Get last "column" after .str.split() operation on column in pandas DataFrame

I have a column in a pandas DataFrame that I would like to split on a single space. The splitting is simple enough with DataFrame.str.split(' '), but I can't make a new column from the last entry. When I .str.split() the column I get a list of arrays and I don't know how to manipulate this to get a new column for my DataFrame.

Here is an example. Each entry in the column contains 'symbol data price' and I would like to split off the price (and eventually remove the "p"... or "c" in half the cases).

import pandas as pd
temp = pd.DataFrame({'ticker' : ['spx 5/25/2001 p500', 'spx 5/25/2001 p600', 'spx 5/25/2001 p700']})
temp2 = temp.ticker.str.split(' ')

which yields

0    ['spx', '5/25/2001', 'p500']
1    ['spx', '5/25/2001', 'p600']
2    ['spx', '5/25/2001', 'p700']

But temp2[0] just gives one list entry's array and temp2[:][-1] fails. How can I convert the last entry in each array to a new column? Thanks!

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You could use the tolist method as an intermediary:

In [99]: import pandas as pd


In [100]: d1 = pd.DataFrame({'ticker' : ['spx 5/25/2001 p500', 'spx 5/25/2001 p600', 'spx 5/25/2001 p700']})


In [101]: d1.ticker.str.split().tolist()
Out[101]:
[['spx', '5/25/2001', 'p500'],
['spx', '5/25/2001', 'p600'],
['spx', '5/25/2001', 'p700']]

From which you could make a new DataFrame:

In [102]: d2 = pd.DataFrame(d1.ticker.str.split().tolist(),
.....:                   columns="symbol date price".split())


In [103]: d2
Out[103]:
symbol       date price
0    spx  5/25/2001  p500
1    spx  5/25/2001  p600
2    spx  5/25/2001  p700

For good measure, you could fix the price:

In [104]: d2["price"] = d2["price"].str.replace("p","").astype(float)


In [105]: d2
Out[105]:
symbol       date  price
0    spx  5/25/2001    500
1    spx  5/25/2001    600
2    spx  5/25/2001    700

PS: but if you really just want the last column, apply would suffice:

In [113]: temp2.apply(lambda x: x[2])
Out[113]:
0    p500
1    p600
2    p700
Name: ticker
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最佳答案

Do this:

In [43]: temp2.str[-1]
Out[43]:
0    p500
1    p600
2    p700
Name: ticker

So all together it would be:

>>> temp = pd.DataFrame({'ticker' : ['spx 5/25/2001 p500', 'spx 5/25/2001 p600', 'spx 5/25/2001 p700']})
>>> temp['ticker'].str.split(' ').str[-1]
0    p500
1    p600
2    p700
Name: ticker, dtype: object
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https://pandas.pydata.org/pandas-docs/stable/text.html

s2 = pd.Series(['a_b_c', 'c_d_e', np.nan, 'f_g_h'])
s2.str.split('_').str.get(1)

or 

s2.str.split('_').str[1]
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Using Pandas 0.20.3:

In [10]: import pandas as pd
...: temp = pd.DataFrame({'ticker' : ['spx 5/25/2001 p500', 'spx 5/25/2001 p600', 'spx 5/25/2001 p700']})
...:


In [11]: temp2 = temp.ticker.str.split(' ', expand=True)  # the expand=True return a DataFrame


In [12]: temp2
Out[12]:
0          1     2
0  spx  5/25/2001  p500
1  spx  5/25/2001  p600
2  spx  5/25/2001  p700


In [13]: temp3 = temp.join(temp2[2])


In [14]: temp3
Out[14]:
ticker     2
0  spx 5/25/2001 p500  p500
1  spx 5/25/2001 p600  p600
2  spx 5/25/2001 p700  p700
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If you are looking for a one-liner (like I came here for), this should do nicely:

temp2 = temp.ticker.str.split(' ', expand = True)[-1]

You can also trivially modify this answer to assign this column back to the original DataFrame as follows: 

temp['last_split'] = temp.ticker.str.split(' ', expand = True)[-1]

Which I imagine is a popular use case here.


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