熊猫数据框分组列

Question

我有一个数据框 df，我需要按部门名称列分组

输入

Employee Name	Department Name	Subjects	Billable	Hours	Date
Anu	CS	Java	Yes	8	01-03-2021
Anu	CS	Python	Yes	9	02-03-2021
Anu	CS	SQL	No	6	03-03-2021
Anu	CS	React	Yes	5	03-03-2021
Anu	CS	.Net	No	8	04-03-2021
Bala	CS	SQL	No	5	01-03-2021
Bala	CS	Python	Yes	4	01-03-2021
Bala	CS	Java	Yes	2	02-03-2021
Bala	CS	.Net	No	8	03-03-2021
Bala	CS	React	Yes	7	04-03-2021

代码

df = pd.pivot_table(df,index=['Department Name','Employee Name','Billable'],columns=['Subjects'],values='Hours',aggfunc={'Hours': np.sum})

# Resetting index
df = df.reset_index ()
list_column = df.columns

# Adding new columns and calculation
total = df.sum(axis=1)
df.insert(len(df.columns), column='Total', value=total)

available_col = len(df.columns)
Utilization_col = len(df.columns)
utilization_row = len(df.columns)

# Adding Available column
available = 168
df.insert(len(df.columns), column='Available', value=available)

# Adding Utilization column
utilization = (total / available)
df.insert(len(df.columns), column='Utilization', value=utilization)

# Filter dataframe using groupby
df1 = df.groupby(['Department Name','Employee Name'], sort=False ).sum(min_count=1)
df1['Available'] = available

# Adding Billable Utilization column and Non-billable Utilization column
df['Billable'] = np.where(df['Billable'] == 'Billable', 'Billable Utilization','Non Billable Utilization')

df2 = (df.groupby(['Employee Name', 'Billable Status'])[list_column].sum().sum(axis=1).unstack().div(available).mul(100)).round(2)

df = df1.join(df2).reset_index()
df.index = df.index

# Round the column value
df['Total'] = df['Total'].round(2)

df = df.groupby(['Department Name','Employee Name'], as_index=False).sum(min_count=1)

我的输出

预期输出

注意：

我尝试使用reset_index，但是groupby函数不起作用。

Answer 1

我尝试了以下功能，我能够得到你想要的输出

def func(x): 
for i in range(1, x['Department Name'].size):
        x['Department Name'].iloc[i] = ''
return x;

df['Department Name'] = df['Department Name'].apply(str)
df = df.groupby('Department 
Name').apply(func).set_index('Department Name')
df.head()

证明