如何在熊猫数据框中的列之间进行条件计算？

Question

下面是一个简单的数据框

import pandas as pd 
import numpy as np 
dates = pd. date_range(' 20130101' , periods=14) 
data = pd.DataFrame({'a':[1,0,0,1,0,0,0,1,1,0,0,1,0,0],'b':[0,0,1,0,0,1,0,0,0,0,1,0,1,0]},index=dates) 

现在我想添加列“C”，符合下列条件都在一起。

1. if a = 1, c = 1
2. if b = 1, c = 0
3. if a = 0 and b = 0, c = c.shift(1) 约束：存在的a = 1和b = 1没有的情况下在同一时间。

这是一个简单的问题，但很难解决......

什么好主意？

Answer 1

IIUC你需要：

data['c'] = np.where(data.a == 1, 1, 
      np.where(data.b == 1, 0, np.nan)) 
print (data) 
      a b c 
2013-01-01 1 0 1.0 
2013-01-02 0 0 NaN 
2013-01-03 0 1 0.0 
2013-01-04 1 0 1.0 
2013-01-05 0 0 NaN 
2013-01-06 0 1 0.0 
2013-01-07 0 0 NaN 
2013-01-08 1 0 1.0 
2013-01-09 1 0 1.0 
2013-01-10 0 0 NaN 
2013-01-11 0 1 0.0 
2013-01-12 1 0 1.0 
2013-01-13 0 1 0.0 
2013-01-14 0 0 NaN 

话，我不知道是否需要bfill或ffill：

data['c'] = data['c'].bfill() 
print (data) 
      a b c 
2013-01-01 1 0 1.0 
2013-01-02 0 0 0.0 
2013-01-03 0 1 0.0 
2013-01-04 1 0 1.0 
2013-01-05 0 0 0.0 
2013-01-06 0 1 0.0 
2013-01-07 0 0 1.0 
2013-01-08 1 0 1.0 
2013-01-09 1 0 1.0 
2013-01-10 0 0 0.0 
2013-01-11 0 1 0.0 
2013-01-12 1 0 1.0 
2013-01-13 0 1 0.0 
2013-01-14 0 0 NaN 

---

data['c'] = data['c'].ffill() 
print (data) 
      a b c 
2013-01-01 1 0 1.0 
2013-01-02 0 0 1.0 
2013-01-03 0 1 0.0 
2013-01-04 1 0 1.0 
2013-01-05 0 0 1.0 
2013-01-06 0 1 0.0 
2013-01-07 0 0 0.0 
2013-01-08 1 0 1.0 
2013-01-09 1 0 1.0 
2013-01-10 0 0 1.0 
2013-01-11 0 1 0.0 
2013-01-12 1 0 1.0 
2013-01-13 0 1 0.0 
2013-01-14 0 0 0.0 

Answer 2

替代

data.assign(
    c=np.where(v.sum(1, keepdims=1), (np.diff(v[:, ::-1]) + 1)/2, np.nan) 
).ffill() 

      a b c 
2013-01-01 1 0 1.0 
2013-01-02 0 0 1.0 
2013-01-03 0 1 0.0 
2013-01-04 1 0 1.0 
2013-01-05 0 0 1.0 
2013-01-06 0 1 0.0 
2013-01-07 0 0 0.0 
2013-01-08 1 0 1.0 
2013-01-09 1 0 1.0 
2013-01-10 0 0 1.0 
2013-01-11 0 1 0.0 
2013-01-12 1 0 1.0 
2013-01-13 0 1 0.0 
2013-01-14 0 0 0.0