如何在 pandas 中的另一列中对一列中的字符串进行切片

问题描述

df=pd.DataFrame({'A':['abcde','fghij','klmno','pqrst'], 'B':[1,2,3,4]})

我想按列B对A列进行切片，例如:abcde[:1]=a, klmno[:3]=klm但是两个陈述都失败了:

I want to slice column A by column B eg: abcde[:1]=a, klmno[:3]=klmbut two statements all failed：

df['new_column']=df.A.map(lambda x: x.str[:df.B])
df['new_column']=df.apply(lambda x: x.A[:x.B])

和

df['new_column']=df['A'].str[:df['B']]

new_column返回NaN

尝试获取new_column:

      A    B  new_column
0   abcde  1     a
1   fghij  2     fg
2   klmno  3     klm
3   pqrst  4     pqrs

非常感谢您

推荐答案

您需要在axis=1 .apply.html"rel =" noreferrer> apply 方法来遍历行:

You need axis=1 in the apply method to loop through rows:

df['new_column'] = df.apply(lambda r: r.A[:r.B], axis=1)
df
#       A   B   new_column
#0  abcde   1   a
#1  fghij   2   fg
#2  klmno   3   klm
#3  pqrst   4   pqrs

---

一种惯用性较低但通常更快的解决方案是使用zip:

df['new_column'] = [A[:B] for A, B in zip(df.A, df.B)]
df

#       A   B   new_column
#0  abcde   1   a
#1  fghij   2   fg
#2  klmno   3   klm
#3  pqrst   4   pqrs

---

%timeit df.apply(lambda r: r.A[:r.B], axis=1)
# 1000 loops, best of 3: 440 µs per loop

%timeit [A[:B] for A, B in zip(df.A, df.B)]
# 10000 loops, best of 3: 27.6 µs per loop

这篇关于如何在 pandas 中的另一列中对一列中的字符串进行切片的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！