Inserting blank row pandas dataframe

Question

i have a columns called 'factor' and each time a name in that column changes, i would like to insert a blank row, is this possible?

for i in range(0, end):
    if df2.at[i + 1, 'factor'] != df2.at[i, 'factor']:

Answer 1

It's inefficient to manually insert rows sequentially in a for loop. As an alternative, you can find the indices where changes occur, construct a new dataframe, concatenate, then sort by index:

df = pd.DataFrame([[1, 1], [2, 1], [3, 2], [4, 2],
                   [5, 2], [6, 3]], columns=['A', 'B'])

switches = df['B'].ne(df['B'].shift(-1))
idx = switches[switches].index

df_new = pd.DataFrame(index=idx + 0.5)
df = pd.concat([df, df_new]).sort_index()

print(df)

       A    B
0.0  1.0  1.0
1.0  2.0  1.0
1.5  NaN  NaN
2.0  3.0  2.0
3.0  4.0  2.0
4.0  5.0  2.0
4.5  NaN  NaN
5.0  6.0  3.0
5.5  NaN  NaN

If necessary, you can use reset_index to normalize the index:

print(df.reset_index(drop=True))

     A    B
0  1.0  1.0
1  2.0  1.0
2  NaN  NaN
3  3.0  2.0
4  4.0  2.0
5  5.0  2.0
6  NaN  NaN
7  6.0  3.0
8  NaN  NaN

Answer 2

Use reindex by Float64Index of edge indices added to 0.5 with union of original index.

df2 = pd.DataFrame({'factor':list('aaabbccdd')})

idx = df2.index.union(df2.index[df2['factor'].shift(-1).ne(df2['factor'])] + .5)[:-1]
print (idx)
Float64Index([0.0, 1.0, 2.0, 2.5, 3.0, 4.0, 4.5, 5.0, 6.0, 6.5, 7.0, 8.0], dtype='float64')

df2 = df2.reindex(idx, fill_value='').reset_index(drop=True)
print (df2)
   factor
0       a
1       a
2       a
3        
4       b
5       b
6        
7       c
8       c
9        
10      d
11      d

If want missing values:

df2 = df2.reindex(idx).reset_index(drop=True)
print (df2)
   factor
0       a
1       a
2       a
3     NaN
4       b
5       b
6     NaN
7       c
8       c
9     NaN
10      d
11      d