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

Question

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!

score 230 · Accepted Answer · edited Jan 13 '20 at 12:05

230

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

edited Jan 13 '20 at 12:05

Erfan

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answered Oct 24 '12 at 16:13

Wes McKinney

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Love the clean solution! – ericmjl Jun 13 '14 at 22:01
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from the author of "Pandas" :) – kmonsoor Jun 18 '14 at 19:47
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I love this solution, but how does it work? Meaning, what is happening "behind the scenes" that allows `str` followed by brackets to select a specific element from the list? – Kevin Markham Feb 27 '16 at 15:49
I had the same question @Kevin Markham. Actually, temp2 is a series of arrays of shape (3,). You can access the various part by indexing like temp2[0][-1] giving p500. What I don't figure out is how you pass from this to temp2.str[-1]. .str will make a stringMethods object again from the series of array, but i don't get why it will retrieve columns with indexing instead of rows (temp2[-1] give the last line), and temp2.str[:] outputs the same shape than temp2. Would you mind enlightening us, please Wes McKinney? – Ando Jurai Jul 04 '16 at 08:43
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I got slightly confused by this, the one-liner is d1.ticker.str.split().str[-1]. Not what you'd expect... – citynorman Feb 04 '17 at 17:56
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This should be the accepted answer, although I agree with the others that this is odd (but useful) design. – Alexander Jul 04 '17 at 11:55
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@KevinMarkham: Here's how it works: `str` works not only for strings but also for lists to some extent. So if you had a string Series `foo` then `foo.str[0]` would take the first character of each string, and `foo.str[-1]` would take the last. But since `str` also works (partially) on lists too, `temp2.str[-1]` takes the last element of each list in the Series. A string, after all, is a sequence of characters, similar to a list. – John Zwinck Nov 08 '17 at 10:31

score 47 · Answer 2 · answered Sep 20 '12 at 01:43

47

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

answered Sep 20 '12 at 01:43

DSM

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This just helped me add a log file in pandas which was too scary and messy to even touch before (single column of data with a lot of information per row). – trench Jan 29 '16 at 12:23
All of these approaches have disastrous performance compared with Wes McKinney's answer. – John Zwinck Nov 08 '17 at 10:32
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@JohnZwinck: wow, a performance-only related downvote on a five-year old answer about functionality which had only been introduced about [two months before](https://github.com/pandas-dev/pandas/issues/1656)? That's.. rigorous, I'll give you that! – DSM Nov 08 '17 at 16:06
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But that's the point of SE: Outdated answers should be shown less prominently. As that's not possible here if OP doesn't change accepted solution, only warning to future users can be difference in votes.. – FooBar Jan 31 '19 at 13:01

score 29 · Answer 3 · answered Jul 07 '17 at 17:52

29

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]

answered Jul 07 '17 at 17:52

James Holland

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you can use -1 to get last element like accessing last element from the list `s2.str.split('_').str.get(-1)` – chanduthedev May 26 '20 at 10:46

score 6 · Answer 4 · answered Nov 13 '17 at 17:12

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

score 5 · Answer 5 · answered Sep 25 '19 at 14:22

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.

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

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