pandas-python中的If语句

我有这样一个数据帧：

TechId  Rating  Category LateNight  Weekend  Weekdays 
  001     5      Teacher      1         1       0
  002     4      Student      0         1       1
  003     3      Driver       1         0       1
  004     1      Teacher      1         0       1
  005     1      Student      1         1       0
  006     0      Teacher      1         1       1
  007     2      Teacher      0         0       1

我想知道Category=Teacher在工作日是否可用，它应该在输出中显示TechID 003006&007

我试过这个密码

Category = int(input("Enter the Trade ID you want technicians in:"))
Latenight = int(input("Press 1 if you want technicians in Latenight/else press 0:"))
Weekends = int(input("Press 1 if you want technicians in Weekends/else press 0:"))
WeekDays = int(input("Press 1 if you want technicians in WeekDays/else press 0:"))

df2 = df1_new[(df1_new['Category'] == Category) &(df1_new['Weekends'] == Weekends) & (df1_new['LateNight'] == Latenight)]
df2 = df1_new[(df1_new['Category'] == Category) &(df1_new['WeekDays'] == WeekDays) & (df1_new['LateNight'] == Latenight)]
df2 = df1_new[(df1_new['Category'] == Category) &(df1_new['WeekDays'] == WeekDays) & (df1_new['LateNight'] == Latenight) & (df1_new['Weekends'] == Weekends)]
df2_new= df2.sort_values(by=['divcount'], ascending=False)
df2_new.head()

Enter the Category you want : Teacher      
Press 1 if you want technicians in Latenight/else press 0:0
Press 1 if you want technicians in Weekends/else press 0:0
Press 1 if you want technicians in WeekDays/else press 0:1

但这只显示TechID 007行。某些TechID可能在周末和工作日都可用。这也应予以考虑。

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有人能帮我吗？

当答案为0时，请求应忽略筛选器，而不是请求0值。一种简单的方法是对0使用

isin（[0,1]）

，对1使用

isin（[1]）

：

Category = int(input("Enter the Trade ID you want technicians in:"))
Latenight = [1] if int(input("Press 1 if you want technicians in Latenight/else press 0:")) else [0,1]
Weekends =[1] if  int(input("Press 1 if you want technicians in Weekends/else press 0:")) else [0,1]
WeekDays = [1] if int(input("Press 1 if you want technicians in WeekDays/else press 0:")) else [0,1]

df2 = df1_new[(df1_new['Category'] == Category) &(df1_new['WeekDays'].isin(WeekDays)) & (df1_new['LateNight'].isin(Latenight)) & (df1_new['Weekends'].isin(Weekends))]

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当答案为0时，请求应忽略筛选器，而不是请求0值。一种简单的方法是对0使用

isin（[0,1]）

，对1使用

isin（[1]）

：

Category = int(input("Enter the Trade ID you want technicians in:"))
Latenight = [1] if int(input("Press 1 if you want technicians in Latenight/else press 0:")) else [0,1]
Weekends =[1] if  int(input("Press 1 if you want technicians in Weekends/else press 0:")) else [0,1]
WeekDays = [1] if int(input("Press 1 if you want technicians in WeekDays/else press 0:")) else [0,1]

df2 = df1_new[(df1_new['Category'] == Category) &(df1_new['WeekDays'].isin(WeekDays)) & (df1_new['LateNight'].isin(Latenight)) & (df1_new['Weekends'].isin(Weekends))]

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首先，你的答案是004、006和007

一个简单的方法是：

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df1_new[（df1_new['Category']='Teacher'）和（df1_new['Weekdays']==1）]

首先，你的答案是004、006和007

一个简单的方法是：

df1_new[（df1_new['Category']='Teacher'）和（df1_new['Weekdays']==1）]

尝试以下方法：

工作日可供选择的教师：

df2 = df.loc[(df['Category'] == 'Teacher') & (df['Weekdays'] == 1)]
df2

df2 = df.loc[(df['Category'] == 'Teacher') & (df['Weekdays'] & df['LateNight']==1)]
df2

输出：

TechId  Rating  Category    LateNight   Weekend Weekdays
004         1   Teacher         1           0       1
006         0   Teacher         1           1       1
007         2   Teacher         0           0       1

TechId  Rating  Category    LateNight   Weekend Weekdays
004         1   Teacher         1           0       1
006         0   Teacher         1           1       1

教师在工作日和深夜可用：

df2 = df.loc[(df['Category'] == 'Teacher') & (df['Weekdays'] == 1)]
df2

df2 = df.loc[(df['Category'] == 'Teacher') & (df['Weekdays'] & df['LateNight']==1)]
df2

输出：

TechId  Rating  Category    LateNight   Weekend Weekdays
004         1   Teacher         1           0       1
006         0   Teacher         1           1       1
007         2   Teacher         0           0       1

TechId  Rating  Category    LateNight   Weekend Weekdays
004         1   Teacher         1           0       1
006         0   Teacher         1           1       1

试试这个：

工作日可供选择的教师：

df2 = df.loc[(df['Category'] == 'Teacher') & (df['Weekdays'] == 1)]
df2

df2 = df.loc[(df['Category'] == 'Teacher') & (df['Weekdays'] & df['LateNight']==1)]
df2

输出：

TechId  Rating  Category    LateNight   Weekend Weekdays
004         1   Teacher         1           0       1
006         0   Teacher         1           1       1
007         2   Teacher         0           0       1

TechId  Rating  Category    LateNight   Weekend Weekdays
004         1   Teacher         1           0       1
006         0   Teacher         1           1       1

教师在工作日和深夜可用：

df2 = df.loc[(df['Category'] == 'Teacher') & (df['Weekdays'] == 1)]
df2

df2 = df.loc[(df['Category'] == 'Teacher') & (df['Weekdays'] & df['LateNight']==1)]
df2

输出：

TechId  Rating  Category    LateNight   Weekend Weekdays
004         1   Teacher         1           0       1
006         0   Teacher         1           1       1
007         2   Teacher         0           0       1

TechId  Rating  Category    LateNight   Weekend Weekdays
004         1   Teacher         1           0       1
006         0   Teacher         1           1       1

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我希望技术人员在深夜和周末或深夜和周末或仅在工作日提供服务。根据要求。我希望技术人员在深夜和周末或深夜和周末或仅在工作日提供服务。根据需要。