如何使用np.哪里嵌套在数据框架与熊猫?

Question

我想实现一个逻辑来添加一个基于以下标准的自定义标签:

if df[(df['value1'] ==0) & (df['value2']==1)] then label1
if df[(df['value1'] ==0) & (df['value2']==0)] then label2
if df[(df['value1'] ==1) & (df['value2']==1)] then label3
if df[(df['value1'] ==1) & (df['value2']==0)] then label4

:

label_class | other columns
label1      |...
label1      |...
label3      |...
label2      |...

我尝试了np。在哪里，但我不确定如何正确地做嵌套。

Answer 1

使用numpy.select:

m1 = (df['value1'] ==0) & (df['value2']==1)
m2 = (df['value1'] ==0) & (df['value2']==0)
m3 = (df['value1'] ==1) & (df['value2']==1)
m4 = (df['value1'] ==1) & (df['value2']==0)
labels = ['label1', 'label2', 'label3', 'label4']
df['label_class'] = np.select([m1, m2, m3, m4], labels)

另一个想法是通过所有组合和标签创建辅助DataFrame，然后通过左连接添加到DataFrame:

df1 = pd.DataFrame({'value1':[0,0,1,1], 'value2':[1,0,1,0], 'label_class':labels})
df = df.merge(df1, on=['value1','value2'], how='left')

两列映射的想法:

d = {(0, 1): 'label1', (0, 0): 'label2', (1, 1): 'label3', (1, 0): 'label4'}
df['label_class'] = df.set_index(['value1','value2']).index.map(d)

Answer 2

np.where的语法如下np.where(condition, value_if_true, value_if_false)

如果是你，你可以这样做:

np.where(df[(df['value1'] ==0) & (df['value2']==1)], 'label1',
np.where(if df[(df['value1'] ==0) & (df['value2']==0)], 'label2',
np.where(if df[(df['value1'] ==1) & (df['value2']==1)], 'label3',
np.where(if df[(df['value1'] ==1) & (df['value2']==0)], 'label4', None))))