Insert a row to pandas dataframe

Question

Just assign row to a particular index, using loc:

 df.loc[-1] = [2, 3, 4]  # adding a row
 df.index = df.index + 1  # shifting index
 df = df.sort_index()  # sorting by index

And you get, as desired:

    A  B  C
 0  2  3  4
 1  5  6  7
 2  7  8  9

See in Pandas documentation Indexing: Setting with enlargement.

Not sure how you were calling concat() but it should work as long as both objects are of the same type. Maybe the issue is that you need to cast your second vector to a dataframe? Using the df that you defined the following works for me:

df2 = pd.DataFrame([[2,3,4]], columns=['A','B','C'])
pd.concat([df2, df])

One way to achieve this is

>>> pd.DataFrame(np.array([[2, 3, 4]]), columns=['A', 'B', 'C']).append(df, ignore_index=True)
Out[330]: 
   A  B  C
0  2  3  4
1  5  6  7
2  7  8  9

Generally, it's easiest to append dataframes, not series. In your case, since you want the new row to be "on top" (with starting id), and there is no function pd.prepend(), I first create the new dataframe and then append your old one.

ignore_index will ignore the old ongoing index in your dataframe and ensure that the first row actually starts with index 1 instead of restarting with index 0.

Typical Disclaimer: Cetero censeo ... appending rows is a quite inefficient operation. If you care about performance and can somehow ensure to first create a dataframe with the correct (longer) index and then just inserting the additional row into the dataframe, you should definitely do that. See:

>>> index = np.array([0, 1, 2])
>>> df2 = pd.DataFrame(columns=['A', 'B', 'C'], index=index)
>>> df2.loc[0:1] = [list(s1), list(s2)]
>>> df2
Out[336]: 
     A    B    C
0    5    6    7
1    7    8    9
2  NaN  NaN  NaN
>>> df2 = pd.DataFrame(columns=['A', 'B', 'C'], index=index)
>>> df2.loc[1:] = [list(s1), list(s2)]

So far, we have what you had as df:

>>> df2
Out[339]: 
     A    B    C
0  NaN  NaN  NaN
1    5    6    7
2    7    8    9

But now you can easily insert the row as follows. Since the space was preallocated, this is more efficient.

>>> df2.loc[0] = np.array([2, 3, 4])
>>> df2
Out[341]: 
   A  B  C
0  2  3  4
1  5  6  7
2  7  8  9

I put together a short function that allows for a little more flexibility when inserting a row:

def insert_row(idx, df, df_insert):
    dfA = df.iloc[:idx, ]
    dfB = df.iloc[idx:, ]

    df = dfA.append(df_insert).append(dfB).reset_index(drop = True)

    return df

which could be further shortened to:

def insert_row(idx, df, df_insert):
    return df.iloc[:idx, ].append(df_insert).append(df.iloc[idx:, ]).reset_index(drop = True)

Then you could use something like:

df = insert_row(2, df, df_new)

where 2 is the index position in df where you want to insert df_new.

It is pretty simple to add a row into a pandas DataFrame:

1. Create a regular Python dictionary with the same columns names as your Dataframe;

2. Use pandas.append() method and pass in the name of your dictionary, where .append() is a method on DataFrame instances;

3. Add ignore_index=True right after your dictionary name.

We can use numpy.insert. This has the advantage of flexibility. You only need to specify the index you want to insert to.

s1 = pd.Series([5, 6, 7])
s2 = pd.Series([7, 8, 9])

df = pd.DataFrame([list(s1), list(s2)],  columns =  ["A", "B", "C"])

pd.DataFrame(np.insert(df.values, 0, values=[2, 3, 4], axis=0))

    0   1   2
0   2   3   4
1   5   6   7
2   7   8   9

For np.insert(df.values, 0, values=[2, 3, 4], axis=0), 0 tells the function the place/index you want to place the new values.