[python] Pandas - Get first row value of a given column

This seems like a ridiculously easy question... but I'm not seeing the easy answer I was expecting.

So, how do I get the value at an nth row of a given column in Pandas? (I am particularly interested in the first row, but would be interested in a more general practice as well).

For example, let's say I want to pull the 1.2 value in Btime as a variable.

Whats the right way to do this?

df_test =

  ATime   X   Y   Z   Btime  C   D   E
0    1.2  2  15   2    1.2  12  25  12
1    1.4  3  12   1    1.3  13  22  11
2    1.5  1  10   6    1.4  11  20  16
3    1.6  2   9  10    1.7  12  29  12
4    1.9  1   1   9    1.9  11  21  19
5    2.0  0   0   0    2.0   8  10  11
6    2.4  0   0   0    2.4  10  12  15

To get e.g the value from column 'test' and row 1 it works like

df[['test']].values[0][0]

as only df[['test']].values[0] gives back a array

In a general way, if you want to pick up the first N rows from the J column from pandas dataframe the best way to do this is:

data = dataframe[0:N][:,J]

Another way of getting the first row and preserving the index:

x = df.first('d') # Returns the first day. '3d' gives first three days.

Another way to do this:

first_value = df['Btime'].values[0]

This way seems to be faster than using .iloc:

In [1]: %timeit -n 1000 df['Btime'].values[20]
5.82 µs ± 142 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [2]: %timeit -n 1000 df['Btime'].iloc[20]
29.2 µs ± 1.28 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

1. df.iloc[0].head(1) - First data set only from entire first row.
2. df.iloc[0] - Entire First row in column.

To access a single value you can use the method iat that is much faster than iloc:

df['Btime'].iat[0]

Output:

1.2

Note that the answer from @unutbu will be correct until you want to set the value to something new, then it will not work if your dataframe is a view.

In [4]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])
In [5]: df['bar'] = 100
In [6]: df['bar'].iloc[0] = 99
/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/pandas-0.16.0_19_g8d2818e-py2.7-macosx-10.9-x86_64.egg/pandas/core/indexing.py:118: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame

See the the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self._setitem_with_indexer(indexer, value)

Another approach that will consistently work with both setting and getting is:

In [7]: df.loc[df.index[0], 'foo']
Out[7]: 'A'
In [8]: df.loc[df.index[0], 'bar'] = 99
In [9]: df
Out[9]:
  foo  bar
0   A   99
2   B  100
1   C  100