I tried:
x=pandas.DataFrame(...)
s = x.take([0], axis=1)
And s
gets a DataFrame, not a Series.
>>> 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'>
>>>
===========================================================================
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.
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
You can get the first column as a Series by following code:
x[x.columns[0]]
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
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
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