I have the following code I think is highly inefficient. Is there a better way to do this type common recoding in pandas?
我认为以下代码效率很低。有没有更好的方法在熊猫中进行这种类型的常见重新编码？
df['F'] = 0
df['F'][(df['B'] >=3) & (df['C'] >=4.35)] = 1
df['F'][(df['B'] >=3) & (df['C'] < 4.35)] = 2
df['F'][(df['B'] < 3) & (df['C'] >=4.35)] = 3
df['F'][(df['B'] < 3) & (df['C'] < 4.35)] = 4
2 个解决方案
#1
11
Use numpy.select and cache boolean masks to variables for better performance:
使用numpy.select并将boolean mask缓存到变量以获得更好的性能：
m1 = df['B'] >= 3
m2 = df['C'] >= 4.35
m3 = df['C'] < 4.35
m4 = df['B'] < 3
df['F'] = np.select([m1 & m2, m1 & m3, m4 & m2, m4 & m3], [1,2,3,4], default=0)
#2
3
In your specific case, you can make use of the fact that booleans are actually integers (False == 0, True == 1) and use simple arithmetic:
在您的具体情况下，您可以利用布尔实际上是整数（False == 0，True == 1）并使用简单算术的事实：
df['F'] = 1 + (df['C'] < 4.35) + 2 * (df['B'] < 3)
Note that this will ignore any NaN's in your B and C columns, these will be assigned as being above your limit.
请注意，这将忽略B和C列中的任何NaN，这些将被指定为高于您的限制。
#1
11
Use numpy.select and cache boolean masks to variables for better performance:
使用numpy.select并将boolean mask缓存到变量以获得更好的性能：
m1 = df['B'] >= 3
m2 = df['C'] >= 4.35
m3 = df['C'] < 4.35
m4 = df['B'] < 3
df['F'] = np.select([m1 & m2, m1 & m3, m4 & m2, m4 & m3], [1,2,3,4], default=0)
#2
3
In your specific case, you can make use of the fact that booleans are actually integers (False == 0, True == 1) and use simple arithmetic:
在您的具体情况下，您可以利用布尔实际上是整数（False == 0，True == 1）并使用简单算术的事实：
df['F'] = 1 + (df['C'] < 4.35) + 2 * (df['B'] < 3)
Note that this will ignore any NaN's in your B and C columns, these will be assigned as being above your limit.
请注意，这将忽略B和C列中的任何NaN，这些将被指定为高于您的限制。