Python pandas: fill a dataframe row by row

问题:

The simple task of adding a row to a pandas.DataFrame object seems to be hard to accomplish. There are 3 stackoverflow questions relating to this, none of which give a working answer.

Here is what I'm trying to do. I have a DataFrame of which I already know the shape as well as the names of the rows and columns.

>>> df = pandas.DataFrame(columns=['a','b','c','d'], index=['x','y','z'])
>>> df
     a    b    c    d
x  NaN  NaN  NaN  NaN
y  NaN  NaN  NaN  NaN
z  NaN  NaN  NaN  NaN

Now, I have a function to compute the values of the rows iteratively. How can I fill in one of the rows with either a dictionary or a pandas.Series ? Here are various attempts that have failed:

>>> y = {'a':1, 'b':5, 'c':2, 'd':3} 
>>> df['y'] = y
AssertionError: Length of values does not match length of index

Apparently it tried to add a column instead of a row.

>>> y = {'a':1, 'b':5, 'c':2, 'd':3} 
>>> df.join(y)
AttributeError: 'builtin_function_or_method' object has no attribute 'is_unique'

Very uninformative error message.

>>> y = {'a':1, 'b':5, 'c':2, 'd':3} 
>>> df.set_value(index='y', value=y)
TypeError: set_value() takes exactly 4 arguments (3 given)

Apparently that is only for setting individual values in the dataframe.

>>> y = {'a':1, 'b':5, 'c':2, 'd':3} 
>>> df.append(y)
Exception: Can only append a Series if ignore_index=True

Well, I don't want to ignore the index, otherwise here is the result:

>>> df.append(y, ignore_index=True)
     a    b    c    d
0  NaN  NaN  NaN  NaN
1  NaN  NaN  NaN  NaN
2  NaN  NaN  NaN  NaN
3    1    5    2    3

It did align the column names with the values, but lost the row labels.

>>> y = {'a':1, 'b':5, 'c':2, 'd':3} 
>>> df.ix['y'] = y
>>> df
                                  a                                 b  \
x                               NaN                               NaN
y  {'a': 1, 'c': 2, 'b': 5, 'd': 3}  {'a': 1, 'c': 2, 'b': 5, 'd': 3}
z                               NaN                               NaN

                                  c                                 d
x                               NaN                               NaN
y  {'a': 1, 'c': 2, 'b': 5, 'd': 3}  {'a': 1, 'c': 2, 'b': 5, 'd': 3}
z                               NaN                               NaN

That also failed miserably.

So how do you do it ?

回答1:

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

In [8]: df.loc['y'] = pandas.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

回答2:

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)

回答3:

This is a simpler version

df = 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>']`

回答4:

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