Is it possible to insert a row at an arbitrary position in a dataframe using pandas?

I have a DataFrame object similar to this one:

       onset    length
1      2.215    1.3
2     23.107    1.3
3     41.815    1.3
4     61.606    1.3
...

What I would like to do is insert a row at a position specified by some index value and update the following indices accordingly. E.g.:

       onset    length
1      2.215    1.3
2     23.107    1.3
3     30.000    1.3  # new row
4     41.815    1.3
5     61.606    1.3
...

What would be the best way to do this?

Answers:

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Method 1

You could slice and use concat to get what you want.

line = DataFrame({"onset": 30.0, "length": 1.3}, index=[3])
df2 = concat([df.iloc[:2], line, df.iloc[2:]]).reset_index(drop=True)

This will produce the dataframe in your example output. As far as I’m aware, concat is the best method to achieve an insert type operation in pandas, but admittedly I’m by no means a pandas expert.

Method 2

I find it more readable to sort rather than slice and concatenate.

line = DataFrame({"onset": 30.0, "length": 1.3}, index=[2.5])
df = df.append(line, ignore_index=False)
df = df.sort_index().reset_index(drop=True)

Method 3

I think it’s even easier without concat or append:

df.loc[2.5] = 30.0, 1.3
df = df.sort_index().reset_index(drop=True)

(Supposing that the index is as provided, starting from 1)

Method 4

line = DataFrame({"onset": 30.0, "length": 1.3}, index=[3])
df2 = concat([df.iloc[:2], line, df.iloc[3:]]).reset_index(drop=True)

this solution is replacing that index values i want to just add one index without replacing the index values.

Method 5

If you want to keep the original indexes this might work beter:

df = pd.DataFrame(dict(x=[0, 1, 2, 3, 4]))
df_update = pd.DataFrame(dict(x=[10, 11, 12]), index=[3, 4, 5])

# concat df_update first
df = pd.concat([df_update, df], axis=0)

# drop duplicates, updates will be prioritized
df = df.iloc[df.index.drop_duplicates()]

# sort to regain order
df.sort_index(inplace=True)

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0