How to get the first column of a pandas DataFrame as a Series?

Question

I tried:

x=pandas.DataFrame(...)
s = x.take([0], axis=1)

And s gets a DataFrame, not a Series.

Answer 1

df[df.columns[i]]

where i is the position/number of the column(starting from 0).

So, i = 0 is for the first column.

You can also get the last column using i = -1

Answer 2

Isn't this the simplest way?

By column name:

In [20]: df = pd.DataFrame({'x' : [1, 2, 3, 4], 'y' : [4, 5, 6, 7]})
In [21]: df
Out[21]:
    x   y
0   1   4
1   2   5
2   3   6
3   4   7

In [23]: df.x
Out[23]:
0    1
1    2
2    3
3    4
Name: x, dtype: int64

In [24]: type(df.x)
Out[24]:
pandas.core.series.Series

Answer 3

>>> import pandas as pd
>>> df = pd.DataFrame({'x' : [1, 2, 3, 4], 'y' : [4, 5, 6, 7]})
>>> df
   x  y
0  1  4
1  2  5
2  3  6
3  4  7
>>> s = df.ix[:,0]
>>> type(s)
<class 'pandas.core.series.Series'>
>>>

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UPDATE

If you're reading this after June 2017, ix has been deprecated in pandas 0.20.2, so don't use it. Use loc or iloc instead. See comments and other answers to this question.

Answer 4

From v0.11+, ... use df.iloc.

In [7]: df.iloc[:,0]
Out[7]: 
0    1
1    2
2    3
3    4
Name: x, dtype: int64

Answer 5

This works great when you want to load a series from a csv file

x = pd.read_csv('x.csv', index_col=False, names=['x'],header=None).iloc[:,0]
print(type(x))
print(x.head(10))


<class 'pandas.core.series.Series'>
0    110.96
1    119.40
2    135.89
3    152.32
4    192.91
5    177.20
6    181.16
7    177.30
8    200.13
9    235.41
Name: x, dtype: float64

Answer 6

You can get the first column as a Series by following code:

x[x.columns[0]]