我的数据框叫做＆＃39; df＆＃39;我想用一个数据框中的列范围内的值替换另一列中的相应值。
6＆lt; = age＆lt; 11然后1
11＆lt; = age＆lt; 16然后2
16＆lt; = age＆lt; 21然后3
21＆lt; = age然后4
age
86508 12.0
86509 6.0
86510 7.0
86511 8.0
86512 10.0
86513 15.0
86514 15.0
86515 16.0
86516 20.0
86517 23.0
86518 23.0
86519 7.0
86520 18.0
结果
age stage
86508 12.0 2
86509 6.0 1
86510 7.0 1
86511 8.0 1
86512 10.0 1
86513 15.0 2
86514 15.0 2
86515 16.0 2
86516 20.0 3
86517 23.0 4
86518 23.0 4
86519 7.0 1
86520 18.0 3
感谢。
答案 0 :(得分：5)
使用pd.cut()：
In [37]: df['stage'] = pd.cut(df.age, bins=[0,11,16,21,300], labels=[1,2,3,4])
In [38]: df
Out[38]:
age stage
86508 12.0 2
86509 6.0 1
86510 7.0 1
86511 8.0 1
86512 10.0 1
86513 15.0 2
86514 15.0 2
86515 16.0 2
86516 20.0 3
86517 23.0 4
86518 23.0 4
86519 7.0 1
86520 18.0 3
或more generic solution provided by @ayhan：
In [39]: df['stage'] = pd.cut(df.age, bins=[0, 11, 16, 21, np.inf], labels=False, right=True) + 1
In [40]: df
Out[40]:
age stage
86508 12.0 2
86509 6.0 1
86510 7.0 1
86511 8.0 1
86512 10.0 1
86513 15.0 2
86514 15.0 2
86515 16.0 2
86516 20.0 3
86517 23.0 4
86518 23.0 4
86519 7.0 1
86520 18.0 3
答案 1 :(得分：4)
使用np.searchsorted
a = np.array([-np.inf, 6, 11, 16, 21, np.inf])
df.assign(stage=a.searchsorted(df.age, side='right') - 1)
age stage
86508 12.0 2
86509 6.0 1
86510 7.0 1
86511 8.0 1
86512 10.0 1
86513 15.0 2
86514 15.0 2
86515 16.0 3
86516 20.0 3
86517 23.0 4
86518 23.0 4
86519 7.0 1
86520 18.0 3
<强>时序
小数据
%%timeit
a = np.array([-np.inf, 6, 11, 16, 21, np.inf])
df.assign(stage=a.searchsorted(df.age, side='right') - 1)
1000 loops, best of 3: 288 µs per loop
%%timeit
df.assign(stage=pd.cut(df.age, bins=[0,11,16,21,300], labels=[1,2,3,4]))
1000 loops, best of 3: 668 µs per loop