Pandas基於兩個現有變量創建新變量

[英]Pandas creating a new variable based on two existing variables

本文翻译自 RJL 查看原文 2018-06-14 61 pandas/ python

I have the following code I think is highly inefficient. Is there a better way to do this type common recoding in pandas?

我認為以下代碼效率很低。有沒有更好的方法在熊貓中進行這種類型的常見重新編碼？

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

Use numpy.select and cache boolean masks to variables for better performance:

使用numpy.select並將boolean mask緩存到變量以獲得更好的性能：

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

In your specific case, you can make use of the fact that booleans are actually integers (False == 0, True == 1) and use simple arithmetic:

在您的具體情況下，您可以利用布爾實際上是整數（False == 0，True == 1）並使用簡單算術的事實：

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.

請注意，這將忽略B和C列中的任何NaN，這些將被指定為高於您的限制。

注意！

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