CODEWITHSUNDEEP
pythonpandasdataframe
samba
samba

Reputation: 3101

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

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.

Upvotes: 45

Views: 122819

Answers (5)

andrewliam.dev
andrewliam.dev

Reputation: 19

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!

Upvotes: 1

gunesevitan
gunesevitan

Reputation: 965

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

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

Upvotes: 1

Bharath M Shetty
Bharath M Shetty

Reputation: 30605

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

Upvotes: 17

MaxU - stand with Ukraine
MaxU - stand with Ukraine

Reputation: 210842

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)

Upvotes: 74

Kim Miller
Kim Miller

Reputation: 886

If it's the first row you need:

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

Upvotes: 4

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