Assuming I have a pandas dataframe such as
df_p = pd.DataFrame(
{'name_array':
[[20130101, 320903902, 239032902],
[20130101, 3253453, 239032902],
[65756, 4342452, 32425432523]],
'name': ['a', 'a', 'c']} )
I want to extract the series which contains the flatten arrays in each row whilst preserving the order
The expected result is a pandas.core.series.Series
This question is not a duplicate because my expected output is a pandas Series, and not a dataframe.
The solutions using melt are slower than OP's original method, which they shared in the answer here, especially after the speedup from my comment on that answer.
I created a larger dataframe to test on:
df = pd.DataFrame({'name_array': np.random.rand(1000, 3).tolist()})
And timing the two solutions using melt on this dataframe yield:
In [16]: %timeit pd.melt(df.name_array.apply(pd.Series).reset_index(), id_vars=['index'],value_name='name_array').drop('variable', axis=1).sort_values('index')
173 ms ± 5.68 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [17]: %timeit df['name_array'].apply(lambda x: pd.Series([i for i in x])).melt().drop('variable', axis=1)['value']
175 ms ± 4.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
The OP's method with the speedup I suggested in the comments:
In [18]: %timeit pd.Series(np.concatenate(df['name_array']))
18 ms ± 887 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
And finally, the fastest solution as provided here but modified to provide a series instead of dataframe output:
In [14]: from itertools import chain
In [15]: %timeit pd.Series(list(chain.from_iterable(df['name_array'])))
402 µs ± 4.68 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
This last method is faster than melt() by 3 orders of magnitude and faster than np.concatenate() by 2 orders of magnitude.
This is the solution I've figured out. Don't know if there are more efficient ways.
df_p = pd.DataFrame(
{'name_array':
[[20130101, 320903902, 239032902],
[20130101, 3253453, 239032902],
[65756, 4342452, 32425432523]],
'name': ['a', 'a', 'c']} )
data = pd.DataFrame( {'column':np.concatenate(df_p['name_array'].values)} )['column']
output:
[0 20130101
1 320903902
2 239032902
3 20130101
4 3253453
5 239032902
6 65756
7 4342452
8 32425432523
Name: column, dtype: int64]
[] around the data, since you're just putting the new values into a list for no reason. Also, OP asked for a series and you're creating a dataframe and then indexing it with the column name to get a series---you should just be able to use the Series() constructor itself without the middle-man :). Edit: lol didn't realize you were OP.pd.Series(np.concatenate(df_p['name_array']))You can use pd.melt:
pd.melt(df_p.name_array.apply(pd.Series).reset_index(),
id_vars=['index'],
value_name='name_array') \
.drop('variable', axis=1) \
.sort_values('index')
OUTPUT:
index name_array
0 20130101
0 320903902
0 239032902
1 20130101
1 3253453
1 239032902
2 65756
2 4342452
2 32425432523
you can flatten list of column's lists, and then create series of that, in this way:
pd.Series([element for row in df_p.name_array for element in row])
namecolumn is irrelevant?chain.from_iterableshould work for you---you just need to pass that into the constructor ofSeriesinstead ofDataFrame. So:pd.Series(list(chain.from_iterable(df['name_array'])))