CODEWITHSUNDEEP
pythonpandasdataframesplitdelimiter
skimchi1993
skimchi1993

Reputation: 229

pandas split by last delimiter

I have the following column in a dataframe with different outputs:

col1
MLB|NBA|NFL
MLB|NBA
NFL|NHL|NBA|MLB

I would like to use the split function to split the column by the last pipe always so something like this:

col1           col2
MLB|NBA        NFL
MLB            NBA
NFL|NHL|NBA    MLB

Upvotes: 6

Views: 4016

Answers (1)

cs95
cs95

Reputation: 402323

With Series.str.rsplit, limiting the number of splits.

df['col1'].str.rsplit('|', n=1, expand=True).rename(columns=lambda x: f'col{x + 1}')

If the above throws you a SyntaxError, it means you're on a python version older than 3.6 (shame on you!). Use instead

df.col1.str.rsplit('|', 1, expand=True)\
  .rename(columns=lambda x: 'col{}'.format(x + 1))

          col1 col2
0      MLB|NBA  NFL
1          MLB  NBA
2  NFL|NHL|NBA  MLB

There's also the faster loopy str.rsplit equivalent.

pd.DataFrame(
    [x.rsplit('|', 1) for x in df['col1'].tolist()], 
    columns=['col1', 'col2']
) 
          col1 col2
0      MLB|NBA  NFL
1          MLB  NBA
2  NFL|NHL|NBA  MLB

P.S., yes, the second solution is faster:

df = pd.concat([df] * 100000, ignore_index=True)

%timeit df['col1'].str.rsplit('|', n=1, expand=True)
%timeit pd.DataFrame([x.rsplit('|', 1) for x in df['col1'].tolist()])

473 ms ± 13.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
128 ms ± 1.29 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Upvotes: 12

Related Questions