Question

在具有不同产出的数据框架内,我有以下一栏:

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

我愿利用分管职能,将一栏按最后管道分开,这样:

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

Answer 1

缩略语

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)

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