Python pandas: fill a dataframe row by row

Question

df['y'] will set a column

since you want to set a row, use .loc

Note that .ix is equivalent here, yours failed because you tried to assign a dictionary to each element of the row y probably not what you want; converting to a Series tells pandas that you want to align the input (for example you then don't have to to specify all of the elements)

In [6]: import pandas as pd

In [7]: df = pd.DataFrame(columns=['a','b','c','d'], index=['x','y','z'])

In [8]: df.loc['y'] = pd.Series({'a':1, 'b':5, 'c':2, 'd':3})

In [9]: df
Out[9]: 
     a    b    c    d
x  NaN  NaN  NaN  NaN
y    1    5    2    3
z  NaN  NaN  NaN  NaN

My approach was, but I can't guarantee that this is the fastest solution.

df = pd.DataFrame(columns=["firstname", "lastname"])
df = df.append({
     "firstname": "John",
     "lastname":  "Johny"
      }, ignore_index=True)

This is a simpler version

import pandas as pd
df = pd.DataFrame(columns=('col1', 'col2', 'col3'))
for i in range(5):
   df.loc[i] = ['<some value for first>','<some value for second>','<some value for third>']`

If your input rows are lists rather than dictionaries, then the following is a simple solution:

import pandas as pd
list_of_lists = []
list_of_lists.append([1,2,3])
list_of_lists.append([4,5,6])

pd.DataFrame(list_of_lists, columns=['A', 'B', 'C'])
#    A  B  C
# 0  1  2  3
# 1  4  5  6