一种想法是使用np.vecorize
,但是因为使用字符串性能只比您的解决方案好一点:
def fun (a,b):
return a.startswith(b) or b.startswith(a)
f = np.vectorize(fun)
a = pd.Series(f(df['a'],df['b']), index=df.index)
print (a)
0 True
1 False
dtype: bool
df = pd.DataFrame(data={'a':['x','yy'], 'b':['xyz','uvw']})
df = pd.concat([df] * 10000, ignore_index=True)
In [132]: %timeit pd.Series(index=df.index, data=[a.startswith(b) or b.startswith(a) for a,b in df[['a', 'b']].to_numpy()])
42.3 ms ± 516 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [133]: %timeit pd.Series(f(df['a'],df['b']), index=df.index)
9.81 ms ± 119 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [134]: %timeit pd.Series(index=df.index, data=[a.startswith(b) or b.startswith(a) for a,b in zip(df['a'],df['b'])])
14.1 ms ± 262 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
#sammywemmy solution
In [135]: %timeit pd.Series([any((a.startswith(b), b.startswith(a))) for a, b in df.to_numpy()], index=df.index)
46.3 ms ± 683 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)