pandas dataframe how to replace None values in a column with list values or remove none values or pd.np.nan

Question

data = {
'list_id' : [[50, None],[20, 68],[10, 7],[73, 4, 26, 3],[50, None],[68, 20, 61, 62],[68, None]]
}

df = pd.DataFrame.from_dict(data)
print (df)

I tried the below steps,

expected Output

    data = {
'list_id' : [[50],[20, 68],[10, 7],[73, 4, 26, 3],[50],[68, 20, 61, 62],[68]]
}
df = pd.DataFrame.from_dict(data)
print (df)
            list_id
0              [50]
1          [20, 68]
2           [10, 7]
3    [73, 4, 26, 3]
4              [50]
5  [68, 20, 61, 62]
6              [68]



 df[['list_id']] = df['list_id'].apply(lambda el: [  f'' if x is None else x for x in el])
 df[['list_id']] = df['list_id'].apply(lambda el: [  f'' if x is None for x in el])

Need to replace None values in column with list elements, either as an empty '' string or None being removed, not sure about np.nan..

jezrael · Accepted Answer · 2019-12-16 11:10:08Z

2

For remove None or NaNs values use notna in list comprehension:

df['list_id'] = df['list_id'].apply(lambda el: [x for x in el if pd.notna(x)])
print (df)
            list_id
0              [50]
1          [20, 68]
2           [10, 7]
3    [73, 4, 26, 3]
4              [50]
5  [68, 20, 61, 62]
6              [68]

For remove only Nones compare by None with not:

df['list_id'] = df['list_id'].apply(lambda el: [x for x in el if x is not None])

answered Dec 16, 2019 at 11:10

jezrael

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