根据字符串长度对Dataframe列进行切片

Question

我想从字符串长度> 4的Dataframe列中的字符串中删除前3个字符

否则，它们应保持不变。

例如

bloomberg_ticker_y

AIM9
DJEM9 # (should be M9)
FAM9
IXPM9 # (should be M9)

我可以按长度过滤字符串：

merged['bloomberg_ticker_y'].str.len() > 4

并切片字符串：

merged['bloomberg_ticker_y'].str[-2:]

但不确定如何将它们放在一起并将其应用于我的数据框

任何帮助，将不胜感激。

Answer 1

您可以使用列表推导：

df = pd.DataFrame({'bloomberg_ticker_y' : ['AIM9', 'DJEM9', 'FAM9', 'IXPM9']})

df['new'] = [x[-2:] if len(x)>4 else x for x in df['bloomberg_ticker_y']]

输出：

  bloomberg_ticker_y   new
0               AIM9  AIM9
1              DJEM9    M9
2               FAM9  FAM9
3              IXPM9    M9

Answer 2

您可以使用numpy.where该条件来根据字符串长度选择切片。

np.where(df['bloomberg_ticker_y'].str.len() > 4, 
         df['bloomberg_ticker_y'].str[3:], 
         df['bloomberg_ticker_y'])
# array(['AIM9', 'M9', 'FAM9', 'M9'], dtype=object)

df['bloomberg_ticker_sliced'] = (
   np.where(df['bloomberg_ticker_y'].str.len() > 4, 
            df['bloomberg_ticker_y'].str[3:], 
            df['bloomberg_ticker_y']))
df
  bloomberg_ticker_y bloomberg_ticker_sliced
0               AIM9                    AIM9
1              DJEM9                      M9
2               FAM9                    FAM9
3              IXPM9                      M9

如果您喜欢基于向量化map的解决方案，那是

df['bloomberg_ticker_y'].map(lambda x: x[3:] if len(x) > 4 else x)

0    AIM9
1      M9
2    FAM9
3      M9
Name: bloomberg_ticker_y, dtype: object

Answer 3

看到了各种各样的答案，因此决定在速度方面进行比较：

# Create big size test dataframe
df = pd.DataFrame({'bloomberg_ticker_y' : ['AIM9', 'DJEM9', 'FAM9', 'IXPM9']})
df = pd.concat([df]*100000)
df.shape

#Out
(400000, 1)

CS95＃1 np.where

%%timeit 
np.where(df['bloomberg_ticker_y'].str.len() > 4, 
         df['bloomberg_ticker_y'].str[3:], 
         df['bloomberg_ticker_y'])

结果：

163 ms ± 12.8 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

CS95 #2 vectorized map based solution

%%timeit 
df['bloomberg_ticker_y'].map(lambda x: x[3:] if len(x) > 4 else x)

Result:

86 ms ± 7.31 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Yatu DataFrame.mask

%%timeit
df.bloomberg_ticker_y.mask(df.bloomberg_ticker_y.str.len().gt(4), 
                           other=df.bloomberg_ticker_y.str[-2:])

Result:

187 ms ± 18.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Vlemaistre list comprehension

%%timeit
[x[-2:] if len(x)>4 else x for x in df['bloomberg_ticker_y']]

Result:

84.8 ms ± 4.85 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

pault str.replace with regex

%%timeit
df["bloomberg_ticker_y"].str.replace(r".{3,}(?=.{2}$)", "")

Result:

324 ms ± 17.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Cobra DataFrame.apply

%%timeit
df.apply(lambda x: (x['bloomberg_ticker_y'][3:] if len(x['bloomberg_ticker_y']) > 4 else x['bloomberg_ticker_y']) , axis=1)

Result:

6.83 s ± 387 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Conclusion

• Fastest method is list comprehension closely followed by vectorized map based solution.

• Slowest method is DataFrame.apply by far (as expected) followed by str.replace with regex