How to get column by number in Pandas?

Question

What\'s the difference between:

Maand[\'P_Sanyo_Gesloten\']
Out[119]: 
Time
2012-08-01 00:00:11    0
2012-08-01 00:05:10    0
2012-08-01 00:10:11    0
2012-0         


        

Answer 1

One is a column (aka Series), while the other is a DataFrame:

In [1]: df = pd.DataFrame([[1,2], [3,4]], columns=['a', 'b'])

In [2]: df
Out[2]:
   a  b
0  1  2
1  3  4

The column 'b' (aka Series):

In [3]: df['b']
Out[3]:
0    2
1    4
Name: b, dtype: int64

The subdataframe with columns (position) in [1]:

In [4]: df[[1]]
Out[4]:
   b
0  2
1  4

Note: it's preferable (and less ambiguous) to specify whether you're talking about the column name e.g. ['b'] or the integer location, since sometimes you can have columns named as integers:

In [5]: df.iloc[:, [1]]
Out[5]:
   b
0  2
1  4

In [6]: df.loc[:, ['b']]
Out[6]:
   b
0  2
1  4

In [7]: df.loc[:, 'b']
Out[7]:
0    2
1    4
Name: b, dtype: int64

Answer 2

The following is taken from http://pandas.pydata.org/pandas-docs/dev/indexing.html. There are a few more examples... you have to scroll down a little

In [816]: df1

           0         2         4         6
0   0.569605  0.875906 -2.211372  0.974466
2  -2.006747 -0.410001 -0.078638  0.545952
4  -1.219217 -1.226825  0.769804 -1.281247
6  -0.727707 -0.121306 -0.097883  0.695775
8   0.341734  0.959726 -1.110336 -0.619976
10  0.149748 -0.732339  0.687738  0.176444

Select via integer slicing

In [817]: df1.iloc[:3]

          0         2         4         6
0  0.569605  0.875906 -2.211372  0.974466
2 -2.006747 -0.410001 -0.078638  0.545952
4 -1.219217 -1.226825  0.769804 -1.281247

In [818]: df1.iloc[1:5,2:4]

          4         6
2 -0.078638  0.545952
4  0.769804 -1.281247
6 -0.097883  0.695775
8 -1.110336 -0.619976

Select via integer list

In [819]: df1.iloc[[1,3,5],[1,3]]

           2         6
2  -0.410001  0.545952
6  -0.121306  0.695775
10 -0.732339  0.176444

Answer 3

Another way is to select a column with the columns array:

In [5]: df = pd.DataFrame([[1,2], [3,4]], columns=['a', 'b'])

In [6]: df
Out[6]: 
   a  b
0  1  2
1  3  4

In [7]: df[df.columns[0]]
Out[7]: 
0    1
1    3
Name: a, dtype: int64