在熊猫数据框中插入一行

I have a dataframe:

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

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

   A  B  C
0  5  6  7
1  7  8  9

[2 rows x 3 columns]

and I need to add a first row [2, 3, 4] to get:

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

I’ve tried append() and concat() functions but can’t find the right way how to do that.

How to add/insert series to dataframe?

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.

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.

this might seem overly simple but its incredible that a simple insert new row function isn’t built in. i’ve read a lot about appending a new df to the original, but i’m wondering if this would be faster.

df.loc[0] = [row1data, blah...]
i = len(df) + 1
df.loc[i] = [row2data, blah...]

Below would be the best way to insert a row into pandas dataframe without sorting and reseting an index:

import pandas as pd

df = pd.DataFrame(columns=['a','b','c'])

def insert(df, row):
    insert_loc = df.index.max()

    if pd.isna(insert_loc):
        df.loc[0] = row
    else:
        df.loc[insert_loc + 1] = row

insert(df,[2,3,4])
insert(df,[8,9,0])
print(df)

concat() seems to be a bit faster than last row insertion and reindexing. In case someone would wonder about the speed of two top approaches:

In [x]: %%timeit
     ...: df = pd.DataFrame(columns=['a','b'])
     ...: for i in range(10000):
     ...:     df.loc[-1] = [1,2]
     ...:     df.index = df.index + 1
     ...:     df = df.sort_index()

17.1 s ± 705 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [y]: %%timeit
     ...: df = pd.DataFrame(columns=['a', 'b'])
     ...: for i in range(10000):
     ...:     df = pd.concat([pd.DataFrame([[1,2]], columns=df.columns), df])

6.53 s ± 127 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

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.

You can simply append the row to the end of the DataFrame, and then adjust the index.

For instance:

df = df.append(pd.DataFrame([[2,3,4]],columns=df.columns),ignore_index=True)
df.index = (df.index + 1) % len(df)
df = df.sort_index()

Or use concat as:

df = pd.concat([pd.DataFrame([[1,2,3,4,5,6]],columns=df.columns),df],ignore_index=True)

The simplest way add a row in a pandas data frame is:

DataFrame.loc[ location of insertion ]= list( )

Example :

DF.loc[ 9 ] = [ ´Pepe’ , 33, ´Japan’ ]

NB: the length of your list should match that of the data frame.