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.

Asked By: Yariv

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Source

Answer 1

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

x[x.columns[0]]

Answered By: HYRY

Answer 2

>>> 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.

Answered By: herrfz

Answer 3

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

Answered By: Jeff

Answer 4

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

Answered By: SamJ

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

Answered By: Christopher Pfeifer

Answer 6

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

Answered By: BlackList96

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

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