How to get index value of a set array in Pandas

Question

If I have a DataFrame with a column showing a set array, how do I get specific values in the set array?

df:

0    Array
1    {878,999,858}

Desired:

1st index: 878
2nd index: 999
3rd index: 858

What I tried:

df['Array'].values[0]
>> {878,999,858}

When I tried to add df['Array'].values[0].index.values[2] to get 2nd index, it doesn't allow me to do so.

Thanks.

Sets aren't ordered objects, so it doesn't make sense to index them by order. There is no guarantee that placing three items into the set will result in them being shown in the same order. If you just want the elements, you can convert to a list and index into that, but be aware that the order may change. — chthonicdaemon
– chthonicdaemon, Commented Mar 12, 2021 at 3:47
Ordering doesn't really matter at this point, really just trying to reduce the values to 1 items in the set — Mick
– Mick, Commented Mar 12, 2021 at 3:48
if it is list you can use df['Array'].str[1] for second index — Epsi95
– Epsi95, Commented Mar 12, 2021 at 3:48
Ordering doesn't really matter at this point, then how are you defining 2nd index. There is no 2nd index in set — Epsi95
– Epsi95, Commented Mar 12, 2021 at 3:49

Paulo Marques · Accepted Answer · 2021-03-12 04:00:45Z

1

It might have better ways to do it but...

Creating test dataframe

>>> import pandas as pd
>>> df = pd.DataFrame({"A": [{878,999,858}]})

>>> df
                 A
0  {858, 878, 999}

>>> df.loc[0]["A"]
{858, 878, 999}

>>> list(df.loc[0]["A"])
[858, 878, 999]

>>> list(df.loc[0]["A"])[0]
858
>>> list(df.loc[0]["A"])[1]
878
>>> list(df.loc[0]["A"])[2]
999

answered Mar 12, 2021 at 4:00

Paulo Marques

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suppose in my computer the set is { 878, 858, 999} so the result will be 878, 858,999 instead 858, 878, 999.