Add a new row to a Pandas DataFrame with specific index name

Question

I'm trying to add a new row to the DataFrame with a specific index name 'e'.

    number   variable       values
a    NaN       bank          true   
b    3.0       shop          false  
c    0.5       market        true   
d    NaN       government    true

I have tried the following but it's creating a new column instead of a new row.

new_row = [1.0, 'hotel', 'true']
df = df.append(new_row)

Still don't understand how to insert the row with a specific index. Will be grateful for any suggestions.

Possible duplicate of Pandas: Appending a row to a dataframe and specify its index label — Zero
– Zero, Commented Oct 7, 2017 at 15:34
@Zero I have read the answers in the link but they're discussing adding random values there. — samba
– samba, Commented Oct 7, 2017 at 16:12
Usually table-oriented APIs distinguish Update and Insert methods, but Pandas mixes them: Existing index labels are updated, new labels are inserted, assuming you don't need duplicates. You're left with convoluted concatenations to mimic the missing method. — mins
– mins, Commented May 9, 2024 at 10:14

MaxU - stand with Ukraine · Accepted Answer · 2018-11-11 14:36:14Z

75

You can use df.loc[_not_yet_existing_index_label_] = new_row.

Demo:

In [3]: df.loc['e'] = [1.0, 'hotel', 'true']

In [4]: df
Out[4]:
   number    variable values
a     NaN        bank   True
b     3.0        shop  False
c     0.5      market   True
d     NaN  government   True
e     1.0       hotel   true

PS using this method you can't add a row with already existing (duplicate) index value (label) - a row with this index label will be updated in this case.

UPDATE:

This might not work in recent Pandas/Python3 if the index is a DateTimeIndex and the new row's index doesn't exist.

it'll work if we specify correct index value(s).

Demo (using pandas: 0.23.4):

In [17]: ix = pd.date_range('2018-11-10 00:00:00', periods=4, freq='30min')

In [18]: df = pd.DataFrame(np.random.randint(100, size=(4,3)), columns=list('abc'), index=ix)

In [19]: df
Out[19]:
                      a   b   c
2018-11-10 00:00:00  77  64  90
2018-11-10 00:30:00   9  39  26
2018-11-10 01:00:00  63  93  72
2018-11-10 01:30:00  59  75  37

In [20]: df.loc[pd.to_datetime('2018-11-10 02:00:00')] = [100,100,100]

In [21]: df
Out[21]:
                       a    b    c
2018-11-10 00:00:00   77   64   90
2018-11-10 00:30:00    9   39   26
2018-11-10 01:00:00   63   93   72
2018-11-10 01:30:00   59   75   37
2018-11-10 02:00:00  100  100  100

In [22]: df.index
Out[22]: DatetimeIndex(['2018-11-10 00:00:00', '2018-11-10 00:30:00', '2018-11-10 01:00:00', '2018-11-10 01:30:00', '2018-11-10 02:00:00'], dtype='da
tetime64[ns]', freq=None)

edited Nov 11, 2018 at 14:36

answered Oct 7, 2017 at 15:14

MaxU - stand with Ukraine

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10 Comments

Bharath M Shetty Over a year ago

Wow super simple. Wish I had used that. Its all about timings

MaxU - stand with Ukraine Over a year ago

@Bharathshetty, yeah, i use this method if i need to add a single row, if i need to add 2+ rows - i;m using your method (df.append(another_DF))

Bharath M Shetty Over a year ago

I added that in my answer. :)

Zero Over a year ago

df.append(pd.Series(new_row, index=df.columns, name='e') -- series should do for single row.

MaxU - stand with Ukraine Over a year ago

@yeliabsalohcin, it'll work - please see the updated answer

|

Markus Dutschke · Accepted Answer · 2019-07-31 12:09:59Z

17

Use append by converting list a dataframe in case you want to add multiple rows at once i.e

df = df.append(pd.DataFrame([new_row],index=['e'],columns=df.columns))

Or for single row (Thanks @Zero)

df = df.append(pd.Series(new_row, index=df.columns, name='e'))

Output:

  number    variable values
a     NaN        bank   True
b     3.0        shop  False
c     0.5      market   True
d     NaN  government   True
e     1.0       hotel   true

edited Jul 31, 2019 at 12:09

Markus Dutschke

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answered Oct 7, 2017 at 15:14

Bharath M Shetty

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3 Comments

Zero Over a year ago

df.append(pd.Series(new_row, index=df.columns, name='e') series should do.

Bharath M Shetty Over a year ago

@Zero Series was my very first thought confused a bit with name and index. So went to DataFrame approach. I updated my answer. I wanted to be first to answer so.

yeliabsalohcin Over a year ago

This works in the case of a pandas dataframe with a DateTimeIndex when trying to add a row with a new datetime which doesn't exist in the index.

Kim Miller · Accepted Answer · 2018-10-18 18:14:19Z

4

If it's the first row you need:

df = Dataframe(columns=[number, variable, values])
df.loc['e', [number, variable, values]] = [1.0, 'hotel', 'true']

answered Oct 18, 2018 at 18:14

Kim Miller

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Comments

gunesevitan · Accepted Answer · 2021-01-06 14:30:21Z

1

df.loc['e', :] = [1.0, 'hotel', 'true']

should be the correct implementation in case of conflicting index and column names.

answered Jan 6, 2021 at 14:30

gunesevitan

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Comments

andrewliam.dev · Accepted Answer · 2022-08-24 20:55:01Z

1

In future versions of Pandas, DataFrame.append(other, ignore_index=False, verify_integrity=False, sort=False) will be deprecated.

Source: Pandas Documentation

The documentation recommends using .concat().

It would look like this (if you wanted an empty row with only the added index name:

df = pd.concat([df, pd.Series(index=['New index label'], dtype=str)])

If you wanted to add data use this:

df = pd.concat([df, pd.Series(data, index=['New index label'], dtype=str)])

Hope that helps!

edited Aug 24, 2022 at 20:55

answered Aug 24, 2022 at 20:36

andrewliam.dev

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