python - 根据 Python 中其他列的条件创建新列

Question

我想在Python中做一些与this one中的问题非常相似的事情R 用户。我的目的是创建一个新列，其值是根据其他列的条件创建的

例如:

d = {'year': [2010, 2011,2013, 2014], 'PD': [0.5, 0.8, 0.9, np.nan], 'PD_thresh': [0.7, 0.8, 0.9, 0.7]}
df_temp = pd.DataFrame(data=d)

现在我想创建一个条件:

伪代码:

if for year X the value of PD is greater or equal to the value of PD_thresh
then set 0 in a new column y_pseudo 

otherwise set 1 

我的预期结果是这样的:

df_temp 
Out[57]: 
   year   PD  PD_thresh  y_pseudo
0  2010  0.5        0.7       0.0
1  2011  0.6        0.7       0.0
2  2013  0.9        0.8       1.0
3  2014  NaN        0.7       NaN

Answer 1

使用numpy.select与 isna和 ge :

m1 = df_temp['PD'].isna()
m2 = df_temp['PD'].ge(df_temp['PD_thresh'])

df_temp['y_pseudo'] = np.select([m1, m2], [np.nan, 1], default=0)
print (df_temp)
   year   PD  PD_thresh  y_pseudo
0  2010  0.5        0.7       0.0
1  2011  0.6        0.8       0.0
2  2013  0.9        0.9       1.0
3  2014  NaN        0.7       NaN

另一个解决方案是将 True/False 映射到 1/0 的掩码转换为整数，并通过 notna 仅设置非缺失行。 :

m2 = df_temp['PD'].ge(df_temp['PD_thresh'])
m3 = df_temp['PD'].notna()

df_temp.loc[m3, 'y_pseudo'] = m2[m3].astype(int)
print (df_temp)
   year   PD  PD_thresh  y_pseudo
0  2010  0.5        0.7       0.0
1  2011  0.6        0.8       0.0
2  2013  0.9        0.9       1.0
3  2014  NaN        0.7       NaN