python - 以矢量化方式切片 Pandas 字符串

Question

我正在尝试以矢量化方式对字符串进行切片，答案是 NaN。尽管如果序列索引(例如 str[:1])恒定，则工作正常。任何帮助

df = pd.DataFrame({'NAME': ['abc','xyz','hello'], 'SEQ': [1,2,1]}) #
df['SUB'] = df['NAME'].str[:df['SEQ']]

输出为

  NAME  SEQ  SUB
0    abc    1  NaN
1    xyz    2  NaN
2  hello    1  NaN

Answer 1

不幸的是，矢量化解决方案不存在。

使用apply使用 lambda 函数:

df['SUB'] = df.apply(lambda x: x['NAME'][:x['SEQ']], axis=1)

或 zip 与 列表理解 以获得更好的性能:

df['SUB'] = [x[:y] for x, y in zip(df['NAME'], df['SEQ'])]

---

print (df)

    NAME  SEQ SUB
0    abc    1   a
1    xyz    2  xy
2  hello    1   h

时间:

df = pd.DataFrame({'NAME': ['abc','xyz','hello'], 'SEQ': [1,2,1]})
df = pd.concat([df] * 1000, ignore_index=True)


In [270]: %timeit df["SUB"] = df.groupby("SEQ").NAME.transform(lambda g: g.str[: g.name])
4.23 ms ± 222 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [271]: %timeit df['SUB'] = df.apply(lambda x: x['NAME'][:x['SEQ']], axis=1)
104 ms ± 2.02 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [272]: %timeit df['SUB'] = [x[:y] for x, y in zip(df['NAME'], df['SEQ'])]
785 µs ± 22.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)