熊猫-如何根据多列的条件创建具有3个输出的列

Question

I have a DataFrame df:

def fake_data():
     return{'Name': fake.name(), 
         'Gender': random.choice(sex_list),
         'Address': fake.street_address(), 
         'Nationality': 'Zimbabwean', 
         'Account_Type': random.choice(accounts_list), 
         'Age': random.randint(0, 2), 
         'Education': random.random() > 0.5, 
         'Employment': random.randint(0, 2),
         'Salary': random.randint(0, 2),
         'Employer_Stability': random.random() > 0.5,
         'Consistency': random.random() > 0.5,
         'Balance': random.randint(0, 2),
         'Residential_Status': random.random() > 0.5
      }

I want to create a column Service_Level that is 0 or 1 or 2 depending on the conditions of the columns;

columns = ['Age','Education', 'Employment', 'Salary', 'Employer_Stability', 'Consistency', 'Balance', 'Residential_Status']

I have tried creating the ['Service_Level'] = 0 with the following, after reading some answers here;

df['Service_Level'] = np.where((df['Age']==0)&(df['Education']==False)&(df['Employment']==0)&(df['Salary']==0)&(df['Employer_Stability']==False)&(df['Consistency']==False)&(df['Balance']==0)&(df['Residential_Status']==False),
                               (df['Age'])|(df['Education'])|(df['Employment'])|(df['Salary'])|(df['Employer_Stability'])|(df['Consistency'])|(df['Balance'])|(df['Residential_Status']), 0)

Then this for ['Service_Level'] = 1

df['Service_Level'] = np.where((df['Age']==1)&(df['Education']==True)&(df['Employment']==1)&(df['Salary']==1)&(df['Employer_Stability']==False)&(df['Consistency']==True)&(df['Balance']==1)&(df['Residential_Status']==True),
                               (df['Age'])|(df['Education'])|(df['Employment'])|(df['Salary'])|(df['Employer_Stability'])|(df['Consistency'])|(df['Balance'])|(df['Residential_Status']), 1)

Then this for ['Service_Level'] = 2

df['Service_Level'] = np.where((df['Age']==2)&(df['Education']==True)&(df['Employment']==2)&(df['Salary']==2)&(df['Employer_Stability']==True)&(df['Consistency']==True)&(df['Balance']==2)&(df['Residential_Status']==True),
                               (df['Age'])|(df['Education'])|(df['Employment'])|(df['Salary'])|(df['Employer_Stability'])|(df['Consistency'])|(df['Balance'])|(df['Residential_Status']), 2)

Unfortunately, I can't figure out how to join these conditions so that I get either 0 or 1 or 2.

If it works, what happens to the states that do not follow those exact conditions? I would like then to also produce and output

Answer 1

You might need to use slicing in conjunction with np.where (which by the way takes three argument, condition, val1(if condion is true), val2)

Your first statement

df['Service_Level'] = np.where(condtion_1, 0, 1)

This will result in df['Service_Level'] with 0s for the rows that met with the first condition and 1 otherwise.

Now you mask the data to get only the rows where service_level is not 0

df[df['Service_Level'] !=0] 

On this dataframe you can apply the second condition with

np.where(condition_2, 1,2) 

to assign 1 to df['Service_Level'] where the condition is true and assign 2 to rest of the rows.

EDIT:

You can use np.where with second condtion inside the first one like this.

df['Service_Level'] = np.where(cond_1, 0, (np.where(cond_2, 1,2)))

For better readability, you may want to first save the conditions as cond_1 etc and use them in np.where