (X, Y)] can be used if you have a MultiIndex (see below for an example), but this is not the case here.

df = pd.DataFrame(None, index=range(2),
                  columns=pd.MultiIndex.from_product([['A', 'B'], [1, 2]]))
df.loc[:, ('A', 1)]

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